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Solve equations with variables on both sides

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Difficulty: 1 – 5

5519968

In \(5x + 2 = 3x + 10\), which operation correctly moves all variable terms to the left side in one step? A) Add \(3x\) to both sides. B) Subtract \(3x\) from both sides. C) Subtract \(5x\) from the left side only. D) Add \(10\) to both sides.

Hints

- Look for an operation that removes the variable term from the right side. - Any valid equation transformation must apply the same operation to both sides.

Solution

1. To collect the variable terms on the left, apply the same operation to both sides. 2. Subtracting \(3x\) from both sides leaves \(2x + 2 = 10\). 3. The correct choice is B.

Answer

B) Subtract \(3x\) from both sides.

5519978

For \(2x + 7 = x + 11\), subtract \(x\) from both sides. What equation results?

Hints

- Apply the stated subtraction to both sides of the equation. - Simplify only the variable terms that are affected by subtracting \(x\).

Solution

1. Subtract \(x\) from each side: \(2x - x + 7 = x - x + 11\). 2. Simplify to get \(x + 7 = 11\).

Answer

\(x + 7 = 11\)

5546688

Jordan is solving \(-2x+7=3x-8\). Jordan says, “I can add \(2x\) to the left side only to remove the \(-2x\).” What is wrong with Jordan's step? Then write the correct equation that results from adding \(2x\) to both sides.

Hints

- Think about what must stay true when one side of an equation is changed. - The goal is to remove the negative variable term without changing the solution set. - Apply the chosen operation to both sides before simplifying.

Solution

1. An operation performed on only one side generally changes the solution set, so Jordan's proposed step would not keep the equation equivalent. 2. Add \(2x\) to both sides: \(-2x+7+2x=3x-8+2x\). 3. Simplify to \(7=5x-8\).

Answer

Jordan must perform the same operation on both sides to keep the equation equivalent. The correct resulting equation is \(7=5x-8\).

5125328

Solve each linear equation. a) \(8x + 13 = 5x + 31\) b) \(14 - 6y = 2y - 18\) c) \(11z - 7 = 3z + 25\)

Hints

- Move all variable terms to one side. - Move constant terms to the other side. - Use inverse operations on both sides. - Divide by the final coefficient of the variable.

Solution

1. For the first equation, subtract \(5x\): \(3x+13=31\). Then \(3x=18\), so \(x=6\). 2. For the second equation, add \(6y\): \(14=8y-18\). Then \(32=8y\), so \(y=4\). 3. For the third equation, subtract \(3z\): \(8z-7=25\). Then \(8z=32\), so \(z=4\).

Answer

a) \(x = 6\) b) \(y = 4\) c) \(z = 4\)

5125448

Simplify each side as much as possible, then solve. a) \(14x - 5 + 3x = 2x + 40\) b) \(12 - 4y + 18 = 10y - 12 - 2y\)

Hints

- Combine variable terms and constant terms separately on each side. - Move all variable terms to one side and constants to the other. - Divide by the final coefficient of the variable.

Solution

1. For the first equation, combine the left side: \(17x-5=2x+40\). Subtract \(2x\): \(15x-5=40\). Add \(5\) and divide by \(15\): \(x=3\). 2. For the second equation, combine both sides: \(30-4y=8y-12\). Add \(4y\) and \(12\): \(42=12y\). Divide by \(12\): \(y=3.5=\frac{7}{2}\).

Answer

a) \(x = 3\) b) \(y = 3.5 = \frac{7}{2}\)

5125568

Simplify both sides, solve \(18x - 15 - 7x = 3x + 25\), and give the solution set \(S\).

Hints

- Combine the variable terms on the left first. - Move all variable terms to one side and constants to the other. - Divide by the final coefficient of \(x\).

Solution

1. Combine like terms on the left: \(11x - 15 = 3x + 25\). 2. Subtract \(3x\): \(8x - 15 = 25\). 3. Add \(15\): \(8x = 40\). 4. Divide by \(8\): \(x = 5\). 5. Therefore, \(S = \{5\}\).

Answer

\(S = \{5\}\)

5136368

Solve \(0.6x + 4 = 1.1x - 3.5\).

Hints

- Move all variable terms to one side. - Move constants to the other side. - Divide by the final decimal coefficient.

Solution

1. Subtract \(0.6x\): \(4 = 0.5x - 3.5\). 2. Add \(3.5\): \(7.5 = 0.5x\). 3. Divide by \(0.5\): \(x = 15\).

Answer

\(x = 15\)

5140958

Solve \(1.5x - 2 = 0.5(x + 6)\).

Hints

- Distribute the decimal factor before moving any variable terms. - Collect the \(x\)-terms on one side of the equation. - Use the remaining constant equation to isolate \(x\).

Solution

1. Distribute on the right: \(1.5x-2=0.5x+3\). 2. Subtract \(0.5x\): \(x-2=3\). 3. Add \(2\): \(x=5\).

Answer

\(x=5\)

5140978

Let \(T_1=\frac{1}{2}(k+12)\) and \(T_2=2k-18\). Find \(k\) when \(T_1=T_2\).

Hints

- Equality of \(T_1\) and \(T_2\) means their two expressions should be set equal. - Distribute the one-half factor before collecting the \(k\)-terms. - Keep the variable terms on one side and constants on the other.

Solution

1. Set the expressions equal: \(\frac{1}{2}(k+12)=2k-18\). 2. Distribute: \(0.5k+6=2k-18\). 3. Subtract \(0.5k\) and add \(18\): \(24=1.5k\). 4. Divide by \(1.5\): \(k=16\).

Answer

\(k=16\)

5141238

Solve each equation. a) \(7x - 12 = 2x + 13\) b) \(3(x - 4) = 2x - 3\) c) \(10y - (2y + 4) = 6y\)

Hints

- Move variable terms to one side. - Distribute or remove parentheses before combining like terms. - Simplify each side before isolating the variable.

Solution

1. For the first equation, subtract \(2x\): \(5x-12=13\). Add \(12\) and divide by \(5\): \(x=5\). 2. For the second equation, distribute: \(3x-12=2x-3\). Subtract \(2x\) and add \(12\): \(x=9\). 3. For the third equation, remove the parentheses: \(10y-2y-4=6y\), so \(8y-4=6y\). Subtract \(6y\): \(2y-4=0\), so \(y=2\).

Answer

a) \(x = 5\) b) \(x = 9\) c) \(y = 2\)

5143748

Solve \(13x - 8 - 5x + 2 = 4x + 18\), then check your solution.

Hints

- Combine variable terms and constants on each side first. - Move variable terms to one side and constants to the other. - Substitute your result into the original equation to check it.

Solution

1. Combine like terms on the left: \(8x - 6 = 4x + 18\). 2. Subtract \(4x\): \(4x - 6 = 18\). 3. Add \(6\): \(4x = 24\). 4. Divide by \(4\): \(x = 6\). 5. Check: The left side is \(13 \cdot 6 - 8 - 5 \cdot 6 + 2 = 42\), and the right side is \(4 \cdot 6 + 18 = 42\).

Answer

\(x = 6\)

5153798

Solve each equation over the rational numbers \(\mathbb{Q}\). a) \(18 - (3x + 6) = 2x - 3\) b) \(5(x - 1) = 2x + 7\)

Hints

- Handle the negative sign before parentheses carefully. - Simplify both sides before moving terms. - Move variable terms to one side and constants to the other.

Solution

1. For the first equation, remove the parentheses: \(18-3x-6=2x-3\), so \(12-3x=2x-3\). Add \(3x\) and \(3\): \(15=5x\), so \(x=3\). 2. For the second equation, distribute: \(5x-5=2x+7\). Subtract \(2x\) and add \(5\): \(3x=12\), so \(x=4\).

Answer

a) \(x = 3\) b) \(x = 4\)

5237438

Lucas has four times as much money in his piggy bank as his younger sister Mia. Lucas has exactly \(\$42\) more than Mia. How much money does Mia have? Solve using a variable-on-both-sides equation. Your response must show the original equation and the first equivalent equation after all variable terms have been collected on one side.

Hints

- Use one variable for Mia's amount. - Write Lucas's amount in two different ways. - Keep the equation equivalent when moving a variable term from one side.

Solution

1. Let \(x\) be the amount Mia has. 2. Lucas's amount can be written as both \(4x\) and \(x+42\), so write \(4x=x+42\). 3. Subtract \(x\) from both sides: \(3x=42\). 4. Divide by \(3\): \(x=14\).

Answer

Original equation: \(4x=x+42\). After collecting variable terms: \(3x=42\). Mia has \(\$14\).

5237448

A crate contains apples and pears. The apples weigh three times as much as the pears. If \(2\,\text{lb}\) of apples were removed and \(5\,\text{lb}\) of pears were added, the pears would still weigh \(3\,\text{lb}\) less than the remaining apples. How many pounds of pears were originally in the crate? Solve using a variable-on-both-sides equation. Your response must show the original equation and the first equivalent equation after all variable terms have been collected on one side.

Hints

- Express the original apple weight in terms of the pear weight. - Update both weights after the stated changes. - Translate “\(3\,\text{lb}\) less” carefully before collecting variable terms.

Solution

1. Let \(b\) be the original pear weight, in pounds. The apple weight is \(3b\). 2. After the changes, the pear weight is \(b+5\), and the apple weight is \(3b-2\). 3. The pears are \(3\,\text{lb}\) lighter, so \(b+5=3b-5\). 4. Subtract \(b\) from both sides: \(5=2b-5\). 5. Add \(5\): \(10=2b\), so \(b=5\).

Answer

Original equation: \(b+5=3b-5\). After collecting variable terms: \(5=2b-5\). The pears originally weighed \(5\,\text{lb}\).

5239588

Let \(A=3(x+4)\) and \(B=10-x\). Find the value of \(x\) for which \(A=B\).

Hints

- Equality of \(A\) and \(B\) means their two expressions should be set equal. - Distribute before collecting the variable terms. - Move all \(x\)-terms to one side before isolating \(x\).

Solution

1. Set the expressions equal: \(3(x+4)=10-x\). 2. Distribute: \(3x+12=10-x\). 3. Add \(x\) and subtract \(12\): \(4x=-2\). 4. Divide by \(4\): \(x=-\frac{1}{2}\).

Answer

\(x=-\frac{1}{2}\)

5239898

One water tank contains three times as much water as a second tank. After \(15.5\,\text{gal}\) is removed from the first tank and \(28.5\,\text{gal}\) is added to the second tank, the two tanks contain equal amounts. How much water was originally in each tank? Solve using a variable-on-both-sides equation. Your response must show the original equation and the first equivalent equation after all variable terms have been collected on one side.

Hints

- Use a variable for the smaller original amount. - Write both tank amounts after the removal and addition. - Set the changed amounts equal before collecting variable terms.

Solution

1. Let \(x\) be the original amount in the second tank, in gallons. The first tank contains \(3x\). 2. After the changes, the first tank contains \(3x-15.5\), and the second contains \(x+28.5\). 3. Set the new amounts equal: \(3x-15.5=x+28.5\). 4. Subtract \(x\) from both sides: \(2x-15.5=28.5\). 5. Add \(15.5\): \(2x=44\), so \(x=22\). 6. The first tank originally contained \(66\,\text{gal}\).

Answer

Original equation: \(3x-15.5=x+28.5\). After collecting variable terms: \(2x-15.5=28.5\). The tanks originally contained \(66\,\text{gal}\) and \(22\,\text{gal}\), respectively.

5366397

Two lines intersect. One of the four angles is exactly one-fourth of the sum of the other three angles. Find the measure of that angle.

Hints

- Recall the sum of all angles around a point. - Express the sum of the other three angles in terms of the unknown angle. - Translate “one-fourth of” into multiplication by a fraction.

Solution

1. Let \(x\) degrees be the angle measure. The four angles around the intersection add to \(360^\circ\). 2. The sum of the other three angles is \(360 - x\). 3. Translate the relationship into an equation: \(x = \frac{1}{4}(360 - x)\). 4. Multiply by \(4\): \(4x = 360 - x\). 5. Add \(x\): \(5x = 360\). Divide by \(5\): \(x = 72\).

Answer

The angle measures \(72^\circ\).

5100588

Seven times a number \(x\) is 10 greater than three times the number that is 2 greater than \(x\). Find \(x\).

Hints

- Translate each phrase into an algebraic expression before writing the equation. - How can you represent “seven times a number” and “2 greater than \(x\)”? - If one quantity is 10 greater than another, where should the 10 appear in the equation? - After distributing, decide how to bring the variable terms together. - Check whether your solution makes the two verbal quantities differ by exactly \(10\).

Solution

1. Translate the statement into an equation: \(7x=3(x+2)+10\). 2. Distribute and combine like terms: \(7x=3x+16\). 3. Subtract \(3x\) from both sides: \(4x=16\). 4. Divide by \(4\): \(x=4\).

Answer

\(x = 4\)

5107278

A bookshelf is \(\frac13\) full. After \(12\) more books are added, it is \(\frac12\) full. Let \(x\) be the number of books the shelf holds when full. a) Write an equation with \(x\) on both sides that models the situation. b) Solve your equation algebraically. Your work must show equivalent equations that remove the fractional coefficients and then isolate \(x\). c) Interpret the solution in the context.

Hints

- Express the initial and final numbers of books as fractions of the same unknown capacity. - Use the least common multiple of the denominators to clear the fractions. - After clearing fractions, collect the variable terms on one side. - Check that adding \(12\) to one-third of your answer gives one-half of it.

Solution

1. Initially the shelf holds \(\frac13x\) books. After adding \(12\) books, it holds \(\frac12x\), so \( \frac13x+12=\frac12x. \) 2. Multiply both sides by \(6\): \(2x+72=3x\). 3. Subtract \(2x\) from both sides: \(72=x\). 4. Therefore, the shelf holds \(72\) books when completely full.

Answer

a) \(\frac13x+12=\frac12x\) b) \(x=72\) c) The shelf's full capacity is \(72\) books.

5119978

Two friends compare their savings. Lucas has \(\$85\) and adds \(\$12\) each month. Sarah has \(\$205\), but she uses \(\$8\) from her savings each month to pay part of her phone bill. After how many months will they have the same amount? How much will each person have then? Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Describe how each person’s savings changes each month. - Write one expression for Lucas’s savings and another for Sarah’s savings after \(x\) months. - Set the two expressions equal to find when the amounts match.

Solution

1. Let \(x\) be the number of months. Write an equation for equal savings: \(85 + 12x = 205 - 8x\). 2. Add \(8x\) to both sides: \(85 + 20x = 205\). 3. Subtract \(85\) from both sides: \(20x = 120\). 4. Divide both sides by \(20\): \(x = 6\). 5. Substitute \(6\) into either expression: \(85 + 12 \cdot 6 = 157\). 6. After 6 months, each friend will have \(\$157\).

Answer

Original equation: \(85+12x=205-8x\). After collecting variable terms: \(85+20x=205\). After \(6\) months, each friend has \(\$157\).

5122168

The formula \(C=\frac{F-32}{1.8}\) converts a temperature from degrees Fahrenheit to degrees Celsius. Is there a temperature at which the Celsius and Fahrenheit readings have the same numerical value? Write and solve an equation to find it.

Hints

- Use one variable for the numerical value that appears on both temperature scales. - Substitute that same variable for both \(C\) and \(F\) in the formula. - Clear the decimal denominator before collecting the variable terms.

Solution

1. Let \(x\) be the common numerical value, so \(C=x\) and \(F=x\). 2. Substitute into the conversion formula: \(x=\frac{x-32}{1.8}\). 3. Multiply both sides by \(1.8\): \(1.8x=x-32\). 4. Subtract \(x\): \(0.8x=-32\). Divide by \(0.8\): \(x=-40\). 5. Therefore, the two scales have the same numerical reading at \(-40\) degrees.

Answer

\(-40^\circ\text{C}=-40^\circ\text{F}\)

5125078

Lucas and Sarah are saving for bicycles. Lucas already has \(\$60\) and adds \(\$4\) each week. Sarah has \(\$10\) and adds \(\$9\) each week. a) Write a linear model for each person's savings after \(w\) weeks. b) Use the models to determine after how many weeks they will have the same amount and state that amount.

Hints

- In each model, identify the starting amount and the amount added each week. - The weekly amount is the coefficient of \(w\), and the starting amount is the constant term. - To find when the balances match, compare the two model outputs for the same value of \(w\).

Solution

1. Lucas starts with \(60\) dollars and adds \(4\) dollars per week, so \(L(w)=60+4w\). 2. Sarah starts with \(10\) dollars and adds \(9\) dollars per week, so \(S(w)=10+9w\). 3. Set the models equal to find when the balances match: \(60+4w=10+9w\). 4. Subtract \(10\) and \(4w\): \(50=5w\), so \(w=10\). 5. Evaluate either model: \(L(10)=60+4\cdot10=100\). Each person has \(\$100\).

Answer

a) Lucas: \(L(w)=60+4w\); Sarah: \(S(w)=10+9w\) b) After \(10\) weeks, each has \(\$100\).

5125338

Solve each equation by distributing and combining like terms. a) \(4(x - 3) = 2x + 10\) b) \(15 - (3a + 4) = 5a - 21\) c) \(2(3t + 4) = 5(t - 2)\)

Hints

- Distribute each factor to every term inside its parentheses. - A minus sign before parentheses changes both signs inside. - Combine like terms on each side before moving variable terms.

Solution

1. For the first equation, distribute: \(4x-12=2x+10\). Then \(2x=22\), so \(x=11\). 2. For the second equation, remove the parentheses: \(15-3a-4=5a-21\), so \(11-3a=5a-21\). Then \(32=8a\), so \(a=4\). 3. For the third equation, distribute: \(6t+8=5t-10\). Subtract \(5t\) and \(8\): \(t=-18\).

Answer

a) \(x = 11\) b) \(a = 4\) c) \(t = -18\)

5125348

Solve each equation. a) \(\frac{3}{4}x - 5 = \frac{1}{2}x + 2\) b) \(0.4(5x - 10) = 1.5x + 7\) c) \(\frac{2x - 4}{3} = x - 5\)

Hints

- Use a common denominator when combining fractional coefficients. - Distribute before combining variable terms. - Multiplying both sides by a denominator can clear a fraction. - Keep decimal points and signs aligned carefully.

Solution

1. For the first equation, subtract \(\frac{1}{2}x\): \(\frac{1}{4}x-5=2\). Then \(\frac{1}{4}x=7\), so \(x=28\). 2. For the second equation, distribute: \(2x-4=1.5x+7\). Then \(0.5x=11\), so \(x=22\). 3. For the third equation, multiply both sides by \(3\): \(2x-4=3x-15\). Add \(15\) and subtract \(2x\): \(11=x\), so \(x=11\).

Answer

a) \(x = 28\) b) \(x = 22\) c) \(x = 11\)

5125378

Solve each equation using equivalent operations. a) \(0.5(4x - 8) + 3x = 6 - (x - 2)\) b) \(\frac{1}{4}(8y + 12) - 5 = 2(y - 1) - y\) c) \(2(3m + 1) - 4(m - 2) = 5m - 2\)

Hints

- Treat decimal and fractional factors like any other factors when distributing. - A minus sign before parentheses acts like a factor of \(-1\). - Simplify both sides fully before moving variable terms. - Verify each result in the original equation.

Solution

1. For the first equation, distribute: \(2x-4+3x=6-x+2\), so \(5x-4=8-x\). Then \(6x=12\), so \(x=2\). 2. For the second equation, distribute: \(2y+3-5=2y-2-y\), so \(2y-2=y-2\). Then \(y=0\). 3. For the third equation, distribute: \(6m+2-4m+8=5m-2\), so \(2m+10=5m-2\). Then \(12=3m\), so \(m=4\).

Answer

a) \(x = 2\) b) \(y = 0\) c) \(m = 4\)

5125398

Solve \(\frac{2}{3}y - 4 = \frac{1}{6}y + 1\).

Hints

- Multiply by a common multiple of the denominators to clear the fractions. - Look for the least positive number that both denominators divide evenly into. - Move variable terms to one side and constants to the other.

Solution

1. Multiply both sides by the least common denominator, \(6\): \(4y - 24 = y + 6\). 2. Subtract \(y\): \(3y - 24 = 6\). 3. Add \(24\): \(3y = 30\). 4. Divide by \(3\): \(y = 10\).

Answer

\(y = 10\)

5125428

Solve \(2(3a - 5) + 4 = 10 - (a + 2)\) for \(a\).

Hints

- Distribute the factor to each term in the parentheses. - A minus sign before parentheses changes both signs inside. - Simplify both sides before moving variable terms.

Solution

1. Distribute on the left: \(6a - 10 + 4\). 2. Remove the parentheses on the right: \(10 - a - 2\). 3. Simplify: \(6a - 6 = 8 - a\). 4. Add \(a\): \(7a - 6 = 8\). 5. Add \(6\): \(7a = 14\). Divide by \(7\): \(a = 2\).

Answer

\(a = 2\)

5125438

Solve \(1.5(4y - 2) = 3(y + 2) + 0.5y\).

Hints

- Distribute carefully with decimal factors. - Combine like terms on each side before moving terms. - Keep decimal points aligned during subtraction and division.

Solution

1. Distribute on the left: \(6y - 3\). 2. Distribute and combine on the right: \(3y + 6 + 0.5y = 3.5y + 6\). 3. The equation is \(6y - 3 = 3.5y + 6\). 4. Subtract \(3.5y\): \(2.5y - 3 = 6\). 5. Add \(3\): \(2.5y = 9\). Divide by \(2.5\): \(y = 3.6\).

Answer

\(y = 3.6\)

5125458

Solve each equation. Pay close attention when removing parentheses. a) \(4(x - 5) = 2(x + 3)\) b) \(25 - (3z + 4) = 2(z - 2)\)

Hints

- Distribute each factor to every term inside its parentheses. - A minus sign before parentheses changes both signs. - Simplify each side before moving variable terms.

Solution

1. For the first equation, distribute: \(4x-20=2x+6\). Subtract \(2x\), add \(20\), and divide by \(2\): \(x=13\). 2. For the second equation, remove the parentheses and distribute: \(25-3z-4=2z-4\), so \(21-3z=2z-4\). Add \(3z\) and \(4\): \(25=5z\), so \(z=5\).

Answer

a) \(x = 13\) b) \(z = 5\)

5125468

Max tried to solve \(2(x + 4) - 3 = 15 - x\), but made an error. Step 1: \(2x + 4 - 3 = 15 - x\) Step 2: \(2x + 1 = 15 - x\) Step 3: \(3x = 14\) Step 4: \(x = \frac{14}{3}\) Explain Max’s error in Step 1, then find the correct solution.

Hints

- Check whether the outside factor was applied to every term inside the parentheses. - Redo the distribution before comparing later steps. - Solve the corrected equation and verify the result.

Solution

1. Max did not distribute \(2\) to both terms inside the parentheses. The correct expansion is \(2x + 8\), not \(2x + 4\). 2. Correct the equation: \(2x + 8 - 3 = 15 - x\). 3. Combine like terms: \(2x + 5 = 15 - x\). 4. Add \(x\): \(3x + 5 = 15\). 5. Subtract \(5\): \(3x = 10\). Divide by \(3\): \(x = \frac{10}{3}\).

Answer

Max failed to multiply \(4\) by \(2\) when distributing. The correct solution is \(x = \frac{10}{3}\).

5125518

Two students solve different equations. Alex solves \(3(a + 4) = 2a + 18\). Sofia solves \(2(a - 3) = 6\). Do the equations have the same solution set? Show your reasoning by solving both equations.

Hints

- Solve each equation independently. - Distribute or divide to remove the parentheses. - Compare the final solution sets, not just the appearance of the equations.

Solution

1. For Alex’s equation, distribute: \(3a + 12 = 2a + 18\). Subtract \(2a\), then subtract \(12\): \(a = 6\). Its solution set is \(S = \{6\}\). 2. For Sofia’s equation, divide by \(2\): \(a - 3 = 3\). Add \(3\): \(a = 6\). Its solution set is \(S = \{6\}\). 3. Since both solution sets are \(\{6\}\), the equations have the same solution set.

Answer

Yes. Both equations have solution set \(S = \{6\}\).

5125548

Analyze each number relationship by writing and solving an equation. a) Adding \(5\) to twice a number gives the same result as subtracting \(3\) from three times the number. What is the number? b) Is there a number that is exactly \(4\) less than five times itself? Justify your answer by solving an equation.

Hints

- The phrase “the same result” indicates an equation. - How can you represent twice a number and three times a number? - If one quantity is \(4\) less than another, how can you relate the two expressions?

Solution

1. For the first relationship, write \(2x+5=3x-3\). Subtract \(2x\): \(5=x-3\). Add \(3\): \(x=8\). 2. For the second relationship, write \(x=5x-4\). Subtract \(x\): \(0=4x-4\). Add \(4\) and divide by \(4\): \(x=1\). 3. Because the second equation has the solution \(x=1\), such a number exists.

Answer

a) The number is \(8\). b) Yes. The number is \(1\).

5125558

Solve each number riddle by writing an equation. a) One-fourth of a number plus \(0.5\) equals twice the number minus \(3\). b) Three times the sum of a number and \(4\) is \(2\) less than five times the number. What is the number?

Hints

- In part b, the word “sum” signals that parentheses are needed. - You may clear the fraction in part a before collecting the variable terms. - Pay attention to which expression is \(2\) less than the other in part b.

Solution

1. For the first riddle, write \(\frac{1}{4}x+0.5=2x-3\). Multiply both sides by \(4\): \(x+2=8x-12\). Then \(14=7x\), so \(x=2\). 2. For the second riddle, write \(3(x+4)=5x-2\). Distribute: \(3x+12=5x-2\). Then \(14=2x\), so \(x=7\).

Answer

a) \(x = 2\) b) \(x = 7\)

5125578

Solve \(4(3y - 5) - (2y + 7) = 3(y + 1)\) and give the solution set \(S\).

Hints

- Distribute each factor to every term inside its parentheses. - A minus sign before parentheses changes both signs. - Simplify both sides before moving variable terms.

Solution

1. Distribute and remove parentheses: \(12y - 20 - 2y - 7 = 3y + 3\). 2. Combine like terms: \(10y - 27 = 3y + 3\). 3. Subtract \(3y\): \(7y - 27 = 3\). 4. Add \(27\): \(7y = 30\). 5. Divide by \(7\): \(y = \frac{30}{7}\). 6. Therefore, \(S = \{\frac{30}{7}\}\).

Answer

\(S = \{\frac{30}{7}\}\)

5125628

A rectangular poster is \(10\,\text{in.}\) wide. If the width is increased by \(5\,\text{in.}\) and the height is decreased by \(2\,\text{in.}\), the area increases by \(10\,\text{in.}^2\). Let \(h\) be the original height. Write an equation with \(h\) on both sides that compares the new area with the original area, solve it algebraically, and report the original height.

Hints

- Write separate expressions for the original and new areas. - Translate “increases by \(10\,\text{in.}^2\)” as a relationship between those area expressions. - Make sure the variable appears in both area expressions before solving. - Collect the variable terms on one side after distributing.

Solution

1. The original area is \(10h\). 2. The new width is \(15\,\text{in.}\), and the new height is \(h-2\), so the new area is \(15(h-2)\). 3. The new area is \(10\,\text{in.}^2\) greater: \(15(h-2)=10h+10\). 4. Distribute: \(15h-30=10h+10\). 5. Subtract \(10h\) and add \(30\): \(5h=40\), so \(h=8\).

Answer

Equation: \(15(h-2)=10h+10\) Original height: \(8\,\text{in.}\)

5125718

Tim is 8 years old, and his younger brother Leo is 2 years old. Write and solve an equation to determine how many years from now Tim will be exactly twice Leo’s age. Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Both brothers get one year older each year. - Use a variable for the number of years from now. - Write an expression for each brother’s future age. - How can you represent “twice Leo’s age” algebraically?

Solution

1. Let \(x\) be the number of years from now. 2. Tim’s age will be \(8 + x\), and Leo’s age will be \(2 + x\). 3. Write the equation \(8 + x = 2(2 + x)\). 4. Distribute: \(8 + x = 4 + 2x\). 5. Subtract \(x\) and then subtract \(4\): \(x = 4\).

Answer

Original equation: \(8+x=2(2+x)\). After distribution and collecting variable terms: \(8=4+x\). In \(4\) years, Tim will be twice Leo's age.

5125728

Ms. Meyer is \(36\) years old, and her son Jonas is \(6\) years old. How many years from now will Ms. Meyer be exactly three times Jonas’s age? Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Use one variable for the number of years that pass. - Both ages increase by that same number of years. - Translate “three times” into an equation before solving.

Solution

1. Let \(x\) be the number of years from now. 2. Their future ages are \(36+x\) and \(6+x\), so write \(36+x=3(6+x)\). 3. Distribute: \(36+x=18+3x\). 4. Subtract \(x\) and then subtract \(18\): \(18=2x\), so \(x=9\).

Answer

Original equation: \(36+x=3(6+x)\). After distribution and collecting variable terms: \(36=18+2x\). Ms. Meyer will be three times Jonas's age in \(9\) years.

5125738

Grandpa Henry is \(50\) years old, and his granddaughter Marie is \(10\) years old. In how many years will Grandpa Henry be exactly three times Marie’s age? Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Use one variable for the number of years in the future. - Add that same variable to both current ages. - Translate “three times” into an equation and then collect the variable terms.

Solution

1. Let \(y\) be the number of years from now. 2. Their future ages are \(50+y\) and \(10+y\), so write \(50+y=3(10+y)\). 3. Distribute: \(50+y=30+3y\). 4. Subtract \(y\) and then subtract \(30\): \(20=2y\), so \(y=10\).

Answer

Original equation: \(50+y=3(10+y)\). After distribution and collecting variable terms: \(50=30+2y\). Grandpa Henry will be three times Marie's age in \(10\) years.

5125868

A hiking trail has three sections. The first section is one-third of the trail’s total length, the second section is one-fifth of the total length, and the last section is \(7\) miles long. Write and solve an equation to find the trail’s total length. Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Let the variable represent the total trail length. - Combine the fractional parts of the first two sections. - What fraction of the whole trail remains for the last section? - The remaining fraction of the total equals \(7\) miles.

Solution

1. Let \(x\) be the trail’s total length in miles. 2. Write the equation \(\frac{1}{3}x + \frac{1}{5}x + 7 = x\). 3. Use a common denominator: \(\frac{5}{15}x + \frac{3}{15}x + 7 = x\). 4. Combine like terms: \(\frac{8}{15}x + 7 = x\). 5. Subtract \(\frac{8}{15}x\) from both sides: \(7 = \frac{7}{15}x\). 6. Multiply by \(\frac{15}{7}\): \(x = 7 \cdot \frac{15}{7} = 15\).

Answer

Original equation: \(\frac{1}{3}x+\frac{1}{5}x+7=x\). After collecting variable terms: \(7=\frac{7}{15}x\). The hiking trail is \(15\) miles long.

5126198

Marie and Sophie run on a \(400\)-meter track. Marie runs at a constant \(3\,\text{m/s}\), and Sophie runs at \(3.8\,\text{m/s}\). After how many seconds has Sophie lapped Marie exactly once? Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- A one-lap lead means the runners’ distances differ by exactly \(400\) meters. - Write each runner’s distance as rate times time. - Put the same time variable in both distance expressions and solve the resulting equation.

Solution

1. Let \(t\) be the number of seconds they have been running. 2. Sophie has lapped Marie exactly once when Sophie’s distance is \(400\) meters greater, so write \(3.8t=3t+400\). 3. Subtract \(3t\): \(0.8t=400\). 4. Divide by \(0.8\): \(t=500\).

Answer

Original equation: \(3.8t=3t+400\). After collecting variable terms: \(0.8t=400\). Sophie laps Marie after \(500\) seconds.

5126208

Luke and Julia ride bicycles to a lake \(12\) miles away. Luke starts at \(3{:}00\) p.m. at \(12\,\text{mph}\). Julia starts on the same route \(10\) minutes later at \(18\,\text{mph}\). a) Write an equation for the time \(t\), in hours after Luke starts, when Julia catches him. b) Find the time and distance from the starting point when she catches him. c) Do both riders reach the lake by \(4{:}00\) p.m.? Justify your answer. For part a), use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Convert the delayed start to hours before comparing the riders' travel times. - Express Julia's travel time in terms of \(t\), accounting for her later start. - At the catch, both riders have traveled the same distance. - Calculate each full-trip travel time separately for part c).

Solution

1. Convert the delayed start: \(10\) minutes is \(\frac{1}{6}\) hour. Luke travels \(12t\) miles, and Julia travels \(18\left(t - \frac{1}{6}\right)\) miles. At the catch, their distances are equal: \(12t = 18\left(t - \frac{1}{6}\right)\). 2. Distribute and solve: \(12t = 18t - 3\), so \(6t = 3\) and \(t = 0.5\) hour. The catch occurs at \(3{:}30\) p.m., \(12 \cdot 0.5 = 6\) miles from the start. 3. Luke takes \(12 \div 12 = 1\) hour and arrives at \(4{:}00\) p.m. Julia takes \(12 \div 18 = \frac{2}{3}\) hour, or \(40\) minutes, and arrives at \(3{:}50\) p.m. Thus both reach the lake by \(4{:}00\) p.m.

Answer

a) Original equation: \(12t=18\left(t-\frac{1}{6}\right)\). After distribution and collecting variable terms: \(6t=3\). b) Julia catches Luke at \(3{:}30\) p.m., \(6\) miles from the start. c) Yes. Julia arrives at \(3{:}50\) p.m., and Luke arrives at \(4{:}00\) p.m.

5128308

Two equations are equivalent when they have the same solution set. Determine whether these equations are equivalent. Equation I: \(4(y + 2) = 2y - 6\) Equation II: \(0.5y + 10 = 3\)

Hints

- Solve each equation independently. - Isolate \(y\) using inverse operations. - Compare the complete solution sets.

Solution

1. Solve Equation I: \(4y + 8 = 2y - 6\). Then \(2y = -14\), so \(y = -7\) and \(S_1 = \{-7\}\). 2. Solve Equation II: \(0.5y + 10 = 3\). Then \(0.5y = -7\), so \(y = -14\) and \(S_2 = \{-14\}\). 3. Because \(S_1 \ne S_2\), the equations are not equivalent.

Answer

No. Equation I has solution \(y = -7\), and Equation II has solution \(y = -14\).

5139378

Solve \(5(x - 3) - 2(x + 1) = 4 - (x - 3)\).

Hints

- Distribute each factor and handle negative signs before parentheses carefully. - Simplify both sides before moving terms. - Move variable terms to one side and constants to the other.

Solution

1. Distribute and remove parentheses: \(5x - 15 - 2x - 2 = 4 - x + 3\). 2. Combine like terms: \(3x - 17 = 7 - x\). 3. Add \(x\): \(4x - 17 = 7\). 4. Add \(17\): \(4x = 24\). 5. Divide by \(4\): \(x = 6\).

Answer

\(x = 6\)

5139508

Two wire frames use the same total length of wire. One frame is a square with side length \(z + 2\) inches. The other is an equilateral triangle with side length \(2z - 1\) inches. Find the value of \(z\).

Hints

- Write a perimeter expression for each shape. - Use the fact that all sides of a square are equal and all sides of an equilateral triangle are equal. - Set the two perimeter expressions equal. - Solve the equation with the variable on both sides.

Solution

1. The square's perimeter is \(4(z + 2)\). 2. The equilateral triangle's perimeter is \(3(2z - 1)\). 3. Set the equal perimeters equal: \(4(z + 2) = 3(2z - 1)\). 4. Distribute: \(4z + 8 = 6z - 3\). 5. Add \(3\) and subtract \(4z\): \(11 = 2z\). 6. Divide by \(2\): \(z = 5.5\).

Answer

The value of \(z\) is \(5.5\).

5140838

Solve \(5(2x - 4) - 3x = 16 - (x + 8)\).

Hints

- A minus sign before parentheses changes both signs inside. - Simplify both sides before moving terms. - Move variable terms to one side and constants to the other.

Solution

1. Distribute and remove parentheses: \(10x - 20 - 3x = 16 - x - 8\). 2. Combine like terms: \(7x - 20 = 8 - x\). 3. Add \(x\): \(8x - 20 = 8\). 4. Add \(20\): \(8x = 28\). 5. Divide by \(8\): \(x = \frac{7}{2} = 3.5\).

Answer

\(x = \frac{7}{2} = 3.5\)

5142398

Write three different linear equations that each have \(x = -6\) as the solution and meet the stated form. 1. An equation of the form \(x + a = b\) 2. An equation of the form \(cx = d\) 3. An equation with \(x\) on both sides of the equal sign

Hints

- Start with the required solution and choose numbers that make both sides equal. - Substitute \(-6\) to test each equation. - For the final equation, include a nonzero \(x\)-term on each side.

Solution

1. For the first form, choose \(a = 10\). Since \(-6 + 10 = 4\), one equation is \(x + 10 = 4\). 2. For the second form, choose \(c = 5\). Since \(5(-6) = -30\), one equation is \(5x = -30\). 3. For an equation with \(x\) on both sides, one example is \(2x = x - 6\). Subtracting \(x\) gives \(x = -6\). 4. Substituting \(-6\) makes both sides equal in all three equations.

Answer

1. One answer is \(x + 10 = 4\). 2. One answer is \(5x = -30\). 3. One answer is \(2x = x - 6\).

5143768

Solve \(\frac{3}{4}(8x - 12) - 2x = 5(x - 4) + 1\), then check your solution.

Hints

- Distribute the fractional and whole-number factors. - Simplify both sides before moving variable terms. - Check the solution in the original equation.

Solution

1. Distribute: \(6x - 9 - 2x = 5x - 20 + 1\). 2. Combine like terms: \(4x - 9 = 5x - 19\). 3. Subtract \(4x\): \(-9 = x - 19\). 4. Add \(19\): \(x = 10\). 5. Check: The left side is \(\frac{3}{4}(8 \cdot 10 - 12) - 2 \cdot 10 = 31\). The right side is \(5(10 - 4) + 1 = 31\).

Answer

\(x = 10\)

5153538

Solve \(2(4x - 3) = 5(x + 2) + 2\).

Hints

- Distribute on both sides first. - Combine like terms before moving terms. - Move variable terms to one side and constants to the other.

Solution

1. Distribute: \(8x - 6 = 5x + 10 + 2\). 2. Combine like terms on the right: \(8x - 6 = 5x + 12\). 3. Subtract \(5x\): \(3x - 6 = 12\). 4. Add \(6\): \(3x = 18\). 5. Divide by \(3\): \(x = 6\).

Answer

\(x = 6\)

5153738

Solve \(\frac{3}{4}x + \frac{1}{2} = \frac{1}{3}x - 1\).

Hints

- Clear the fractions using a common multiple of all denominators. - Multiply every term by the same number. - Move variable terms to one side after clearing the fractions.

Solution

1. Multiply every term by the least common denominator, \(12\): \(9x + 6 = 4x - 12\). 2. Subtract \(4x\): \(5x + 6 = -12\). 3. Subtract \(6\): \(5x = -18\). 4. Divide by \(5\): \(x = -\frac{18}{5} = -3.6\).

Answer

\(x = -\frac{18}{5} = -3.6\)

5153748

Solve \(\frac{2}{5}(x + 3) = \frac{1}{2}x - \frac{1}{10}\).

Hints

- Clear the fractions with the least common denominator. - Distribute after the denominators are removed. - Move variable terms to one side.

Solution

1. Multiply every term by \(10\): \(4(x + 3) = 5x - 1\). 2. Distribute: \(4x + 12 = 5x - 1\). 3. Subtract \(4x\): \(12 = x - 1\). 4. Add \(1\): \(x = 13\).

Answer

\(x = 13\)

5224888

The longer side of a rectangle is exactly three times the shorter side. If the longer side is shortened by \(5\,\text{in.}\) and the shorter side is lengthened by \(5\,\text{in.}\), the new figure is a square. Find the original side lengths and perimeter of the rectangle. Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Express the longer side in terms of the shorter side. - Write the two changed side lengths. - A square has equal side lengths, so set those expressions equal. - Use the original dimensions to calculate the perimeter.

Solution

1. Let \(s\) inches be the shorter side. The longer side is \(3s\) inches. 2. After the changes, the side lengths are \(s + 5\) and \(3s - 5\). 3. Because the new figure is a square, set the side lengths equal: \(3s - 5 = s + 5\). 4. Subtract \(s\) and add \(5\): \(2s = 10\), so \(s = 5\). 5. The original longer side is \(3 \cdot 5\,\text{in.} = 15\,\text{in.}\). 6. The original perimeter is \(2(5 + 15) = 40\,\text{in.}\).

Answer

Original equation: \(3s-5=s+5\). After collecting variable terms: \(2s-5=5\). The original side lengths are \(5\,\text{in.}\) and \(15\,\text{in.}\), and the perimeter is \(40\,\text{in.}\).

5227768

Consider \(10(x - 7) = 5(x - 7)\). Find the value of \(x\) that makes the equation true. Briefly explain how you can solve it without distributing.

Hints

- Identify the expression that appears as a factor on both sides. - When can two different multiples of the same number be equal? - What value must the common factor have? - Solve the resulting one-step equation.

Solution

1. Both sides contain the common factor \(x - 7\), so the equation has the form \(10A = 5A\), where \(A = x - 7\). 2. Because \(10\) and \(5\) are different, the products can be equal only when the common factor is \(0\). 3. Set \(x - 7 = 0\). 4. Add \(7\): \(x = 7\). 5. Check: \(10(7 - 7) = 0\) and \(5(7 - 7) = 0\).

Answer

\(x = 7\). The unequal coefficients produce equal products only when the common factor \(x - 7\) is \(0\).

5228848

A square and an equilateral triangle have the same perimeter. Each side of the triangle is \(2.5\,\text{in.}\) longer than a side of the square. Find the side length of each shape.

Hints

- Write a perimeter expression for each shape. - Express the triangle's side in terms of the square's side. - Set the two perimeters equal. - Solve the equation with the variable on both sides.

Solution

1. Let \(s\) inches be the side length of the square. The triangle's side length is \(s + 2.5\) inches. 2. Set the perimeters equal: \(4s = 3(s + 2.5)\). 3. Distribute: \(4s = 3s + 7.5\). 4. Subtract \(3s\): \(s = 7.5\). 5. The triangle's side length is \(7.5\,\text{in.} + 2.5\,\text{in.} = 10\,\text{in.}\).

Answer

The square''s side length is \(7.5\,\text{in.}\), and the triangle''s side length is \(10\,\text{in.}\).

5230148

Solve the equation step by step, then check your solution. \(45 - (3x - 12) = 15 - (x - 4)\)

Hints

- Distribute each negative sign through its parentheses. - Combine like terms on each side. - Move variable terms to one side and constants to the other. - Substitute your result into both sides to check it.

Solution

1. Distribute the subtraction on both sides: \(45 - 3x + 12 = 15 - x + 4\). 2. Combine like terms: \(57 - 3x = 19 - x\). 3. Add \(3x\): \(57 = 19 + 2x\). 4. Subtract \(19\): \(38 = 2x\). 5. Divide by \(2\): \(x = 19\). 6. Check: the left side is \(45 - (3 \cdot 19 - 12) = 0\), and the right side is \(15 - (19 - 4) = 0\).

Answer

\(x = 19\)

5230788

Solve each equation for \(x\). 1) \(3(x + 4) = 2(2x - 1)\) 2) \(5(x - 1.2) = 3(x + 2)\) 3) \(0.5(6x - 10) = 2x + 7\)

Hints

- Distribute before moving terms. - Move all variable terms to one side. - Move constants to the other side. - Simplify each side before solving.

Solution

1. For the first equation, distribute on both sides: \(3x+12=4x-2\). Subtract \(3x\) and add \(2\): \(14=x\), so \(x=14\). 2. For the second equation, distribute: \(5x-6=3x+6\). Subtract \(3x\) and add \(6\): \(2x=12\), so \(x=6\). 3. For the third equation, distribute \(0.5\): \(3x-5=2x+7\). Subtract \(2x\) and add \(5\): \(x=12\).

Answer

1) \(x = 14\) 2) \(x = 6\) 3) \(x = 12\)

5230828

Solve the equation and check your answer. \(3(2x - 4) = 2(x + 5) - 2\)

Hints

- Distribute on both sides first. - Simplify each side before moving terms. - Move variable terms to one side and constants to the other. - Check by substituting your result into the original equation.

Solution

1. Distribute on both sides: \(6x - 12 = 2x + 10 - 2\). 2. Combine like terms on the right: \(6x - 12 = 2x + 8\). 3. Subtract \(2x\): \(4x - 12 = 8\). 4. Add \(12\): \(4x = 20\). 5. Divide by \(4\): \(x = 5\). 6. Check: the left side is \(3(2 \cdot 5 - 4) = 18\), and the right side is \(2(5 + 5) - 2 = 18\).

Answer

\(x = 5\)

5230848

For what value of \(y\) do expressions \(A\) and \(B\) have the same value? \(A = 5(3y - 4) - 2(7y - 8)\) \(B = 4(y + 2) - 3(2y - 1)\)

Hints

- Equal expression values can be represented with an equation. - Simplify each expression separately. - Move variable terms to one side and constants to the other. - Check by evaluating both original expressions at your result.

Solution

1. Set the expressions equal: \(5(3y - 4) - 2(7y - 8) = 4(y + 2) - 3(2y - 1)\). 2. Simplify the left side: \(15y - 20 - 14y + 16 = y - 4\). 3. Simplify the right side: \(4y + 8 - 6y + 3 = -2y + 11\). 4. Solve \(y - 4 = -2y + 11\). Add \(2y\): \(3y - 4 = 11\). 5. Add \(4\): \(3y = 15\). Divide by \(3\): \(y = 5\).

Answer

\(y = 5\)

5231658

Seven times an integer is \(31\) greater than five times the next integer. Find the two consecutive integers.

Hints

- Represent an integer and the next integer using one variable. - Translate “\(31\) greater than” carefully when writing the equation. - Distribute before collecting the variable terms.

Solution

1. Let \(x\) be the first integer. The next integer is \(x + 1\). 2. Write the equation \(7x = 5(x + 1) + 31\). 3. Distribute and combine constants: \(7x = 5x + 5 + 31 = 5x + 36\). 4. Subtract \(5x\): \(2x = 36\). 5. Divide by \(2\): \(x = 18\). 6. The next integer is \(18 + 1 = 19\).

Answer

The consecutive integers are \(18\) and \(19\).

5231748

Solve \(20 - 3[x - (2x - 5)] = 2x + 1\).

Hints

- Work from the innermost parentheses outward. - Distribute each negative sign to every term inside its parentheses. - Simplify both sides before moving terms.

Solution

1. Simplify the inner parentheses: \(x - (2x - 5) = x - 2x + 5 = -x + 5\). 2. Substitute this result: \(20 - 3(-x + 5) = 2x + 1\). 3. Distribute \(-3\): \(20 + 3x - 15 = 2x + 1\). 4. Combine constants: \(3x + 5 = 2x + 1\). 5. Subtract \(2x\) and then subtract \(5\): \(x = -4\).

Answer

\(x = -4\)

5237428

Luke and Sarah hike at different speeds. Sarah hikes at \(1.5\) times Luke's speed. Luke hikes for \(4\) hours, while Sarah hikes for \(2\) hours. Luke travels \(3\) miles farther than Sarah. a) Write an equation that can be used to find Luke's speed \(v\), in miles per hour. b) Find both hiking speeds. c) How far would Sarah travel if she hiked for \(4\) hours?

Hints

- Write each distance as speed times time. - Express Sarah's speed in terms of Luke's speed. - Use the \(3\)-mile difference to form the equation. - For part c), use Sarah's solved speed with the new time.

Solution

1. Luke travels \(4v\) miles. Sarah's speed is \(1.5v\), so in \(2\) hours she travels \(2(1.5v)=3v\) miles. 2. Luke travels \(3\) miles farther, so \(4v=3v+3\). 3. Subtract \(3v\): \(v=3\). Luke hikes at \(3\,\text{mph}\), and Sarah hikes at \(1.5\cdot3=4.5\,\text{mph}\). 4. In \(4\) hours, Sarah would travel \(4\cdot4.5=18\) miles.

Answer

a) \(4v = 3v + 3\) b) Luke hikes at \(3\,\text{mph}\), and Sarah hikes at \(4.5\,\text{mph}\). c) Sarah would travel \(18\) miles.

5239908

Lucas has \(12\) more trading cards than Marie. If Marie quadruples her number of cards and Lucas triples his number, Marie will have \(10\) more cards than Lucas. How many cards did each person have originally?

Hints

- Express Lucas’s original card count in terms of Marie’s. - Apply the quadrupling and tripling to the entire original amounts. - Translate “Marie has \(10\) more” into an equation. - Use parentheses around Lucas’s original expression.

Solution

1. Let \(x\) be Marie’s original number of cards. 2. Lucas originally has \(x + 12\) cards. 3. After the changes, Marie has \(4x\), and Lucas has \(3(x + 12)\). 4. Since Marie then has \(10\) more cards, write \(4x = 3(x + 12) + 10\). 5. Distribute and combine constants: \(4x = 3x + 36 + 10 = 3x + 46\). 6. Subtract \(3x\): \(x = 46\). 7. Lucas originally had \(46 + 12 = 58\) cards.

Answer

Marie originally had \(46\) cards, and Lucas originally had \(58\) cards.

5239928

Container 1 holds \(10\,\text{gal}\) more water than Container 2. If \(5\,\text{gal}\) is poured from Container 2 into Container 1, Container 1 will contain exactly three times as much water as Container 2. a) Find the original amount of water in each container. b) If \(x\) is the original amount in Container 2, explain what \(x - 5\) represents.

Hints

- Express both original amounts using one variable. - Track how the transfer changes each container’s amount. - The factor of \(3\) applies to the amount remaining in Container 2. - A subtraction from a starting value represents what remains after removal.

Solution

1. Let \(x\) be the original amount in Container 2. Then Container 1 initially contains \(x+10\). 2. After the transfer, Container 2 contains \(x-5\), and Container 1 contains \(x+15\). The condition gives \(x+15=3(x-5)\). 3. Distribute and solve: \(x+15=3x-15\), so \(30=2x\) and \(x=15\). Therefore, Container 2 originally held \(15\,\text{gal}\), and Container 1 originally held \(25\,\text{gal}\). 4. The expression \(x-5\) is the amount remaining in Container 2 after \(5\,\text{gal}\) is poured out.

Answer

a) Container 1 originally held \(25\,\text{gal}\), and Container 2 held \(15\,\text{gal}\). b) \(x - 5\) is the number of gallons remaining in Container 2 after the transfer.

5239948

Two seventh-grade classes are saving for a trip. Class A has \(\$120\), and Class B has \(\$100\). Class A spends some money on snacks. Class B spends \(\$20\) more than Class A. Afterward, Class A has exactly three times as much money left as Class B. How much does each class spend? Solve using a variable-on-both-sides equation. Your response must show the original equation and the first equivalent equation after all variable terms have been collected on one side.

Hints

- Write an expression for the money each class has left. - Keep parentheses around Class B's spending expression until it is simplified. - Translate the “three times as much left” relationship before collecting variable terms.

Solution

1. Let \(x\) be the amount Class A spends. Class B spends \(x+20\). 2. The remaining amounts are \(120-x\) and \(100-(x+20)=80-x\). 3. Write \(120-x=3(80-x)\). 4. Distribute: \(120-x=240-3x\). 5. Add \(3x\) to both sides: \(120+2x=240\). 6. Subtract \(120\): \(2x=120\), so \(x=60\). 7. Class B spends \(60+20=80\) dollars.

Answer

Original equation: \(120-x=3(80-x)\). After collecting variable terms: \(120+2x=240\). Class A spends \(\$60\), and Class B spends \(\$80\).

5240038

Two rain barrels contain different amounts of water. Barrel A holds \(85\,\text{gal}\), and Barrel B holds \(110\,\text{gal}\). Three times as much water is removed from Barrel B as from Barrel A. Afterward, Barrel A contains \(15\,\text{gal}\) more than Barrel B. How many gallons are removed from each barrel? Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Use one variable to describe both removal amounts. - Write expressions for the water remaining in each barrel. - Translate “\(15\,\text{gal}\) more” into an equation comparing the remainders.

Solution

1. Let \(x\) be the amount removed from Barrel A. Then \(3x\) is removed from Barrel B. 2. The remaining amounts are \(85 - x\) and \(110 - 3x\). 3. Since Barrel A has \(15\,\text{gal}\) more afterward, write \(85 - x = (110 - 3x) + 15\). 4. Simplify: \(85 - x = 125 - 3x\). 5. Add \(3x\) and subtract \(85\): \(2x = 40\), so \(x = 20\). 6. Barrel B loses \(3 \cdot 20 = 60\,\text{gal}\).

Answer

Original equation: \(85-x=(110-3x)+15\). After collecting variable terms: \(85+2x=125\). \(20\,\text{gal}\) is removed from Barrel A, and \(60\,\text{gal}\) is removed from Barrel B.

5240048

Two smartphones begin with battery charges of \(4200\,\text{mAh}\) and \(3500\,\text{mAh}\). During use, Battery 1 loses twice as many milliamp-hours as Battery 2. At the end, Battery 2 has \(200\,\text{mAh}\) more charge remaining than Battery 1. How much charge did each battery lose? Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Use a variable for the smaller charge loss. - Subtract each loss from the corresponding starting charge. - Place the extra \(200\,\text{mAh}\) on the side with the smaller remaining charge. - Check that the final remaining charges differ by \(200\,\text{mAh}\).

Solution

1. Let \(x\) be the charge lost by Battery 2. Battery 1 loses \(2x\). 2. The remaining charges are \(4200 - 2x\) and \(3500 - x\). 3. Since Battery 2 has \(200\,\text{mAh}\) more remaining, write \(3500 - x = (4200 - 2x) + 200\). 4. Combine constants: \(3500 - x = 4400 - 2x\). 5. Add \(2x\) and subtract \(3500\): \(x = 900\). 6. Battery 1 loses \(2 \cdot 900 = 1800\,\text{mAh}\).

Answer

Original equation: \(3500-x=(4200-2x)+200\). After collecting variable terms: \(3500+x=4400\). Battery 1 lost \(1800\,\text{mAh}\), and Battery 2 lost \(900\,\text{mAh}\).

5240558

A landscaping team is assigned a park area. The plan is to complete \(400\,\text{ft}^2\) per day. The team actually completes \(500\,\text{ft}^2\) per day. By a date \(2\) days earlier than the planned completion date, the team has completed the entire original assignment plus an additional \(200\,\text{ft}^2\). What was the area of the original assignment?

Hints

- Express the planned area as daily rate times planned days. - Express the actual area using the greater daily rate and \(2\) fewer days. - Relate the actual area to the original assignment plus the additional area. - Use the planned number of days as the variable.

Solution

1. Let \(x\) be the planned number of workdays. 2. The original assigned area is \(400x\,\text{ft}^2\). 3. At the actual rate, the team works \(x - 2\) days and completes \(500(x - 2)\,\text{ft}^2\). 4. The actual area is \(200\,\text{ft}^2\) greater than the original assignment, so write \(400x + 200 = 500(x - 2)\). 5. Distribute and solve: \(400x + 200 = 500x - 1000\), so \(1200 = 100x\) and \(x = 12\). 6. The original area was \(400 \cdot 12 = 4800\,\text{ft}^2\).

Answer

The original assignment covered \(4800\,\text{ft}^2\).

5244198

Two identical water tanks begin with the same amount of water. Water is pumped from the first tank at \(10\,\text{gal/min}\) and from the second tank at \(15\,\text{gal/min}\). The first pump runs \(5\) minutes longer than the second. When both pumps stop, the first tank contains \(50\,\text{gal}\), and the second contains \(30\,\text{gal}\). How much water was originally in each tank? Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Use a variable for the second pump’s running time. - The first pump runs \(5\) minutes longer. - Original amount equals amount removed plus amount remaining. - The two original amounts were equal.

Solution

1. Let \(x\) be the running time of the second pump in minutes. The first pump runs for \(x + 5\) minutes. 2. The original amount in the first tank is \(10(x + 5) + 50\,\text{gal}\). 3. The original amount in the second tank is \(15x + 30\,\text{gal}\). 4. Set the original amounts equal: \(10(x + 5) + 50 = 15x + 30\). 5. Distribute and solve: \(10x + 100 = 15x + 30\), so \(70 = 5x\) and \(x = 14\). 6. Substitute: \(15 \cdot 14 + 30 = 240\,\text{gal}\).

Answer

Original equation: \(10(x+5)+50=15x+30\). After distribution and collecting variable terms: \(70=5x\). Each tank originally contained \(240\,\text{gal}\) of water.

5244238

Two youth groups receive the same amount of money for outings. Group A buys zoo tickets for \(\$15.00\) per person and has \(\$10.00\) left. Group B visits an adventure park that costs \(\$12.00\) per person. Group B has \(5\) more people than Group A and has \(\$7.00\) left. How much money did each group receive?

Hints

- Represent the number of people in one group with a variable. - Write each group's starting amount as ticket cost plus money left. - Set the two expressions equal. - Substitute the group size back into either expression.

Solution

1. Let \(x\) be the number of people in Group A. 2. Group A received \(15x + 10\) dollars. 3. Group B has \(x + 5\) people, so it received \(12(x + 5) + 7\) dollars. 4. Set the equal amounts equal: \(15x + 10 = 12(x + 5) + 7\). 5. Distribute and simplify: \(15x + 10 = 12x + 67\). 6. Subtract \(12x + 10\): \(3x = 57\), so \(x = 19\). 7. The amount received was \(15 \cdot 19 + 10 = 295\) dollars.

Answer

Each group received \(\$295.00\).

5268178

Anna and Ben are training on a long, straight path. Ben has a \(0.3\)-mile head start and runs at a constant speed of \(6\,\text{mph}\). Anna starts from the same point and runs after him at \(8\,\text{mph}\). a) After how many minutes will Anna catch Ben? b) Explain why Anna would never catch Ben if he ran at \(9\,\text{mph}\). For part a), use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Write each runner's distance from the common starting point. - Include Ben's head start in his distance expression. - Set the distances equal for part a), then convert hours to minutes. - For part b), compare the two speeds.

Solution

1. Let \(t\) be the number of hours Anna runs before catching Ben. Anna's distance is \(8t\) miles, and Ben's distance from the common starting point is \(6t+0.3\) miles. 2. Set the distances equal: \(8t=6t+0.3\). Then \(2t=0.3\), so \(t=0.15\) hour. 3. Convert to minutes: \(0.15\cdot60=9\). Anna catches Ben after \(9\) minutes. 4. If Ben runs at \(9\,\text{mph}\), he is faster than Anna. His head start grows rather than shrinks, so Anna can never catch him.

Answer

a) Original equation: \(8t=6t+0.3\). After collecting variable terms: \(2t=0.3\). Anna catches Ben after \(9\) minutes. b) At \(9\,\text{mph}\), Ben runs faster than Anna, so the distance between them continually increases.

5279188

A sightseeing boat travels at \(15\,\text{mph}\) in still water. It takes exactly \(2\) hours to travel downstream between two docks and \(3\) hours to return upstream along the same route. Find the speed of the river current. Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Write the downstream and upstream speeds in terms of the current. - Use distance equals rate times time in each direction. - The distance between the docks is the same both ways. - Distribute before collecting the variable terms.

Solution

1. Let \(c\,\text{mph}\) be the speed of the current. 2. The downstream speed is \(15 + c\), and the upstream speed is \(15 - c\). 3. The distance between the docks is the same in both directions, so \(2(15 + c) = 3(15 - c)\). 4. Distribute: \(30 + 2c = 45 - 3c\). 5. Add \(3c\) and subtract \(30\): \(5c = 15\). 6. Divide by \(5\): \(c = 3\).

Answer

Original equation: \(2(15+c)=3(15-c)\). After distribution and collecting variable terms: \(30+5c=45\). The river current flows at \(3\,\text{mph}\).

5120398

Fence posts are installed with equal spacing. The gap from the first post to the beginning of the fence line and the gap from the last post to the end are the same as the gaps between posts. With \(12\,\text{ft}\) gaps, a certain number of posts is needed. With \(7.5\,\text{ft}\) gaps, \(3\) more posts are needed. Let \(x\) be the number of posts in the first arrangement. Write an equation with \(x\) on both sides by equating the two expressions for the same fence length. Solve that equation algebraically, then find the fence length.

Hints

- Count gaps, including both end gaps, for each arrangement. - Write one length expression for each spacing choice using the same variable. - Because the physical fence length is unchanged, set those two variable expressions equal. - Solve the resulting equation by collecting the variable terms on one side.

Solution

1. With \(x\) posts, the first arrangement has \(x+1\) equal gaps, so its length is \(12(x+1)\). 2. The second arrangement has \(x+3\) posts and \(x+4\) gaps, so its length is \(7.5(x+4)\). 3. Equate the two length expressions: \(12(x+1)=7.5(x+4)\). 4. Distribute: \(12x+12=7.5x+30\). Subtract \(7.5x\) and \(12\): \(4.5x=18\), so \(x=4\). 5. The fence length is \(12(4+1)=60\,\text{ft}\).

Answer

Equation: \(12(x+1)=7.5(x+4)\) Solution: \(x=4\) Fence length: \(60\,\text{ft}\)

5137648

A tractor leaves a farm at \(7{:}00\) a.m. and travels at \(25\,\text{mph}\). At \(7{:}42\) a.m., a truck leaves the same farm in the same direction at \(60\,\text{mph}\). Use a linear equation to find when and how far from the farm the truck catches the tractor. Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Convert the delayed start to hours. - Express the truck's travel time in terms of the tractor's travel time. - Set their distances equal at the catch. - Convert the solved time back to a clock time.

Solution

1. Convert the delayed start: \(42\) minutes is \(\frac{42}{60} = 0.7\) hour. 2. Let \(t\) be the tractor's travel time in hours when the catch occurs. The truck travels for \(t - 0.7\) hours. 3. Their distances are equal at the catch: \(25t = 60(t - 0.7)\). 4. Distribute and solve: \(25t = 60t - 42\), so \(35t = 42\) and \(t = 1.2\) hours. 5. One hour and \(12\) minutes after \(7{:}00\) a.m. is \(8{:}12\) a.m. 6. The distance from the farm is \(25 \cdot 1.2 = 30\) miles.

Answer

Original equation: \(25t=60(t-0.7)\). After distribution and collecting variable terms: \(35t=42\). The truck catches the tractor at \(8{:}12\) a.m., \(30\) miles from the farm.

5225148

Two numbers have a sum of \(100\). When the larger number is decreased by \(10\), the result is twice the smaller number. a) Find the two original numbers. b) Keep the sum at \(100\), but now require the larger number decreased by \(10\) to equal three times the smaller number. Compared with part a, must the larger number increase or decrease? Justify your answer. Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- If the smaller number is \(y\), express the larger number using the known sum. - Translate “twice” and “three times” into equations. - For part b, solve the revised equation and compare the larger number with the result from part a. - Because the sum stays fixed, a decrease in one number causes an equal increase in the other.

Solution

1. Let \(y\) be the smaller number. Then the larger number is \(100-y\), so the first condition gives \((100-y)-10=2y\). 2. Simplify: \(90-y=2y\), so \(90=3y\) and \(y=30\). The larger number is \(100-30=70\). 3. Under the revised condition, solve \(90-y=3y\). Then \(90=4y\), so \(y=22.5\). 4. The larger number is now \(100-22.5=77.5\), which is greater than \(70\). Therefore, the larger number increases.

Answer

a) Original equation: \((100-y)-10=2y\). After collecting variable terms: \(90=3y\). The numbers are \(70\) and \(30\). b) Original equation: \((100-y)-10=3y\). After collecting variable terms: \(90=4y\). The larger number increases to \(77.5\), while the smaller number decreases to \(22.5\).

5239468

A chemist has \(200\,\text{g}\) of a \(10\%\) salt solution. The chemist adds \(x\) grams of a \(40\%\) salt solution to make a mixture that is exactly \(28\%\) salt. a) Write an equation with \(x\) on both sides that equates the amount of salt before and after mixing. b) Solve the equation algebraically. Show at least two equivalent-equation steps after writing the model. c) Interpret the solution.

Hints

- Compute the grams of salt already present. - Express both the added salt and the final total mass using \(x\). - The final salt amount equals \(28\%\) of the final mixture mass. - After distributing, collect the variable terms on one side.

Solution

1. The original solution contains \(0.10\cdot200=20\) grams of salt. The added solution contributes \(0.40x\) grams of salt. 2. The final mixture has mass \(200+x\) grams and is \(28\%\) salt, so \( 20+0.40x=0.28(200+x). \) 3. Distribute: \(20+0.40x=56+0.28x\). 4. Subtract \(0.28x\): \(20+0.12x=56\). 5. Subtract \(20\): \(0.12x=36\). 6. Divide by \(0.12\): \(x=300\). 7. The chemist must add \(300\,\text{g}\) of the \(40\%\) solution.

Answer

a) \(20+0.40x=0.28(200+x)\) b) \(x=300\) c) Add \(300\,\text{g}\) of the \(40\%\) solution.

5239698

Write two different linear equations that each have \(x = -4\) as their only solution. 1. The first equation must contain parentheses and have \(x\) on both sides of the equal sign. 2. The second equation must have \(x\) on both sides of the equal sign and no parentheses. Verify that each equation has \(x = -4\) as its only solution.

Hints

- Begin with the desired solution \(x = -4\). - Apply reversible operations to create equivalent equations with variable terms on both sides. - Include a grouped expression only in the first equation. - Solve each equation you create to verify that it has exactly one solution.

Solution

1. One equation meeting the first condition is \(3(x + 6) = x + 10\). Distributing gives \(3x + 18 = x + 10\), so \(2x = -8\) and \(x = -4\). 2. One equation meeting the second condition is \(2x + 10 = x + 6\). Subtracting \(x\) and then \(10\) gives \(x = -4\). 3. Both equations use variable terms on both sides and each has exactly one solution, \(x = -4\).

Answer

1. \(3(x + 6) = x + 10\) 2. \(2x + 10 = x + 6\) Both equations have \(x = -4\) as their only solution.

5239988

A school has two computer labs. Lab Alpha initially has twice as many laptops as Lab Beta. \(9\) laptops are moved from Alpha to Beta. After the move, Beta has \(\frac{4}{5}\) as many laptops as Alpha. Find the original number of laptops in each lab. Briefly explain why the school’s total number of laptops does not change.

Hints

- Express both original counts using one variable. - Update each count after the transfer of \(9\) laptops. - The fraction \(\frac{4}{5}\) describes a multiplicative relationship between the new counts. - Decide whether any laptops enter or leave the school.

Solution

1. Let \(x\) be the original number of laptops in Lab Beta. Lab Alpha has \(2x\). 2. After the move, Beta has \(x + 9\), and Alpha has \(2x - 9\). 3. Write the equation \(x + 9 = \frac{4}{5}(2x - 9)\). 4. Multiply both sides by \(5\): \(5x + 45 = 4(2x - 9)\). 5. Distribute: \(5x + 45 = 8x - 36\). 6. Add \(36\) and subtract \(5x\): \(81 = 3x\), so \(x = 27\). 7. Lab Alpha originally had \(2 \cdot 27 = 54\) laptops. 8. The total stays the same because laptops are moved within the school; none are added or removed.

Answer

Lab Alpha originally had \(54\) laptops, and Lab Beta had \(27\). The total remains \(81\) because the laptops only change locations.

5240008

A cordless drill originally costs five times as much as a drill-bit set. During a sale, the price of each item is reduced by \(\$12\). After the discounts, the drill costs eight times as much as the drill-bit set. Find the original price of each item.

Hints

- Express the original drill price in terms of the drill-bit set price. - Subtract the discount from both original prices. - Use the new price ratio to write an equation. - Distribute carefully on the right side.

Solution

1. Let \(b\) dollars be the original price of the drill-bit set. The drill originally costs \(5b\) dollars. 2. After the discounts, write \(5b - 12 = 8(b - 12)\). 3. Distribute: \(5b - 12 = 8b - 96\). 4. Add \(96\) and subtract \(5b\): \(84 = 3b\). 5. Divide by \(3\): \(b = 28\). 6. The drill originally costs \(5 \cdot 28 = 140\) dollars.

Answer

The drill-bit set originally cost \(\$28\), and the cordless drill originally cost \(\$140\).

5240028

Two digital photo albums contain pictures. Album A has four times as many photos as Album B. If \(12\) photos are moved from Album A to Album B, Album A then has exactly twice as many photos as Album B. Starting from the original amounts, how many photos must be moved from Album A to Album B so the albums contain the same number?

Hints

- First use the \(12\)-photo transfer to find the original amounts. - Moving photos between albums does not change the total number of photos. - Equal albums must each contain half of the total. - Compare Album A’s original amount with the equal-share amount.

Solution

1. Let \(b\) be the original number of photos in Album B. Album A has \(4b\). 2. Use the first transfer condition: \(4b - 12 = 2(b + 12)\). 3. Distribute and solve: \(4b - 12 = 2b + 24\), so \(2b = 36\) and \(b = 18\). 4. The original amounts are \(18\) photos in Album B and \(4 \cdot 18 = 72\) photos in Album A. 5. There are \(72 + 18 = 90\) photos total, so equal albums would each contain \(90 \div 2 = 45\) photos. 6. Album A must lose \(72 - 45 = 27\) photos.

Answer

A total of \(27\) photos must be moved from Album A to Album B.

5240088

A school choir has boys and girls in the ratio \(3:7\). After \(8\) boys join and \(4\) girls leave, the ratio becomes \(1:2\). How many boys and girls were originally in the choir?

Hints

- Represent the original ratio with a common factor \(x\). - Update each group after students join or leave. - A \(1:2\) ratio means the second group is twice the first. - Write and solve the resulting proportion.

Solution

1. Let the original numbers be \(3x\) boys and \(7x\) girls. 2. After the changes, there are \(3x + 8\) boys and \(7x - 4\) girls. 3. Write the new ratio equation \(\frac{3x + 8}{7x - 4} = \frac{1}{2}\). 4. Cross multiply: \(2(3x + 8) = 7x - 4\). 5. Distribute and solve: \(6x + 16 = 7x - 4\), so \(x = 20\). 6. The original choir had \(3 \cdot 20 = 60\) boys and \(7 \cdot 20 = 140\) girls.

Answer

The choir originally had \(60\) boys and \(140\) girls.

5240528

Tim and Jordan are training for a distance race. Tim begins running at \(5\,\text{mph}\). Jordan starts from the same place \(12\) minutes later and runs at \(6\,\text{mph}\). a) How many hours after Jordan starts will Jordan catch Tim? b) Jordan wants to catch Tim exactly at the \(3\)-mile mark. What speed must Jordan maintain if Tim continues at \(5\,\text{mph}\) and Jordan still starts \(12\) minutes later? For part a), use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Convert the delayed start to hours. - At the catch-up point, both runners have traveled the same distance. - For part b), first find when Tim reaches the \(3\)-mile mark. - Subtract the delayed start from Tim's travel time.

Solution

1. Convert the head start: \(12\) minutes is \(0.2\) hour. Let \(t\) be Jordan's running time in hours. 2. At the catch-up point, both runners have traveled the same distance, so \(6t=5(t+0.2)\). Solving gives \(6t=5t+1\), so \(t=1\). 3. Tim reaches the \(3\)-mile mark in \(\frac{3}{5}=0.6\) hour. 4. Jordan has \(0.6-0.2=0.4\) hour to run \(3\) miles, so the required speed is \(\frac{3}{0.4}=7.5\,\text{mph}\).

Answer

a) Original equation: \(6t=5(t+0.2)\). After distribution and collecting variable terms: \(t=1\). Jordan catches Tim \(1\) hour after Jordan starts. b) Jordan must run at \(7.5\,\text{mph}\).

5240588

Two hiking groups follow the same trail to a lodge. Group A leaves at \(8{:}30\) a.m. and hikes at an average speed of \(2\,\text{mph}\). Group B leaves from the same point at \(10{:}00\) a.m. and hikes at \(4\,\text{mph}\). a) At what time does Group B catch Group A? b) How far from the starting point do the groups meet? c) The lodge is \(7\) miles from the starting point. Determine whether Group B catches Group A before either group reaches the lodge. For part a), use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Find Group A's head start in hours. - Set the two distance expressions equal. - Add the solved travel time to Group B's starting time. - Compare the meeting distance with the distance to the lodge.

Solution

1. Group A has a \(1.5\)-hour head start. Let \(t\) be the number of hours Group B hikes before catching Group A. 2. Equal distances give \(2(t+1.5)=4t\). Solving gives \(2t+3=4t\), so \(t=1.5\). Group B catches Group A at \(11{:}30\) a.m. 3. The meeting distance is \(4\cdot1.5=6\) miles from the starting point. 4. Since \(6<7\), the meeting occurs before the lodge, exactly \(1\) mile before it.

Answer

a) Original equation: \(2(t+1.5)=4t\). After distribution and collecting variable terms: \(3=2t\). Group B catches Group A at \(11{:}30\) a.m. b) The groups meet \(6\) miles from the starting point. c) Yes. The meeting point is \(1\) mile before the lodge.

5241238

A part-time worker agrees to work for \(8\) weeks in exchange for \(\$420\) and a tablet. The worker must leave after \(5\) weeks, so the compensation is adjusted in proportion to the time worked. The worker receives \(\$60\) and keeps the tablet. What value was assigned to the tablet?

Hints

- Represent the tablet’s value with a variable. - Find an expression for the total value of the \(8\)-week agreement. - What fraction of the full work period is represented by \(5\) weeks? - Set the proportional value equal to the cash plus the tablet.

Solution

1. Let \(x\) be the value of the tablet in dollars. 2. The total value of the original \(8\)-week compensation is \(x + 420\). 3. The value earned for \(5\) weeks is \(\frac{5}{8}(x + 420)\). 4. The compensation actually received is worth \(x + 60\), so write \(\frac{5}{8}(x + 420) = x + 60\). 5. Multiply by \(8\): \(5(x + 420) = 8(x + 60)\). 6. Distribute and solve: \(5x + 2100 = 8x + 480\), so \(1620 = 3x\) and \(x = 540\).

Answer

The tablet was valued at \(\$540\).

5241268

At a community center, one-third of the students choose a cooking activity and \(25\%\) choose a makerspace activity. The number choosing dance is three times the difference between the cooking group and the makerspace group. The remaining \(10\) students choose table tennis. a) Write an equation that can be used to find the total number of students \(x\). b) Find the total number of students at the center. Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Rewrite \(25\%\) as a fraction. - Compare the cooking and makerspace group sizes before writing their difference. - All four activity groups must add to the total number of students. - Simplify the expression for the dance group before solving.

Solution

1. Let \(x\) be the total number of students. The cooking group has \(\frac{1}{3}x\) students, the makerspace group has \(\frac{1}{4}x\), and the dance group has \(3\left(\frac{1}{3}x-\frac{1}{4}x\right)\). 2. All four groups add to the total, so \(x=\frac{1}{3}x+\frac{1}{4}x+3\left(\frac{1}{3}x-\frac{1}{4}x\right)+10\). 3. Simplify the dance term to \(\frac{1}{4}x\). Then \(x=\frac{1}{3}x+\frac{1}{4}x+\frac{1}{4}x+10=\frac{5}{6}x+10\). 4. Subtract \(\frac{5}{6}x\): \(\frac{1}{6}x=10\), so \(x=60\).

Answer

a) Original equation: \(x=\frac{1}{3}x+\frac{1}{4}x+3\left(\frac{1}{3}x-\frac{1}{4}x\right)+10\). After simplifying and collecting variable terms: \(\frac{1}{6}x=10\). b) There are \(60\) students at the center.

5244208

Two classes set the same goal for a paper-recycling drive. Class A collects an average of \(12\,\text{lb}\) per day, and Class B collects \(15\,\text{lb}\) per day. Class B started later and has collected for \(2\) fewer days. At one point, Class A is \(40\,\text{lb}\) short of the goal, while Class B is \(46\,\text{lb}\) short. What is the common collection goal?

Hints

- Identify the quantity that is the same for both classes. - If Class A collects for \(x\) days, express Class B’s number of days. - For each class, write collected amount plus amount still needed. - Set the two goal expressions equal.

Solution

1. Let \(x\) be the number of days Class A has collected. Class B has collected for \(x - 2\) days. 2. Class A’s goal can be written as \(12x + 40\). 3. Class B’s goal can be written as \(15(x - 2) + 46\). 4. Set the expressions equal: \(12x + 40 = 15(x - 2) + 46\). 5. Distribute and solve: \(12x + 40 = 15x + 16\), so \(24 = 3x\) and \(x = 8\). 6. The goal is \(12 \cdot 8 + 40 = 136\,\text{lb}\).

Answer

The collection goal is \(136\,\text{lb}\) for each class.

5280748

A cyclist and a runner start at the same time at Point P and travel along a straight route toward Point Q. The cyclist reaches Q and immediately turns back. The cyclist meets the runner exactly \(40\) minutes after the start at a point \(2\) miles from Q. The runner takes exactly \(1\) hour to travel from P to Q. Find the distance from P to Q and the speed of each person. Use a variable-on-both-sides equation. Your response must show the original equation and an equivalent equation after all variable terms have been collected on one side.

Hints

- Use the runner's \(1\)-hour travel time to connect distance and speed. - At the meeting point, the runner is \(2\) miles short of Q. - Convert \(40\) minutes to hours. - The cyclist travels to Q and then \(2\) miles back.

Solution

1. Let \(s\) miles be the distance from P to Q. 2. Because the runner takes \(1\) hour to travel \(s\) miles, the runner's speed is \(s\,\text{mph}\). 3. The meeting occurs after \(40\) minutes, or \(\frac{2}{3}\) hour. By then, the runner has traveled \(s - 2\) miles. 4. Write \(s - 2 = \frac{2}{3}s\). 5. Subtract \(\frac{2}{3}s\): \(\frac{1}{3}s = 2\), so \(s = 6\). 6. The runner's speed is therefore \(6\,\text{mph}\). 7. The cyclist travels \(6 + 2 = 8\) miles in \(\frac{2}{3}\) hour, so the cyclist's speed is \(8 \div \frac{2}{3} = 12\,\text{mph}\).

Answer

Original equation: \(s-2=\frac{2}{3}s\). After collecting variable terms: \(\frac{1}{3}s=2\). The distance from P to Q is \(6\) miles. The runner travels at \(6\,\text{mph}\), and the cyclist travels at \(12\,\text{mph}\).

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