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Solve equations with distribution

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5519948
To begin solving \(2(x + 3) = 10\), which equation correctly applies the distributive property? A) \(2x + 3 = 10\) B) \(2x + 6 = 10\) C) \(x + 6 = 10\) D) \(2x + 5 = 10\)

Hints

- Focus only on the first step needed to remove the parentheses. - Check whether the outside factor is applied to every term inside the parentheses.

Solution

1. The factor \(2\) multiplies both terms inside the parentheses. 2. Therefore, \(2(x + 3)\) rewrites as \(2x + 6\). 3. The correct choice is B.

Answer

B) \(2x + 6 = 10\)
5519958
The rectangle in the diagram has area \(15\) square units. Which equation correctly applies the distributive property to the area equation and begins solving for \(x\)? A) \(3x + 2 = 15\) B) \(3x + 6 = 15\) C) \(x + 6 = 15\) D) \(3x + 5 = 15\)
Figure for problem 551995

Hints

- Use the two side labels in the diagram to form the rectangle's area expression. - Check which choice applies the outside factor to every term inside the parentheses.

Solution

1. The side labels in the diagram give the area equation \(3(x + 2) = 15\). 2. Distributing \(3\) gives \(3x + 6 = 15\). 3. The correct choice is B.

Answer

B) \(3x + 6 = 15\)
5100618
Solve \(3(2b + 3) - 4(2 + b) = 5\).

Hints

- Use the distributive property to remove both sets of parentheses. - Pay close attention to the subtraction before the second set of parentheses. - Combine like terms before isolating the variable.

Solution

1. Distribute: \(6b + 9 - 8 - 4b = 5\). 2. Combine like terms: \(2b + 1 = 5\). 3. Subtract \(1\): \(2b = 4\). 4. Divide by \(2\): \(b = 2\).

Answer

\(b = 2\)
5125358
Solve each equation. a) \(4(x - 3) + 7 = 19\) b) \(12 - 3(y + 2) = 0\) c) \(5z + 2(z - 4) = 13\)

Hints

- Distribute the factor to every term inside the parentheses. - Pay special attention to the negative factor in part b. - Combine like terms before isolating the variable. - Substitute your result to check it.

Solution

1. For the first equation, distribute and combine: \(4x-12+7=19\), so \(4x-5=19\). Add \(5\) and divide by \(4\) to get \(x=6\). 2. For the second equation, distribute: \(12-3y-6=0\), so \(6-3y=0\). Subtract \(6\) and divide by \(-3\) to get \(y=2\). 3. For the third equation, distribute and combine: \(5z+2z-8=13\), so \(7z-8=13\). Add \(8\) and divide by \(7\) to get \(z=3\).

Answer

a) \(x = 6\) b) \(y = 2\) c) \(z = 3\)
5125388
Solve \(7x - (2x + 5) = 15\) using equivalent operations.

Hints

- A minus sign before parentheses changes the signs inside. - Simplify the left side before isolating the variable. - Use inverse operations one step at a time.

Solution

1. Remove the parentheses, including the negative sign: \(7x - 2x - 5 = 15\). 2. Combine like terms: \(5x - 5 = 15\). 3. Add \(5\): \(5x = 20\). 4. Divide by \(5\): \(x = 4\).

Answer

\(x = 4\)
5125418
Solve \(5x - 3(x + 4) = 4\).

Hints

- Distribute \(-3\) to both terms inside the parentheses. - Combine like terms before isolating the variable. - Use inverse operations on both sides.

Solution

1. Distribute the negative factor: \(5x - 3x - 12 = 4\). 2. Combine like terms: \(2x - 12 = 4\). 3. Add \(12\): \(2x = 16\). 4. Divide by \(2\): \(x = 8\).

Answer

\(x = 8\)
5225228
Let \(A=5-2(x+1)\). Find the value of \(x\) that makes \(A=-1\). Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Replace \(A\) with the target value \(-1\) to form an equation. - Distribute \(-2\) to both terms inside the parentheses. - After simplifying, isolate \(x\) with inverse operations.

Solution

1. Set the expression equal to \(-1\): \(5-2(x+1)=-1\). 2. Distribute the negative factor and combine constants: \(5-2x-2=-1\), so \(3-2x=-1\). 3. Subtract \(3\): \(-2x=-4\). Divide by \(-2\): \(x=2\).

Answer

Before distribution: \(5-2(x+1)=-1\). After distribution: \(5-2x-2=-1\). \(x=2\).
5227758
For what value of \(x\) does \(3(x + 5) - 12\) equal \(0\)? Solve step by step. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Use the distributive property to remove the parentheses. - Combine the constant terms before solving. - Isolate the variable term, then divide by its coefficient. - Substitute your answer into the expression to check that it equals \(0\).

Solution

1. Write an equation: \(3(x + 5) - 12 = 0\). 2. Distribute \(3\): \(3x + 15 - 12 = 0\). 3. Combine like terms: \(3x + 3 = 0\). 4. Subtract \(3\): \(3x = -3\). 5. Divide by \(3\): \(x = -1\).

Answer

Before distribution: \(3(x+5)-12=0\). After distribution: \(3x+15-12=0\). \(x=-1\).
5230138
Solve each equation and determine whether the equations have the same solution set. 1) \(18 - (x + 7) = 5\) 2) \(24 = 35 - (x + 5)\) For each equation, use the distributive property before isolating the variable. Your response must show each equation immediately before and immediately after distribution.

Hints

- Distribute the negative sign to every term in the parentheses. - Simplify each side before isolating \(x\). - Two equations have the same solution set when exactly the same values make them true.

Solution

1. For the first equation, distribute the subtraction: \(18-x-7=5\). Combine constants to get \(11-x=5\), so \(-x=-6\) and \(x=6\). 2. For the second equation, distribute the subtraction: \(24=35-x-5\). Combine constants to get \(24=30-x\), so \(-6=-x\) and \(x=6\). 3. Both equations have the same single solution, so they have the same solution set.

Answer

1) Before distribution: \(18-(x+7)=5\). After distribution: \(18-x-7=5\). \(S=\{6\}\). 2) Before distribution: \(24=35-(x+5)\). After distribution: \(24=35-x-5\). \(S=\{6\}\). The equations have the same solution set.
5230818
Solve \(6(x - 4) + 8 = 20\) for \(x\). Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Apply the factor outside the parentheses to each term inside. - Combine the constant terms. - Isolate the variable term before dividing. - Divide by the coefficient of \(x\).

Solution

1. Distribute \(6\): \(6x - 24 + 8 = 20\). 2. Combine constants: \(6x - 16 = 20\). 3. Add \(16\): \(6x = 36\). 4. Divide by \(6\): \(x = 6\).

Answer

Before distribution: \(6(x-4)+8=20\). After distribution: \(6x-24+8=20\). \(x=6\).
5230898
Solve \(0.5(x - 1.2) + 0.4(x + 0.5) = 0.5\) for \(x\).

Hints

- Use the distributive property to remove each set of parentheses. - Combine the terms containing \(x\) and the constants separately. - Isolate the variable term before dividing. - Check by substitution.

Solution

1. Distribute: \(0.5x - 0.6 + 0.4x + 0.2 = 0.5\). 2. Combine like terms: \(0.9x - 0.4 = 0.5\). 3. Add \(0.4\): \(0.9x = 0.9\). 4. Divide by \(0.9\): \(x = 1\).

Answer

\(x = 1\)
5125308
Two students solve \(6(x + 2) = 30\) in different ways. Alex distributes first and then solves for \(x\). Priya divides both sides by \(6\) first. a) Solve the equation using Alex’s method. b) Solve the equation using Priya’s method. c) Compare the methods. When is Priya’s first step especially efficient?

Hints

- Carry out both methods one step at a time. - Compare the number and complexity of the steps. - Look closely at the relationship between \(6\) and \(30\).

Solution

1. Using Alex’s method, distribute: \(6x+12=30\). 2. Subtract \(12\): \(6x=18\). Divide by \(6\): \(x=3\). 3. Using Priya’s method, divide both sides of the original equation by \(6\): \(x+2=5\). 4. Subtract \(2\): \(x=3\). 5. Both methods give the same solution. Dividing first is especially efficient when removing the outside factor immediately simplifies both sides without creating awkward values.

Answer

a) \(x = 3\) b) \(x = 3\) c) Dividing first is efficient when removing the outside factor immediately makes the equation simpler.
5125368
Solve each equation for the indicated variable. a) \(3(x + 4) - 2x = 5(x - 2)\) b) \(8 - (2a + 5) = 3(a + 1)\) c) \(4(k - 1) + 2(3 - k) = 10\)

Hints

- Distribute every factor, including a negative sign before parentheses. - Combine like terms on each side. - Move variable terms to one side before isolating the variable. - Check the zero solution in part b carefully.

Solution

1. For the first equation, distribute and combine: \(3x+12-2x=5x-10\), so \(x+12=5x-10\). Subtract \(x\) and add \(10\): \(22=4x\), so \(x=5.5\). 2. For the second equation, remove the parentheses: \(8-2a-5=3a+3\), so \(3-2a=3a+3\). Subtract \(3\) and add \(2a\): \(0=5a\), so \(a=0\). 3. For the third equation, distribute and combine: \(4k-4+6-2k=10\), so \(2k+2=10\). Subtract \(2\) and divide by \(2\): \(k=4\).

Answer

a) \(x = 5.5\) b) \(a = 0\) c) \(k = 4\)
5139488
A rectangular garden bed has length \(2x + 5\) feet and width \(x + 3\) feet. Its perimeter is \(52\,\text{ft}\). Find the value of \(x\). Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Use the perimeter formula for a rectangle. - Add the expressions for the length and width first. - Set the perimeter expression equal to \(52\). - Use the distributive property when removing parentheses.

Solution

1. Use the perimeter formula for a rectangle: \(2(\text{length} + \text{width}) = \text{perimeter}\). 2. Substitute the given expressions: \(2((2x + 5) + (x + 3)) = 52\). 3. Combine like terms: \(2(3x + 8) = 52\). 4. Distribute: \(6x + 16 = 52\). 5. Subtract \(16\): \(6x = 36\). 6. Divide by \(6\): \(x = 6\).

Answer

Before distribution: \(2(3x+8)=52\). After distribution: \(6x+16=52\). \(x=6\).
5139498
An isosceles triangle has a base of length \(y - 2\) inches. Each of its two congruent sides has length \(2y + 1\) inches. The perimeter is \(35\,\text{in.}\). Find the value of \(y\).

Hints

- Recall what an isosceles triangle tells you about its side lengths. - Add all three side lengths to write the perimeter equation. - Remember that the expression for a congruent side is used twice. - Distribute before combining like terms.

Solution

1. Add the base and the two congruent sides: \((y - 2) + 2(2y + 1) = 35\). 2. Distribute: \(y - 2 + 4y + 2 = 35\). 3. Combine like terms: \(5y = 35\). 4. Divide by \(5\): \(y = 7\).

Answer

The value of \(y\) is \(7\).
5143758
Solve \(5(1.2x - 2) - 0.4x = 18\), then check your solution.

Hints

- Distribute the factor outside the parentheses. - Combine decimal coefficients carefully. - Substitute the result into the original equation to check it.

Solution

1. Distribute: \(6x - 10 - 0.4x = 18\). 2. Combine like terms: \(5.6x - 10 = 18\). 3. Add \(10\): \(5.6x = 28\). 4. Divide by \(5.6\): \(x = 5\). 5. Check: \(5(1.2 \cdot 5 - 2) - 0.4 \cdot 5 = 5(4) - 2 = 18\).

Answer

\(x = 5\)
5153758
Solve \(\frac{2x - 1}{3} - \frac{x + 3}{6} = \frac{1}{2}\).

Hints

- Multiply by a common denominator to clear all fractions. - Apply the subtraction sign to the entire second numerator. - Combine like terms before isolating \(x\).

Solution

1. Multiply every term by \(6\): \(2(2x - 1) - (x + 3) = 3\). 2. Distribute and remove parentheses: \(4x - 2 - x - 3 = 3\). 3. Combine like terms: \(3x - 5 = 3\). 4. Add \(5\): \(3x = 8\). 5. Divide by \(3\): \(x = \frac{8}{3}\).

Answer

\(x = \frac{8}{3}\)
5224878
A rectangle has a perimeter of \(126\,\text{in.}\). Its length is \(13\,\text{in.}\) greater than its width. Find both dimensions. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Represent the width with a variable. - Express the length in terms of the width. - Use the perimeter formula for a rectangle. - Distribute and solve the equation.

Solution

1. Let \(x\) inches be the width. The length is \(x + 13\) inches. 2. Use the perimeter formula: \(126 = 2(x + x + 13)\). 3. Distribute and combine like terms: \(126 = 4x + 26\). 4. Subtract \(26\): \(100 = 4x\). 5. Divide by \(4\): \(x = 25\). 6. The length is \(25\,\text{in.} + 13\,\text{in.} = 38\,\text{in.}\).

Answer

Before distribution: \(126=2(2x+13)\). After distribution: \(126=4x+26\). The width is \(25\,\text{in.}\), and the length is \(38\,\text{in.}\).
5224948
Consider \(4(x - 1.5) = 10\). Solve the equation in two different ways and show that both methods give the same result. a) First use the distributive property. b) First divide both sides by \(4\).

Hints

- Apply the factor outside the parentheses to each term inside. - Dividing both sides by the same nonzero number preserves equality. - Keep both sides balanced at every step. - Compare the final value from each method.

Solution

1. Using the distributive property first gives \(4x-6=10\). Add \(6\): \(4x=16\). Divide by \(4\): \(x=4\). 2. Dividing both sides of the original equation by \(4\) first gives \(x-1.5=10\div4=2.5\). Add \(1.5\): \(x=4\). 3. Both valid methods give the same solution, \(x=4\).

Answer

a) \(x = 4\) b) \(x = 4\)
5224958
A rectangle has a perimeter of \(40\,\text{in.}\). If the shorter side is increased by \(3\,\text{in.}\) and the longer side is decreased by \(3\,\text{in.}\), the new figure is a square. Find the original side lengths. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Use the fact that all sides of a square are equal. - Determine how much longer the original long side is than the short side. - Write a perimeter equation for the original rectangle. - Solve for the shorter side first.

Solution

1. Let \(x\) inches be the shorter side. 2. Because adding \(3\) to the shorter side and subtracting \(3\) from the longer side makes them equal, the original longer side is \(x + 6\) inches. 3. Use the perimeter formula: \(2(x + x + 6) = 40\). 4. Distribute and combine like terms: \(4x + 12 = 40\). 5. Subtract \(12\) and divide by \(4\): \(4x = 28\), so \(x = 7\). 6. The longer side is \(7\,\text{in.} + 6\,\text{in.} = 13\,\text{in.}\).

Answer

Before distribution: \(2(2x+6)=40\). After distribution: \(4x+12=40\). The original side lengths are \(7\,\text{in.}\) and \(13\,\text{in.}\).
5225288
At a field day, \(220\) certificates are awarded: winner certificates, merit certificates, and participation certificates. There are \(20\) more merit certificates than winner certificates. The number of participation certificates is four times the number of merit certificates. Write and solve an equation to find the number of each type of certificate.

Hints

- Let the smallest certificate count be \(x\). - Use parentheses because the factor of \(4\) applies to the entire merit-certificate count. - The three expressions must add to \(220\). - Check that the three results total \(220\).

Solution

1. Let \(x\) be the number of winner certificates. 2. The number of merit certificates is \(x + 20\), and the number of participation certificates is \(4(x + 20)\). 3. Write the total equation: \(x + (x + 20) + 4(x + 20) = 220\). 4. Distribute and combine like terms: \(x + x + 20 + 4x + 80 = 220\), so \(6x + 100 = 220\). 5. Subtract \(100\): \(6x = 120\). Divide by \(6\): \(x = 20\). 6. There are \(20\) winner certificates, \(20 + 20 = 40\) merit certificates, and \(4 \cdot 40 = 160\) participation certificates.

Answer

There are \(20\) winner certificates, \(40\) merit certificates, and \(160\) participation certificates.
5227848
Solve each equation. 1) \(3(a + 5) = 27\) 2) \(4(2b - 3) = 5b + 9\) 3) \(14 - 2(x - 1) = 6\)

Hints

- Use the distributive property before isolating the variable in each equation. - Track negative factors carefully when removing parentheses. - After distributing, combine like terms before using inverse operations. - Substitute each result to check it.

Solution

1. Distribute \(3\): \(3a + 15 = 27\). Subtract \(15\): \(3a = 12\). Divide by \(3\): \(a = 4\). 2. Distribute \(4\): \(8b - 12 = 5b + 9\). Subtract \(5b\): \(3b - 12 = 9\). Add \(12\): \(3b = 21\). Divide by \(3\): \(b = 7\). 3. Distribute \(-2\): \(14 - 2x + 2 = 6\). Combine constants: \(16 - 2x = 6\). Subtract \(16\): \(-2x = -10\). Divide by \(-2\): \(x = 5\).

Answer

1) \(a = 4\) 2) \(b = 7\) 3) \(x = 5\)
5228838
A rectangular garden bed has a perimeter of \(44\,\text{ft}\). Its length is \(2\,\text{ft}\) less than three times its width. Find the length and width. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Represent the width with a variable. - Express the length using the given relationship. - Use the perimeter formula for a rectangle. - Distribute before solving the equation.

Solution

1. Let \(x\) feet be the width. The length is \(3x - 2\) feet. 2. Use the perimeter formula: \(2(x + 3x - 2) = 44\). 3. Combine like terms: \(2(4x - 2) = 44\). 4. Distribute: \(8x - 4 = 44\). 5. Add \(4\): \(8x = 48\). Divide by \(8\): \(x = 6\). 6. The length is \(3 \cdot 6\,\text{ft} - 2\,\text{ft} = 16\,\text{ft}\).

Answer

Before distribution: \(2(4x-2)=44\). After distribution: \(8x-4=44\). The garden bed is \(6\,\text{ft}\) wide and \(16\,\text{ft}\) long.
5229308
Solve each equation using equivalent transformations. a) \(0.6t + 2(0.1t - 0.6) = 2\) b) \(1\frac{1}{2}w - (0.5w + 4) = 6\) c) \(\frac{2}{3}(6x - 9) + 0.5x = 7.5\)

Hints

- Apply each outside factor or subtraction to every term inside its parentheses. - Rewrite fractions or mixed numbers in a convenient equivalent form when useful. - Combine like terms after removing parentheses. - Check each answer by substitution.

Solution

1. For a), distribute \(2\): \(0.6t + 0.2t - 1.2 = 2\). Combine like terms: \(0.8t - 1.2 = 2\). Then \(0.8t = 3.2\), so \(t = 4\). 2. For b), rewrite \(1\frac{1}{2}\) as \(1.5\) and distribute the subtraction: \(1.5w - 0.5w - 4 = 6\). Combine like terms: \(w - 4 = 6\), so \(w = 10\). 3. For c), distribute \(\frac{2}{3}\): \(4x - 6 + 0.5x = 7.5\). Combine like terms: \(4.5x - 6 = 7.5\). Then \(4.5x = 13.5\), so \(x = 3\).

Answer

a) \(t = 4\) b) \(w = 10\) c) \(x = 3\)
5229358
A group of \(32\) people visits a museum. Adult admission costs \(\$10\), and child admission costs \(\$7\). The group pays \(\$254\) in all. How many adults and how many children are in the group? Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Express the number of children using the total number of people. - Multiply each admission price by the corresponding number of people. - Distribute before combining like terms. - Check that the two group sizes add to \(32\).

Solution

1. Let \(x\) be the number of adults. Then the number of children is \(32 - x\). 2. Write the cost equation: \(10x + 7(32 - x) = 254\). 3. Distribute: \(10x + 224 - 7x = 254\). 4. Combine like terms: \(3x + 224 = 254\). 5. Subtract \(224\): \(3x = 30\), so \(x = 10\). 6. The number of children is \(32 - 10 = 22\).

Answer

Before distribution: \(10x+7(32-x)=254\). After distribution: \(10x+224-7x=254\). The group has \(10\) adults and \(22\) children.
5229438
A shipping center uses small, medium, and large packages. A medium package weighs \(2\,\text{lb}\) more than a small package, and a large package weighs twice as much as a small package. A cart holds \(10\) small, \(5\) medium, and \(2\) large packages with a total weight of \(86\,\text{lb}\). Find the weight of each package size.

Hints

- Use the small-package weight as the variable. - Express the other package weights in terms of the small-package weight. - Multiply each package weight by the number of packages of that size. - Add the three weight expressions to equal \(86\,\text{lb}\).

Solution

1. Let \(x\) be the weight of one small package in pounds. 2. A medium package weighs \(x + 2\), and a large package weighs \(2x\). 3. Write the total-weight equation \(10x + 5(x + 2) + 2(2x) = 86\). 4. Distribute and combine like terms: \(10x + 5x + 10 + 4x = 86\), so \(19x + 10 = 86\). 5. Subtract \(10\): \(19x = 76\). Divide by \(19\): \(x = 4\). 6. The medium package weighs \(4 + 2 = 6\,\text{lb}\), and the large package weighs \(2 \cdot 4 = 8\,\text{lb}\).

Answer

A small package weighs \(4\,\text{lb}\), a medium package weighs \(6\,\text{lb}\), and a large package weighs \(8\,\text{lb}\).
5229448
In a triangle, angle \(\beta\) is \(20^\circ\) greater than angle \(\alpha\). Angle \(\gamma\) is three times as large as angle \(\beta\). Find all three angle measures.

Hints

- Recall the sum of the interior angles of a triangle. - Express the other two angles in terms of \(\alpha\). - Pay attention to which angle is tripled. - Distribute before combining like terms.

Solution

1. Let \(x\) degrees be the measure of angle \(\alpha\). 2. Then \(\beta = x + 20\), and \(\gamma = 3(x + 20)\). 3. The interior angles of a triangle add to \(180^\circ\), so \(x + (x + 20) + 3(x + 20) = 180\). 4. Distribute and combine like terms: \(5x + 80 = 180\). 5. Subtract \(80\) and divide by \(5\): \(5x = 100\), so \(x = 20\). 6. Therefore, \(\alpha = 20^\circ\), \(\beta = 40^\circ\), and \(\gamma = 120^\circ\).

Answer

The angle measures are \(\alpha = 20^\circ\), \(\beta = 40^\circ\), and \(\gamma = 120^\circ\).
5230168
At a school carnival, a class sells muffins for \(\$1.50\), pretzels for \(\$1.10\), and juice boxes for \(\$0.80\). The class sells the same number of muffins and pretzels, and it sells \(10\) more juice boxes than muffins. Total sales are \(\$144.00\). How many of each item are sold?

Hints

- Use one variable for the equal numbers of muffins and pretzels. - Express the number of juice boxes in terms of that variable. - Multiply each price by its quantity and add. - Distribute the juice-box price across the parentheses.

Solution

1. Let \(x\) be the number of muffins sold. Then \(x\) pretzels and \(x + 10\) juice boxes are sold. 2. Write the revenue equation: \(1.50x + 1.10x + 0.80(x + 10) = 144\). 3. Distribute and combine like terms: \(1.5x + 1.1x + 0.8x + 8 = 144\), so \(3.4x + 8 = 144\). 4. Subtract \(8\): \(3.4x = 136\). 5. Divide by \(3.4\): \(x = 40\). 6. Therefore, \(40\) muffins, \(40\) pretzels, and \(50\) juice boxes are sold.

Answer

The class sells \(40\) muffins, \(40\) pretzels, and \(50\) juice boxes.
5230838
Solve \(6(2x - 3) - 3(4x + 5) + 4(2x - 1) = 9\) for \(x\).

Hints

- Pay close attention to the negative factor before the second set of parentheses. - Combine all terms containing \(x\). - Combine all constant terms separately.

Solution

1. Distribute each factor: \(12x - 18 - 12x - 15 + 8x - 4 = 9\). 2. Combine like terms: \(8x - 37 = 9\). 3. Add \(37\): \(8x = 46\). 4. Divide by \(8\): \(x = \frac{46}{8} = \frac{23}{4} = 5.75\).

Answer

\(x = \frac{23}{4} = 5.75\)
5230908
Solve \(0.8(x + 2) - 0.3(2x - 4) + 0.5x = 4.2\).

Hints

- Distribute the negative decimal factor carefully. - Combine variable terms and constants separately. - Use subtraction to isolate the variable term. - Divide by the coefficient of \(x\).

Solution

1. Distribute, paying attention to the subtraction: \(0.8x + 1.6 - 0.6x + 1.2 + 0.5x = 4.2\). 2. Combine the variable terms and constants: \(0.7x + 2.8 = 4.2\). 3. Subtract \(2.8\): \(0.7x = 1.4\). 4. Divide by \(0.7\): \(x = 2\).

Answer

\(x = 2\)
5231768
Ms. Nguyen buys \(45\) writing supplies for a class project for a total of \(\$38.60\). The order includes markers that cost \(\$1.20\) each and pencils that cost \(\$0.50\) each. a) Write an equation that can be used to find the number of markers \(x\). b) Find the number of each item. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Express the number of pencils using the total number of items. - Multiply each unit price by its quantity. - Add the two costs to form the total-cost equation. - Check both the item count and total price.

Solution

1. Let \(x\) be the number of markers. Then the number of pencils is \(45-x\), so the cost equation is \(1.20x+0.50(45-x)=38.60\). 2. Distribute and combine like terms: \(1.2x+22.5-0.5x=38.6\), so \(0.7x+22.5=38.6\). 3. Subtract \(22.5\): \(0.7x=16.1\). Divide by \(0.7\): \(x=23\). 4. The number of pencils is \(45-23=22\).

Answer

a) Before distribution: \(1.20x+0.50(45-x)=38.60\). After distribution: \(1.20x+22.50-0.50x=38.60\). b) Ms. Nguyen buys \(23\) markers and \(22\) pencils.
5237508
A school carnival sells \(150\) admission tickets. Child tickets cost \(\$2.00\), and adult tickets cost \(\$5.00\). Total ticket sales are \(\$420.00\). Use a calculation to determine whether more child tickets or adult tickets were sold. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Represent one ticket count with a variable. - Use the total number of tickets to express the other count. - Write an equation for total revenue. - Compare the two quantities after solving.

Solution

1. Let \(x\) be the number of adult tickets. Then the number of child tickets is \(150 - x\). 2. Write the revenue equation: \(5x + 2(150 - x) = 420\). 3. Distribute and combine like terms: \(5x + 300 - 2x = 420\), so \(3x + 300 = 420\). 4. Subtract \(300\): \(3x = 120\), so \(x = 40\). 5. The number of child tickets is \(150 - 40 = 110\). Since \(110 > 40\), more child tickets were sold.

Answer

Before distribution: \(5x+2(150-x)=420\). After distribution: \(5x+300-2x=420\). More child tickets were sold: \(110\) child tickets and \(40\) adult tickets.
5237618
A school snack stand sells sandwiches for \(\$1.80\) and pretzels for \(\$1.20\). During one break, it sells \(60\) snacks for a total of \(\$93.60\). How many sandwiches and how many pretzels were sold? Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Express one item count using the total number of snacks. - Multiply each price by its quantity. - Distribute before combining like terms. - Check both the total item count and revenue.

Solution

1. Let \(x\) be the number of sandwiches. Then the number of pretzels is \(60 - x\). 2. Write the revenue equation: \(1.80x + 1.20(60 - x) = 93.60\). 3. Distribute and combine like terms: \(1.8x + 72 - 1.2x = 93.6\), so \(0.6x + 72 = 93.6\). 4. Subtract \(72\): \(0.6x = 21.6\). 5. Divide by \(0.6\): \(x = 36\). 6. The number of pretzels is \(60 - 36 = 24\).

Answer

Before distribution: \(1.80x+1.20(60-x)=93.60\). After distribution: \(1.80x+72-1.20x=93.60\). The stand sold \(36\) sandwiches and \(24\) pretzels.
5240118
Two pumps drain a pool. Pump A moves \(10\,\text{gal}\) more per minute than Pump B. Pump A runs for \(45\) minutes, and Pump B runs for \(60\) minutes. Together, they move \(9900\,\text{gal}\) of water. What is each pump’s rate in gallons per minute?

Hints

- Use a variable for the lower pump rate. - Amount moved equals rate times time. - Add the amounts moved by both pumps to equal the total. - Keep Pump A’s rate \(10\) greater than Pump B’s.

Solution

1. Let \(x\) be Pump B’s rate in gallons per minute. Pump A’s rate is \(x + 10\). 2. Write the total-volume equation \(45(x + 10) + 60x = 9900\). 3. Distribute and combine like terms: \(45x + 450 + 60x = 9900\), so \(105x + 450 = 9900\). 4. Subtract \(450\): \(105x = 9450\). 5. Divide by \(105\): \(x = 90\). 6. Pump A’s rate is \(90 + 10 = 100\,\text{gal/min}\).

Answer

Pump A moves \(100\,\text{gal/min}\), and Pump B moves \(90\,\text{gal/min}\).
5240128
Two classes plant trees for an environmental project. Class A plants \(3\) more trees per hour than Class B. Class A works for \(4\) hours, and Class B works for \(5\) hours. Together, they plant \(93\) trees. a) Write an equation to find Class B’s hourly rate \(x\). b) How many trees per hour does each class plant? c) How many trees would Class B plant if it worked for the same \(4\) hours as Class A?

Hints

- Let \(x\) represent Class B’s trees per hour. - Multiply each hourly rate by the number of hours worked. - Add the two totals to equal \(93\). - For part c, use Class B’s rate with a new time.

Solution

1. Let Class B’s hourly rate be \(x\). Then Class A’s hourly rate is \(x+3\), so the total-tree equation is \(4(x+3)+5x=93\). 2. Distribute and combine: \(4x+12+5x=93\), so \(9x=81\) and \(x=9\). 3. Class B plants \(9\) trees per hour, and Class A plants \(9+3=12\) trees per hour. 4. In \(4\) hours, Class B would plant \(9\cdot4=36\) trees.

Answer

a) \(4(x + 3) + 5x = 93\) b) Class A plants \(12\) trees per hour, and Class B plants \(9\) trees per hour. c) Class B would plant \(36\) trees in \(4\) hours.
5240298
A class quiz has \(12\) questions. A team earns \(4\) points for each correct answer and loses \(2\) points for each incorrect answer. a) Team Math Masters finishes with \(30\) points. How many questions did the team answer correctly? b) Team Puzzle Solvers finishes with \(6\) points. How many questions did the team answer correctly? For each score equation, use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Start with the total number of questions. - If \(x\) questions are correct, how many are incorrect? - Write one expression that includes points earned and points lost. - Think about how the score changes when one more answer is correct instead of incorrect.

Solution

1. Let \(x\) be the number of correct answers. Then \(12-x\) answers are incorrect, so a team’s score is \(4x-2(12-x)\). 2. For a score of \(30\), solve \(4x-2(12-x)=30\). This simplifies to \(6x-24=30\), so \(6x=54\) and \(x=9\). 3. For a score of \(6\), solve \(4x-2(12-x)=6\). This simplifies to \(6x-24=6\), so \(6x=30\) and \(x=5\).

Answer

a) Before distribution: \(4x-2(12-x)=30\). After distribution: \(4x-24+2x=30\). Team Math Masters answered \(9\) questions correctly. b) Before distribution: \(4x-2(12-x)=6\). After distribution: \(4x-24+2x=6\). Team Puzzle Solvers answered \(5\) questions correctly.
5240308
In an arcade target game, a player gets \(15\) attempts. Each hit earns \(7\) points, and each miss loses \(3\) points. a) How many hits did a player make if the final score was \(55\) points? b) Explain mathematically why a final score of exactly \(50\) points is impossible. For each score equation, use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Write a formula for the score in terms of the number of hits. - Use the fact that there are \(15\) total attempts. - What values can the number of hits reasonably take? - Solve the score equation for \(50\) and interpret the result.

Solution

1. Let \(x\) be the number of hits. Then \(15-x\) attempts are misses, so a score of \(55\) gives \(7x-3(15-x)=55\). 2. Distribute and solve: \(10x-45=55\), so \(10x=100\) and \(x=10\). 3. A score of \(50\) would require \(7x-3(15-x)=50\), which simplifies to \(10x-45=50\). 4. This gives \(x=9.5\). Because the number of hits must be a whole number from \(0\) through \(15\), a score of exactly \(50\) points is impossible.

Answer

a) Before distribution: \(7x-3(15-x)=55\). After distribution: \(7x-45+3x=55\). The player made \(10\) hits. b) Before distribution: \(7x-3(15-x)=50\). After distribution: \(7x-45+3x=50\). This gives \(x=9.5\), which is impossible because the number of hits must be a whole number.
5240608
A semi-truck and a car start at the same time from opposite ends of a \(360\)-mile highway and drive toward each other. The car averages \(20\,\text{mph}\) more than the truck. They meet after exactly \(3\) hours. Find the average speed of each vehicle. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Represent the slower vehicle's speed with a variable. - Write the faster vehicle's speed in terms of that variable. - Use distance equals rate times time for each vehicle. - The two distances add to the full highway length.

Solution

1. Let \(v\,\text{mph}\) be the truck's average speed. The car's average speed is \(v + 20\). 2. In \(3\) hours, the truck travels \(3v\) miles and the car travels \(3(v + 20)\) miles. 3. Their distances add to \(360\), so \(3v + 3(v + 20) = 360\). 4. Distribute and combine like terms: \(3v + 3v + 60 = 360\), so \(6v + 60 = 360\). 5. Subtract \(60\) and divide by \(6\): \(v = 50\). 6. The car's speed is \(50 + 20 = 70\,\text{mph}\).

Answer

Before distribution: \(3v+3(v+20)=360\). After distribution: \(3v+3v+60=360\). The truck averages \(50\,\text{mph}\), and the car averages \(70\,\text{mph}\).
5268068
A research tank contains \(50\,\text{L}\) of salt water with \(1.5\,\text{kg}\) of dissolved salt. An experiment requires a concentration of exactly \(20\,\text{g}\) of salt per liter. Write an equation and find how many liters of pure water must be added. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Convert all salt masses to the same unit. - Adding pure water does not change the amount of salt. - Write the new total volume in terms of \(x\). - Concentration times volume equals the salt mass.

Solution

1. Convert the salt mass: \(1.5\,\text{kg} = 1500\,\text{g}\). 2. Let \(x\) be the number of liters of pure water added. The new volume is \(50 + x\,\text{L}\). 3. The salt mass stays at \(1500\,\text{g}\), so write \(20(50 + x) = 1500\). 4. Distribute and solve: \(1000 + 20x = 1500\), so \(20x = 500\) and \(x = 25\).

Answer

Before distribution: \(20(50+x)=1500\). After distribution: \(1000+20x=1500\). Add \(25\,\text{L}\) of pure water.
5279488
Two rectangular banners are each \(24\,\text{in.}\) high. The longer banner is \(10\,\text{in.}\) longer than the shorter banner. Their combined area is \(1680\,\text{in.}^2\). Find the length of each banner. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Use the same height in both area expressions. - Express the longer banner's length in terms of the shorter banner's length. - Add the two areas and set the sum equal to the combined area. - Distribute before solving.

Solution

1. Let \(l\) inches be the length of the shorter banner. The longer banner is \(l + 10\) inches. 2. Add the two areas: \(24l + 24(l + 10) = 1680\). 3. Distribute and combine like terms: \(48l + 240 = 1680\). 4. Subtract \(240\): \(48l = 1440\). 5. Divide by \(48\): \(l = 30\). 6. The longer banner is \(30\,\text{in.} + 10\,\text{in.} = 40\,\text{in.}\).

Answer

Before distribution: \(24l+24(l+10)=1680\). After distribution: \(24l+24l+240=1680\). The banners are \(30\,\text{in.}\) and \(40\,\text{in.}\) long.
5279698
Solve \(2(3x - 1.5) - 4(x + 2) + 0.5x = 6.5\).

Hints

- Distribute the negative factor carefully. - Combine variable terms and constants separately. - Isolate the variable term before dividing. - Align decimal place values when calculating.

Solution

1. Distribute: \(6x - 3 - 4x - 8 + 0.5x = 6.5\). 2. Combine like terms: \(2.5x - 11 = 6.5\). 3. Add \(11\): \(2.5x = 17.5\). 4. Divide by \(2.5\): \(x = 7\).

Answer

\(x = 7\)
5280308
A wooden beam is \(150\,\text{in.}\) long and is cut into three pieces. The second piece is \(10\,\text{in.}\) longer than the first piece. The third piece is twice as long as the second piece. Find the length of each piece.

Hints

- Use the first piece as the variable. - The third piece is twice the entire length of the second piece. - Add expressions for all three pieces and set the sum equal to \(150\). - Check that your three lengths add to the original beam length.

Solution

1. Let \(x\) be the length of the first piece in inches. 2. The second piece is \(x + 10\), and the third piece is \(2(x + 10)\). 3. The pieces total \(150\,\text{in.}\), so write \(x + (x + 10) + 2(x + 10) = 150\). 4. Distribute and combine like terms: \(x + x + 10 + 2x + 20 = 150\), so \(4x + 30 = 150\). 5. Subtract \(30\): \(4x = 120\). Divide by \(4\): \(x = 30\). 6. The piece lengths are \(30\,\text{in.}\), \(40\,\text{in.}\), and \(80\,\text{in.}\).

Answer

The pieces are \(30\,\text{in.}\), \(40\,\text{in.}\), and \(80\,\text{in.}\) long.
5280378
A freight train leaves Cedar Junction at \(10{:}00\) a.m. and travels toward Lakeview Station at \(50\,\text{mph}\). The stations are \(150\) miles apart. At \(10{:}30\) a.m., a passenger train leaves Lakeview Station toward Cedar Junction at \(75\,\text{mph}\). Use an equation to find how many minutes after its departure the passenger train meets the freight train. Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Account for the freight train's earlier departure. - Write a distance expression for each train. - The two distances add to the distance between the stations. - Convert the solved time from hours to minutes.

Solution

1. Let \(x\) be the passenger train's travel time in hours before the trains meet. 2. The freight train leaves \(0.5\) hour earlier, so it travels for \(x+0.5\) hours. 3. Their distances add to \(150\) miles: \(50(x+0.5)+75x=150\). 4. Distribute and combine like terms: \(50x+25+75x=150\), so \(125x+25=150\). 5. Subtract \(25\): \(125x=125\), so \(x=1\). 6. One hour is \(60\) minutes, so the passenger train travels for \(60\) minutes before the meeting.

Answer

Before distribution: \(50(x+0.5)+75x=150\). After distribution: \(50x+25+75x=150\). The passenger train meets the freight train \(60\) minutes after leaving Lakeview Station.
5280388
Two hiking groups are \(18\) miles apart on the same trail and hike toward each other. Group A hikes at \(4\,\text{mph}\), and Group B hikes at \(5\,\text{mph}\). a) How many hours does it take the groups to meet if they start at the same time? b) Group B now starts \(45\) minutes after Group A. How many minutes does Group B hike before the groups meet? c) Explain without calculating why Group B's hiking time in part b) must be shorter than its hiking time in part a). For part b), use the distributive property before isolating the variable. Your response must show the part-b equation immediately before and immediately after distribution.

Hints

- Add the distances the groups hike toward each other. - Convert the delayed start to hours. - In part b), Group A's hiking time is longer than Group B's. - For part c), focus on the remaining distance when Group B starts.

Solution

1. If both groups start together and meet after \(x\) hours, their combined distance is \(4x+5x=18\). Thus \(9x=18\) and \(x=2\) hours. 2. For the delayed start, convert \(45\) minutes to \(0.75\) hour. Let \(y\) be Group B's hiking time. Group A then hikes for \(y+0.75\) hours. 3. Solve \(4(y+0.75)+5y=18\). This gives \(9y=15\), so \(y=\frac{5}{3}\) hour, or \(100\) minutes. 4. Group A reduces the distance between the groups before Group B begins, so less distance remains to be closed while Group B is hiking. Therefore, Group B's hiking time is shorter in the delayed-start situation.

Answer

a) The groups meet after \(2\) hours. b) Before distribution: \(4(y+0.75)+5y=18\). After distribution: \(4y+3+5y=18\). Group B hikes for \(100\) minutes before the meeting. c) Group A has already reduced the distance between the groups before Group B starts, so Group B needs less hiking time.
5125768
Mr. Smith and Ms. Weber start at opposite ends of a \(17\)-mile hiking trail. Mr. Smith starts at \(2{:}00\) p.m. and walks at \(4\,\text{mph}\). Ms. Weber starts at \(2{:}30\) p.m. and walks toward him at \(6\,\text{mph}\). a) At what time do they meet? b) How far from Mr. Smith's starting point do they meet? Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Use one time variable measured from the earlier start. - Subtract \(0.5\) hour for the later hiker's travel time. - Add the two distances and set the sum equal to the trail length. - Use the earlier hiker's speed to find the meeting location.

Solution

1. Let \(t\) be the number of hours after \(2{:}00\) p.m. when they meet. Mr. Smith travels \(4t\) miles, and Ms. Weber travels \(6(t-0.5)\) miles. 2. Their distances add to \(17\): \(4t+6(t-0.5)=17\). 3. Distribute and solve: \(10t-3=17\), so \(10t=20\) and \(t=2\). They meet at \(4{:}00\) p.m. 4. Mr. Smith travels \(4\cdot2=8\) miles, so the meeting point is \(8\) miles from his starting point.

Answer

a) Before distribution: \(4t+6(t-0.5)=17\). After distribution: \(4t+6t-3=17\). They meet at \(4{:}00\) p.m. b) They meet \(8\) miles from Mr. Smith's starting point.
5280188
A homeowner installs \(28\) solar panels. Type A panels produce \(320\,\text{W}\) each, and Type B panels produce \(400\,\text{W}\) each. Together, the panels produce \(10\,\text{kW}\). a) How many panels of each type are installed? b) The system output will be increased to \(12\,\text{kW}\). How many additional Type B panels are needed if the number of Type A panels stays the same? Use the distributive property before isolating the variable. Your response must show the equation immediately before and immediately after distribution.

Hints

- Convert all power values to watts. - Use one variable for the number of one panel type. - The two panel counts add to \(28\). - In part b, keep the Type A count unchanged and recalculate the required Type B output.

Solution

1. Convert \(10\,\text{kW}\) to \(10{,}000\,\text{W}\). Let \(x\) be the number of Type A panels, so \(28-x\) are Type B panels. 2. Solve \(320x+400(28-x)=10{,}000\). This simplifies to \(-80x=-1200\), so \(x=15\). Therefore, there are \(15\) Type A panels and \(13\) Type B panels. 3. The \(15\) Type A panels produce \(15\cdot320=4800\,\text{W}\). 4. To reach \(12{,}000\,\text{W}\), Type B panels must produce \(12{,}000-4800=7200\,\text{W}\), requiring \(7200\div400=18\) Type B panels. 5. Since \(13\) Type B panels are already installed, \(18-13=5\) additional Type B panels are needed.

Answer

a) Before distribution: \(320x+400(28-x)=10{,}000\). After distribution: \(320x+11{,}200-400x=10{,}000\). There are \(15\) Type A panels and \(13\) Type B panels. b) \(5\) additional Type B panels are needed.

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