Free math worksheets

- Simplify both sides first. - Decide what it means if both sides become the same expression.

1. Distribute the left side: \(4x + 8 = 4x + 8\). 2. Both sides are the same expression, so the equation is true for every value of \(x\). 3. Therefore, the equation has infinitely many solutions.

Infinitely many solutions

- Compare the variable terms on both sides. - Remove matching variable terms and interpret the statement that remains.

1. Subtract \(3x\) from both sides: \(5 = -2\). 2. This statement is false, so no value of \(x\) can make the original equation true. 3. Therefore, the equation has no solution.

No solution

- Simplify each equation as far as possible. - A contradiction means no solution. - An identity means infinitely many solutions. - Do not confuse \(x = 0\) with having no solution.

1. The first equation gives \(3=-2\) after subtracting \(5x\). This contradiction means there is no solution. 2. The second equation simplifies to \(2x+8=2x+8\), an identity true for every real value of \(x\), so there are infinitely many solutions. 3. The third equation gives \(x=0\) after dividing by \(3\), so it has exactly one solution.

(1) No solution (2) Infinitely many solutions (3) Exactly one solution, \(x = 0\)

- Look for properties such as combining like terms, the distributive property, or the identity property. - A single counterexample proves that an equation is not an identity. - Try rewriting one side until it matches the other, or solve the equation to see whether only certain values work.

1. Combining like terms shows \(n+n=2n\) for every \(n\), so the first equation is an identity. 2. The equation \(3x=15\) is true only when \(x=5\), so the second equation is not an identity. 3. The distributive property gives \(4(a+b)=4a+4b\) for all allowable values, so the third equation is an identity. 4. Solving \(y-5=5-y\) gives \(2y=10\), so \(y=5\). The fourth equation is not an identity. 5. The multiplicative identity property gives \(z\cdot1=z\) for every \(z\), so the fifth equation is an identity.

Statements 1), 3), and 5) are identities.

- Simplify each equation until the variable is isolated or eliminated. - A false statement indicates no solution. - An identity indicates infinitely many solutions. - Do not confuse a solution of \(x = 0\) with having no solution.

1. The first equation becomes \(4x-4=4x-1\). Subtracting \(4x\) gives \(-4=-1\), a contradiction, so there is no solution. 2. The second equation becomes \(2x+8=2x+8\), an identity, so every real number is a solution. 3. Adding \(x\) to both sides of the third equation gives \(5=7\), a contradiction, so there is no solution. 4. Dividing the fourth equation by \(3\) gives \(x=0\), so there is exactly one solution. 5. Therefore, equations 1) and 3) are the equations with no solution.

1) No solution 2) All real numbers 3) No solution 4) \(x = 0\) Therefore, equations 1) and 3) have no solution.

- Solve every equation separately. - Write or compare the resulting solution sets. - Equivalent equations must have exactly the same solutions.

1. For the first pair, \(4x-7=13\) gives \(x=5\), and \(2x+5=15\) also gives \(x=5\). Both equations have solution set \(S=\{5\}\), so they are equivalent. 2. For the second pair, \(3(x-2)=12\) gives \(x=6\), while \(x+4=9\) gives \(x=5\). The solution sets differ, so the equations are not equivalent.

a) Equivalent; both have solution set \(S = \{5\}\). b) Not equivalent; the first has \(S = \{6\}\) and the second has \(S = \{5\}\).

- A true statement such as \(0 = 0\) means every value in the domain works. - A false statement such as \(0 = 5\) means no value works. - Distribute and combine like terms before deciding the number of solutions.

1. The first equation simplifies to \(3x+3=3x+3\), an identity true for every rational number, so \(S=\mathbb{Q}\). 2. The second equation gives \(5=-1\) after subtracting \(2x\), a contradiction, so \(S=\varnothing\). 3. The third equation becomes \(4x-10=2x+2\), so \(2x=12\), \(x=6\), and \(S=\{6\}\). 4. The fourth equation gives \(\frac{x}{4}=3\), so \(x=12\) and \(S=\{12\}\).

a) Infinitely many solutions: \(S = \mathbb{Q}\) b) No solution: \(S = \varnothing\) c) One solution: \(S = \{6\}\) d) One solution: \(S = \{12\}\)

- Distribute and combine like terms first. - Notice what happens if the variable terms cancel. - A false numerical statement means there is no solution. - Use the empty-set symbol for the solution set.

1. Distribute: \(6n - 12 - 2n - 2 = 4n + 12\). 2. Combine like terms: \(4n - 14 = 4n + 12\). 3. Subtract \(4n\) from both sides: \(-14 = 12\). 4. This contradiction is never true, so the equation has no solution and \(S = \varnothing\).

The equation has no solution, so \(S = \varnothing\).

- Identical expressions produce infinitely many solutions. - Equal variable terms with unequal constants produce no solution. - For part c, substitute \(5\) into the known side and use the resulting value.

1. For infinitely many solutions, both sides must be identical. One valid right side is \(4x+10\). 2. For no solution, keep the same variable term but change the constant. For example, \(4x+1\) gives \(10=1\) after subtracting \(4x\). 3. To make \(x=5\) the unique solution, evaluate the left side at \(5\): \(4\cdot5+10=30\). Using the right side \(30\) gives \(4x+10=30\), whose only solution is \(5\).

a) One answer is \(4x + 10\). b) One answer is \(4x + 1\). c) One answer is \(30\).

- Zero is a valid number. - Solve both equations and compare the resulting statements. - Distinguish the set \(\{0\}\) from the empty set \(\varnothing\).

1. For \(8x = 0\), divide by \(8\): \(x = 0\). Zero is a number and is a valid solution, so \(S = \{0\}\). 2. For \(x + 1 = x\), subtract \(x\) from both sides: \(1 = 0\). This contradiction is never true, so \(S = \varnothing\). 3. Jonah’s claim is incorrect. A solution of zero means there is exactly one solution; an empty solution set means there is no solution.

Jonah is incorrect. For \(8x = 0\), \(S = \{0\}\). For \(x + 1 = x\), \(S = \varnothing\).

- Compare the slopes of the two lines in each equation. - Simplify any expression with parentheses before comparing. - Recall when two lines are parallel and when they coincide.

1. For the first equation, the lines \(y=2x+5\) and \(y=2x-3\) have the same slope but different \(y\)-intercepts. They are distinct parallel lines, so there is no solution. 2. For the second equation, simplify the right side: \(\frac{1}{2}(6x-2)=3x-1\). Both sides represent the same line, so every real value of \(x\) is a solution. 3. For the third equation, the lines \(y=4x+2\) and \(y=-x+7\) have different slopes. They intersect once, so the equation has exactly one solution.

1. No solution; the lines are parallel and distinct. 2. Infinitely many solutions; both expressions represent the same line. 3. Exactly one solution; the lines have different slopes.

- Move variable terms to one side. - Distribute before combining like terms. - A false numerical statement means the solution set is empty. - State each result as a solution set.

1. For the first equation, subtract \(2x\) and \(7\): \(3x=-15\), so \(x=-5\) and \(S=\{-5\}\). 2. For the second equation, distribute: \(6x+12=6x-1\). Subtract \(6x\) to get \(12=-1\), a contradiction, so \(S=\varnothing\). 3. For the third equation, distribute: \(3x+6=2x+10\). Subtract \(2x\) and \(6\): \(x=4\), so \(S=\{4\}\).

a) \(S = \{-5\}\) b) \(S = \varnothing\) c) \(S = \{4\}\)

- Simplify both sides before comparing them. - Handle minus signs before parentheses carefully. - An identity has infinitely many solutions. - A contradiction has no solution.

1. For equation A, simplify both sides: \(4x-2x-6=2x-6\), so \(2x-6=2x-6\). This identity has infinitely many solutions. 2. For equation B, distribute on the right: \(\frac{2}{3}x+5=\frac{2}{3}x+4\). Subtracting \(\frac{2}{3}x\) gives \(5=4\), so there is no solution. 3. For equation C, simplify the right side: \(3x+7=4x-3\). Solving gives \(x=10\), so there is exactly one solution.

A: Infinitely many solutions B: No solution C: Exactly one solution

- Substitute each value of \(k\) separately. - Simplify the right side first. - Compare the variable coefficients and constants after simplification.

1. Simplify the right side to \(5x+12\). 2. With \(k=5\), the equation is \(5x+12=5x+12\), an identity, so there are infinitely many solutions. 3. With \(k=4\), the equation is \(4x+12=5x+12\). Subtracting \(4x\) and \(12\) gives \(x=0\), so there is exactly one solution. 4. With \(k=5\) and the left constant changed to \(10\), the equation is \(5x+10=5x+12\). Subtracting \(5x\) gives \(10=12\), so there is no solution.

a) Infinitely many solutions b) Exactly one solution, \(x = 0\) c) No solution

- Simplify both sides completely. - Notice what happens when the variable terms cancel. - A false numerical statement means there is no solution.

1. Remove the parentheses on the left: \(6x - 2x - 4 = 4(x + 1)\). 2. Simplify and distribute: \(4x - 4 = 4x + 4\). 3. Subtract \(4x\): \(-4 = 4\). 4. This contradiction is never true, so \(S = \varnothing\).

\(S = \varnothing\)

- Simplify the left side and compare it with the right side. - An identity is true for every value in the stated domain. - A solution set cannot include values outside its domain.

1. Distribute on the left: \(0.5(8x-4)=4x-2\), so the equation becomes \(4x-2=4x-2\), an identity. 2. Over the rational-number domain, every allowable value works, so \(S=\mathbb{Q}\). 3. Over the finite domain \(\{0,1,2\}\), the identity is true for every listed value, so \(S=\{0,1,2\}\).

a) \(S = \mathbb{Q}\) b) \(S = \{0, 1, 2\}\)

- Subtract the common variable term from both sides in each example. - Compare the statements that remain. - A contradiction means no solution, while an identity means infinitely many solutions.

1. For Example A, subtract \(3x\) from both sides: \(4 = 9\). This contradiction is false, so the equation has no solution. 2. For Example B, subtract \(3x\) from both sides: \(4 = 4\). This identity is always true, so every real number is a solution. 3. The student's claim is incorrect. Equal variable terms can lead to no solution when the constants differ, or infinitely many solutions when the constants are equal.

The claim is incorrect. Example A has no solution, so \(S = \varnothing\). Example B is true for every real number, so \(S = \mathbb{R}\).

- Equivalent equations have the same solution set. - Look for a reversible operation applied to both sides. - Apply an operation to every term on each side. - Check whether the proposed second equation is exactly the result of the transformation.

1. Adding \(4\) to both sides of the first equation in the first pair gives \(x=16\), so the equations are equivalent. 2. Dividing both sides of \(6x=24\) by \(3\) gives \(2x=8\), so the second pair is equivalent. 3. Adding \(5\) to both sides of \(x+5=10\) gives \(x+10=15\), not \(x+10=20\), so the third pair is not equivalent. 4. Multiplying every term of \(\frac{x}{2}+3=7\) by \(2\) gives \(x+6=14\), so the fourth pair is equivalent.

1) Equivalent; add \(4\) to both sides. 2) Equivalent; divide both sides by \(3\). 3) Not equivalent. 4) Equivalent; multiply every term by \(2\).

- Ask when the variable terms cancel after moving them to one side. - A true statement such as \(0 = 0\) gives infinitely many solutions. - A false statement such as \(0 = 1\) gives no solution. - For part c, substitute \(x = 5\) and choose constants that make the equation true.

1. For infinitely many solutions, both sides must be identical. One choice is \(a=4\) and \(b=12\). 2. For no solution, the coefficients of \(x\) must match while the constants differ. One choice is \(a=4\) and \(b=10\). 3. To make \(x=5\) the unique solution, substitute \(5\): \(4\cdot5+12=5a+b\), so \(32=5a+b\). Also require \(a\ne4\). 4. Choosing \(a=2\) gives \(b=22\), so \(a=2\), \(b=22\) is one valid example.

a) One example is \(a = 4\), \(b = 12\). b) One example is \(a = 4\), \(b = 10\). c) One example is \(a = 2\), \(b = 22\).

- Distribute on the right side first. - Infinitely many solutions occur when both sides simplify to the same expression. - One solution occurs when the variable coefficients differ. - No solution occurs when equal variable terms leave unequal constants.

1. Distribute on the right: \(4(2x+\square)=8x+4\square\). To match \(8x+20\), require \(4\square=20\), so the box value is \(5\). 2. For exactly one solution, choose a left-side coefficient different from \(8\). For example, \(7x+20=4(2x+5)\) becomes \(7x+20=8x+20\), so \(x=0\). 3. For no solution, keep equal variable coefficients but make the constants different. For example, \(8x+20=4(2x+6)\) becomes \(8x+20=8x+24\), a contradiction.

a) \(5\) b) One example is \(7x + 20 = 4(2x + 5)\), which has solution \(x = 0\). c) One example is \(8x + 20 = 4(2x + 6)\), which has no solution.