Free Math Worksheets

- 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. For (1), subtract \(5x\): \(3 = -2\). This contradiction means there is no solution. 2. For (2), distribute: \(2x + 8 = 2x + 8\). This identity is true for every real value of \(x\), so there are infinitely many solutions. 3. For (3), divide by \(3\): \(x = 0\). There is 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 gives \(n+n=2n\), so statement 1 is true for every \(n\). 2. The equation \(3x=15\) is true only when \(x=5\), so statement 2 is not an identity. 3. The distributive property gives \(4(a+b)=4a+4b\), so statement 3 is an identity. 4. Solving \(y-5=5-y\) gives \(2y=10\), so \(y=5\). Statement 4 is not an identity. 5. The multiplicative identity property gives \(z\cdot 1=z\) for every \(z\), so statement 5 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. For equation 1), distribute: \(4x - 4 = 4x - 1\). Subtract \(4x\): \(-4 = -1\), which is false. There is no solution. 2. For equation 2), distribute: \(2x + 8 = 2x + 8\). This identity is true for every real number, so there are infinitely many solutions. 3. For equation 3), add \(x\): \(5 = 7\), which is false. There is no solution. 4. For equation 4), divide by \(3\): \(x = 0\). There is one solution. 5. Therefore, equations 1) and 3) have 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 pair a), solve \(4x - 7 = 13\): \(4x = 20\), so \(x = 5\). 2. Solve \(2x + 5 = 15\): \(2x = 10\), so \(x = 5\). 3. Both equations have solution set \(S = \{5\}\), so they are equivalent. 4. For pair b), solve \(3(x - 2) = 12\): \(x - 2 = 4\), so \(x = 6\). 5. Solve \(x + 4 = 9\): \(x = 5\). 6. 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 \(S = \{5\}\).

- Choose one coordinate and use the equation to determine the other. - Think about how many rational values you can choose for \(x\). - For whole numbers, list the possible x-values systematically from \(0\) to \(12\).

1. Examples include \((6, 6)\), \((-1, 13)\), \((4.5, 7.5)\), and \((0, 12)\). Each pair has sum \(12\). 2. There are infinitely many rational-number solutions because any rational choice of \(x\) determines exactly one rational value \(y = 12 - x\). 3. For whole-number solutions, \(x\) can be any integer from \(0\) through \(12\), and \(y = 12 - x\). That gives 13 ordered pairs, so the claim is true.

1. One possible set is \((6, 6)\), \((-1, 13)\), \((4.5, 7.5)\), and \((0, 12)\). 2. Infinitely many 3. True; there are 13 whole-number solutions.

- 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. For a), distribute to get \(3x + 3 = 3x + 3\). This identity is true for every rational number, so \(S = \mathbb{Q}\). 2. For b), subtract \(2x\) to get \(5 = -1\). This contradiction is never true, so \(S = \varnothing\). 3. For c), distribute: \(4x - 10 = 2x + 2\). Then \(2x = 12\), so \(x = 6\) and \(S = \{6\}\). 4. For d), subtract \(2\): \(\frac{x}{4} = 3\). Multiply by \(4\): \(x = 12\), so \(S = \{12\}\).

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. Use \(4x + 10\). 2. For no solution, keep the same variable term but use a different constant. For example, use \(4x + 1\). Then subtracting \(4x\) gives \(10 = 1\), a contradiction. 3. For \(x = 5\) to be the only solution, evaluate the left side: \(4 \cdot 5 + 10 = 30\). Use the constant expression \(30\), giving \(4x + 10 = 30\), 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. The lines \(y = 2x + 5\) and \(y = 2x - 3\) have the same slope but different \(y\)-intercepts. They are distinct parallel lines, so the equation has no solution. 2. 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. The equation has infinitely many solutions. 3. 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 a), subtract \(2x\) and \(7\): \(3x = -15\), so \(x = -5\) and \(S = \{-5\}\). 2. For b), distribute: \(6x + 12 = 6x - 1\). Subtract \(6x\): \(12 = -1\). This contradiction means \(S = \varnothing\). 3. For c), 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 A, simplify both sides: \(4x - 2x - 6 = 2x - 6\), so \(2x - 6 = 2x - 6\). This identity has infinitely many solutions. 2. For 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 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: \(3(x + 4) + 2x = 3x + 12 + 2x = 5x + 12\). 2. For a), \(5x + 12 = 5x + 12\) is an identity, so there are infinitely many solutions. 3. For b), \(4x + 12 = 5x + 12\). Subtracting \(4x\) and \(12\) gives \(x = 0\), so there is exactly one solution. 4. For c), \(5x + 10 = 5x + 12\). 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: \(4x - 2 = 4x - 2\). 2. This equation is an identity, so it is true for every value in the domain. 3. For part a), every rational number is a solution, so \(S = \mathbb{Q}\). 4. For part b), every listed domain value is a solution, 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. Add \(4\) to both sides of \(x - 4 = 12\) to obtain \(x = 16\). The equations are equivalent. 2. Divide both sides of \(6x = 24\) by \(3\) to obtain \(2x = 8\). The equations are equivalent. 3. Adding \(5\) to both sides of \(x + 5 = 10\) gives \(x + 10 = 15\), not \(x + 10 = 20\). The equations are not equivalent. 4. Multiply every term of \(\frac{x}{2} + 3 = 7\) by \(2\) to obtain \(x + 6 = 14\). The equations are 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. Choose \(a = 4\) and \(b = 12\). 2. For no solution, the coefficients of \(x\) must match but the constants must differ. One example is \(a = 4\) and \(b = 10\). 3. For \(x = 5\) to be a solution, substitute \(5\): \(4 \cdot 5 + 12 = 5a + b\), so \(32 = 5a + b\). Also require \(a \ne 4\) so the equation has one solution. Choosing \(a = 2\) gives \(b = 22\).

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. For part a), 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 part b), choose a coefficient different from \(8\). For example, use \(7\): \(7x + 20 = 4(2x + 5)\). Then \(7x + 20 = 8x + 20\), so \(x = 0\). Because the variable coefficients differ, the solution is unique. 3. For part c), keep the coefficient of \(x\) equal on both sides but make the constants different. For example, \(8x + 20 = 4(2x + 6)\) becomes \(8x + 20 = 8x + 24\), which has no solution.

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.