Certainly! Here are some practice problems that focus on solving multi-step equations, similar to what you might encounter in a Unit Test for a math unit on this topic.
Practice Problems:
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Solve for \( x \): \[ 3(x + 4) - 2 = 16 \]
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Solve for \( y \): \[ 5y - 3(2y - 4) = 8 \]
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Solve for \( z \): \[ 2(z - 1) + 3(2z + 4) = 26 \]
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Solve for \( a \): \[ 4(a + 3) - 2(a - 1) = 20 \]
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Solve for \( b \): \[ \frac{3b + 2}{4} - 1 = 2 \]
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Solve for \( c \): \[ 7(c + 1) - 3(2c - 5) = 22 \]
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Solve for \( d \): \[ 6(d - 2) + 4 = 2d + 10 \]
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Solve for \( e \): \[ 2(3e + 4) = 5(e + 2) + 2 \]
Solutions
Here are the solutions to the practice problems:
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Problem: \( 3(x + 4) - 2 = 16 \)
Solution: \[ 3x + 12 - 2 = 16 \ 3x + 10 = 16 \ 3x = 6 \ x = 2 \] -
Problem: \( 5y - 3(2y - 4) = 8 \)
Solution: \[ 5y - 6y + 12 = 8 \ -y + 12 = 8 \ -y = -4 \ y = 4 \] -
Problem: \( 2(z - 1) + 3(2z + 4) = 26 \)
Solution: \[ 2z - 2 + 6z + 12 = 26 \ 8z + 10 = 26 \ 8z = 16 \ z = 2 \] -
Problem: \( 4(a + 3) - 2(a - 1) = 20 \)
Solution: \[ 4a + 12 - 2a + 2 = 20 \ 2a + 14 = 20 \ 2a = 6 \ a = 3 \] -
Problem: \( \frac{3b + 2}{4} - 1 = 2 \)
Solution: \[ \frac{3b + 2}{4} = 3 \ 3b + 2 = 12 \ 3b = 10 \ b = \frac{10}{3} \text{ or } 3 \frac{1}{3} \] -
Problem: \( 7(c + 1) - 3(2c - 5) = 22 \)
Solution: \[ 7c + 7 - 6c + 15 = 22 \ c + 22 = 22 \ c = 0 \] -
Problem: \( 6(d - 2) + 4 = 2d + 10 \)
Solution: \[ 6d - 12 + 4 = 2d + 10 \ 6d - 8 = 2d + 10 \ 4d = 18 \ d = \frac{9}{2} \text{ or } 4.5 \] -
Problem: \( 2(3e + 4) = 5(e + 2) + 2 \)
Solution: \[ 6e + 8 = 5e + 10 + 2 \ 6e + 8 = 5e + 12 \ e = 4 \]
Feel free to reach out if you have any questions on these problems or if you need further assistance!