Question

problems 4–7: solve each equation. show your thinking.

1. $2b + 8 - 5b + 3 = -13 + 8b - 5$
2. $2x + 7 - 5x + 8 = 3(5 + 6x) - 12x$
3. $3(3 - 3x) = 2(x + 3) - 30$
4. $\frac{1}{3}(z + 4) - 6 = \frac{2}{3}(5 - z)$

Explanation:

Problem 4

Step1: Combine like terms on both sides

Left: $2b - 5b + 8 + 3 = -3b + 11$
Right: $-13 - 5 + 8b = 8b - 18$
Equation becomes: $-3b + 11 = 8b - 18$

Step2: Move $b$ terms to left, constants to right

$-3b - 8b = -18 - 11$
$-11b = -29$

Step3: Solve for $b$

$b = \frac{-29}{-11} = \frac{29}{11}$

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Problem 5

Step1: Combine like terms, expand right side

Left: $2x - 5x + 7 + 8 = -3x + 15$
Right: $15 + 18x - 12x = 15 + 6x$
Equation becomes: $-3x + 15 = 6x + 15$

Step2: Move $x$ terms to right, constants to left

$15 - 15 = 6x + 3x$
$0 = 9x$

Step3: Solve for $x$

$x = 0$

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Problem 6

Step1: Expand both sides

Left: $9 - 9x$
Right: $2x + 6 - 30 = 2x - 24$
Equation becomes: $9 - 9x = 2x - 24$

Step2: Move $x$ terms to right, constants to left

$9 + 24 = 2x + 9x$
$33 = 11x$

Step3: Solve for $x$

$x = \frac{33}{11} = 3$

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Problem 7

Step1: Eliminate denominators (multiply by 3)

$(z + 4) - 18 = 2(5 - z)$

Step2: Expand and combine like terms

$z + 4 - 18 = 10 - 2z$
$z - 14 = 10 - 2z$

Step3: Move $z$ terms to left, constants to right

$z + 2z = 10 + 14$
$3z = 24$

Step4: Solve for $z$

$z = \frac{24}{3} = 8$

Answer:

1. $b = \frac{29}{11}$
2. $x = 0$
3. $x = 3$
4. $z = 8$