Question

name: colin forcrimer unit 1 review multi-step equations & inequalities part 1: solving equations 1. \\(\frac{2}{5}x - 9 = -17\\) 2. \\(7 = 5 - \frac{1}{3}x\\) 3. \\(19 - 2(3x - 5) = 15\\) 4. \\(8(2x + 5) = 12x - (2x + 14)\\) 5. \\(6 - 4(3x - 2) = 6(1 - 2x) - 1\\) 6. \\(-9(x + 4) = -5x - (4x + 36)\\)

Explanation:

Problem 1: $\frac{2}{5}x - 9 = -17$

Step1: Add 9 to both sides

$\frac{2}{5}x = -17 + 9 = -8$

Step2: Multiply by $\frac{5}{2}$

$x = -8 \times \frac{5}{2} = -20$

Problem 2: $7 = 5 - \frac{1}{3}x$

Step1: Subtract 5 from both sides

$2 = -\frac{1}{3}x$

Step2: Multiply by -3

$x = 2 \times (-3) = -6$

Problem 3: $19 - 2(3x - 5) = 15$

Step1: Expand the bracket

$19 - 6x + 10 = 15$ → $29 - 6x = 15$

Step2: Subtract 29 from both sides

$-6x = -14$

Step3: Divide by -6

$x = \frac{-14}{-6} = \frac{7}{3}$

Problem 4: $8(2x + 5) = 12x - (2x + 14)$

Step1: Expand both sides

$16x + 40 = 12x - 2x - 14$ → $16x + 40 = 10x - 14$

Step2: Subtract 10x from both sides

$6x + 40 = -14$

Step3: Subtract 40 from both sides

$6x = -54$

Step4: Divide by 6

$x = -9$

Problem 5: $6 - 4(3x - 2) = 6(1 - 2x) - 1$

Step1: Expand both sides

$6 - 12x + 8 = 6 - 12x - 1$ → $14 - 12x = 5 - 12x$

Step2: Add 12x to both sides

$14 = 5$ (Contradiction, no solution)

Problem 6: $-9(x + 4) = -5x - (4x + 36)$

Step1: Expand both sides

$-9x - 36 = -5x - 4x - 36$ → $-9x - 36 = -9x - 36$

Step2: Add 9x + 36 to both sides

$0 = 0$ (Identity, infinite solutions)

Answer:

1. $x = -20$
2. $x = -6$
3. $x = \frac{7}{3}$
4. $x = -9$
5. No solution
6. Infinite solutions