Solving Linear Equations: Additional Practice (Set 2)

Distribute on each side, subtract $4x$, then divide by $2$.

$6x - 8 = 4x + 4, \qquad 2x = 12, \qquad x = 6$

Add $2$, then multiply both sides by the reciprocal $\frac{4}{3}$.

$\frac{3}{4}x = 9, \qquad x = \frac{4}{3} \cdot 9 = 12$

Distribute the $-2$, combine like terms, then solve.

$7x - 2x + 6 = 26, \qquad 5x + 6 = 26, \qquad x = 4$

Distribute on each side, then solve.

$3x + 6 = 4x - 2, \qquad 8 = x$

Subtract $4$, then multiply both sides by the reciprocal $\frac{3}{2}$.

$\frac{2}{3}x = 6, \qquad x = \frac{3}{2} \cdot 6 = 9$

Add $4x$ to both sides so the variable stays positive, then solve.

$9 = 1 + 2x, \qquad 8 = 2x, \qquad x = 4$

Distribute both products, combine like terms, then solve.

$8x - 12 - 3x + 3 = 16, \qquad 5x - 9 = 16, \qquad x = 5$

Distribute the $-2$, then solve.

$-2x + 6 = 4, \qquad -2x = -2, \qquad x = 1$

Distribute the $-3$, combine like terms, then solve.

$8x - 6x + 3 = 11, \qquad 2x + 3 = 11, \qquad x = 4$

Multiply both sides by $2$, then subtract $1$.

$x + 1 = 10, \qquad x = 9$

Subtract $3$, then multiply both sides by $4$.

$\frac{x}{4} = 2, \qquad x = 8$

Subtract $4x$ from both sides so the variable stays positive, then finish.

$-7 = 2x - 15, \qquad 8 = 2x, \qquad x = 4$