Solving Two-Step Equations: Additional Practice (Set 2)

Subtract the constant $\frac{3}{2}$ from both sides, then multiply by the divisor $2$.

$\frac{x}{2} = 4 - \frac{3}{2} = \frac{8}{2} - \frac{3}{2} = \frac{5}{2}$

Multiplying by $2$ gives $x = 5$. Check: $\frac{5}{2} + \frac{3}{2} = \frac{8}{2} = 4$, which is true.

Add $5$ to both sides, then divide by $8$.

$8x - 5 + 5 = 11 + 5 \;\Rightarrow\; 8x = 16$

Dividing by $8$ gives $x = 2$. Check: $8(2) - 5 = 16 - 5 = 11$, which is true.

Distribute the $4$ across both terms inside the parentheses, then solve.

$4(x + 1) = 4x + 4 = 2 \;\Rightarrow\; 4x = -2 \;\Rightarrow\; x = -\frac{2}{4} = -\frac{1}{2}$

Check in the original: $4\left(-\frac{1}{2} + 1\right) = 4 \cdot \frac{1}{2} = 2$, which is true.

Add the decimal constant $0.5$ to both sides, then divide by $2$.

$2x - 0.5 + 0.5 = 3.5 + 0.5 \;\Rightarrow\; 2x = 4$

Dividing by $2$ gives $x = 2$. Check: $2(2) - 0.5 = 4 - 0.5 = 3.5$, which is true.

Subtract the decimal constant $1.4$ from both sides, then divide by $6$.

$6x + 1.4 - 1.4 = 5 - 1.4 \;\Rightarrow\; 6x = 3.6$

Dividing by $6$ gives $x = 0.6$. Check: $6(0.6) + 1.4 = 3.6 + 1.4 = 5$, which is true.

The variable term is on the right; subtract the constant $2$ from both sides, then divide by $6$.

$14 - 2 = 6x + 2 - 2 \;\Rightarrow\; 12 = 6x$

Dividing by $6$ gives $x = 2$. Check: $6(2) + 2 = 12 + 2 = 14$, which is true.

Combine the like terms $5x - 3x = 2x$ first, then solve the two-step equation.

$2x + 7 = 1 \;\Rightarrow\; 2x = -6 \;\Rightarrow\; x = -3$

Check in the original: $5(-3) - 3(-3) + 7 = -15 + 9 + 7 = 1$, which is true.

Subtract the constant $3$ from both sides, leaving $\frac{1}{2}x = 5$. Dividing by $\frac{1}{2}$ is the same as multiplying by $2$.

$x = 5 \cdot 2 = 10$

Check: $\frac{1}{2}(10) + 3 = 5 + 3 = 8$, which is true.

Add the constant $2$ to both sides, leaving $\frac{3}{4}x = 9$. Then divide by $\frac{3}{4}$, which means multiply by the reciprocal $\frac{4}{3}$.

$x = 9 \cdot \frac{4}{3} = \frac{36}{3} = 12$

Check: $\frac{3}{4}(12) - 2 = 9 - 2 = 7$, which is true.

Substitute $10$ for $x$ in the original left side, dividing before adding.

$\frac{10}{2} + 4 = 5 + 4 = 9$

The left side equals $9$, the same as the right side, so the solution checks out.

Distributing multiplies the $3$ by every term inside the parentheses, both the $x$ and the $-4$.

$3(x - 4) = 3 \cdot x - 3 \cdot 4 = 3x - 12$

So the next equation is $3x - 12 = 9$. From there, add $12$ and divide by $3$ to get $x = 7$.

Subtract the constant $5$ from both sides, leaving $-\frac{x}{3} = -3$. Multiply both sides by $-3$ to undo the division by $3$ and the negative sign together.

$-3 \cdot \left(-\frac{x}{3}\right) = -3 \cdot (-3) \;\Rightarrow\; x = 9$

Check: $-\frac{9}{3} + 5 = -3 + 5 = 2$, which is true.