Linear Equations in Disguise: Additional Practice (Set 1)

The LCD of $2$ and $4$ is $4$. Multiply every term by $4$.

$2x + x = 36, \qquad 3x = 36, \qquad x = 12$

Cross-multiply.

$2(x + 2) = 3(x - 4), \qquad 2x + 4 = 3x - 12, \qquad 16 = x$

Multiply every term by $10$, then gather the variable.

$3x - 5 = x + 7, \qquad 2x = 12, \qquad x = 6$

Cross-multiply.

$9 \cdot 2 = 3x, \qquad 18 = 3x, \qquad x = 6$

The LCD of $4$ and $2$ is $4$. Multiply every term by $4$.

$3x - 2x = 8, \qquad x = 8$

Clear the inner group first: $x - (3 - x) = 2x - 3$.

$2(2x - 3) = 10, \qquad 4x - 6 = 10, \qquad 4x = 16, \qquad x = 4$

The LCD of $2$ and $3$ is $6$. Multiply every term by $6$, keeping each numerator grouped.

$3(x + 1) + 2(x - 1) = 36, \qquad 3x + 3 + 2x - 2 = 36, \qquad 5x + 1 = 36$

Then $5x = 35$, so $x = 7$.

The numbers reach hundredths, so multiply every term by $100$.

$5x + 10 = 20, \qquad 5x = 10, \qquad x = 2$

The LCD of $3$ and $2$ is $6$. Multiply every term by $6$, keeping $x - 1$ grouped.

$2x - 3(x - 1) = -6, \qquad 2x - 3x + 3 = -6, \qquad -x + 3 = -6, \qquad x = 9$

Multiply every term by $10$, then gather the variable.

$6x = 2x + 12, \qquad 4x = 12, \qquad x = 3$

The two fractions on the left share the denominator $x$, so combine them first.

$\frac{2}{x} + \frac{3}{x} = \frac{5}{x}, \qquad \frac{5}{x} = \frac{5}{2}, \qquad x = 2$

Since $2 \neq 0$, it is a valid solution.

Cross-multiply.

$3(3x + 1) = 2(5x - 1), \qquad 9x + 3 = 10x - 2, \qquad 5 = x$