Linear Equations in Disguise: Practice

The LCD of $3$ and $2$ is $6$. Multiply every term by $6$.

$4x - 3x = 6, \qquad x = 6$

Cross-multiply.

$4(x - 1) = 2(x + 1), \qquad 4x - 4 = 2x + 2, \qquad 2x = 6, \qquad x = 3$

The coefficient reaches hundredths, so multiply both sides by $100$.

$5x = 200, \qquad x = 40$

Multiply both sides by $x + 1$ to clear the denominator.

$8 = 2(x + 1), \qquad 8 = 2x + 2, \qquad 2x = 6, \qquad x = 3$

Since $3 \neq -1$, it is a valid solution.

Cross-multiply.

$x + 1 = 2(x - 2), \qquad x + 1 = 2x - 4, \qquad 5 = x$

Since $5 \neq 2$, the solution is $x = 5$.

Distribute the left side.

$4x + 4 = 4x + 4$

Subtracting $4x$ leaves $4 = 4$, true for every $x$, so there are infinitely many solutions.

Subtract $5x$ from both sides and watch the variable cancel.

$-3 = 2$

That is false, so the equation has no solution.

The LCD of $2$, $3$, and $6$ is $6$. Multiply every term by $6$.

$3x + 2x + x = 36, \qquad 6x = 36, \qquad x = 6$

Multiply both sides by $5$ to clear the denominator.

$2x + 1 = 15, \qquad 2x = 14, \qquad x = 7$

The LCD of $5$ and $10$ is $10$. Multiply every term by $10$.

$2x - 20 = x, \qquad x = 20$

Cross-multiply.

$2(2x - 1) = 3(x + 4), \qquad 4x - 2 = 3x + 12, \qquad x = 14$

Multiply both sides by $x$, then divide by $4$.

$12 = 4x, \qquad x = 3$