Linear Equations in Disguise: Practice

The LCD of $2$ and $5$ is $10$. Multiply every term by $10$, keeping each numerator grouped.

$5(x + 1) - 2(x - 3) = 20, \qquad 5x + 5 - 2x + 6 = 20, \qquad 3x + 11 = 20$

Then $3x = 9$, so $x = 3$.

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

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

So $x = 7$.

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

$25x + 50 = 10x + 140, \qquad 15x = 90, \qquad x = 6$

Subtract $1$ from both sides, then clear the denominator.

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

Since $\tfrac{5}{2} \neq 0$, it is a valid solution.

Cross-multiply.

$x - 3 = 2(x - 5), \qquad x - 3 = 2x - 10, \qquad 7 = x$

Since $7 \neq 5$, the solution is $x = 7$.

Multiply both sides by $x - 3$.

$x = 3$

But $x = 3$ is the excluded value that makes the denominator zero, so it must be rejected. The equation has no solution.

Distribute the left side.

$6x - 3 = 6x - 3$

Subtracting $6x$ leaves $-3 = -3$, true for every $x$, so there are infinitely many solutions.

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

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

Multiply every term by $10$, then distribute.

$4(x + 1) = x + 10, \qquad 4x + 4 = x + 10, \qquad 3x = 6, \qquad x = 2$

Simplify each side.

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

Subtracting $3x$ leaves $3 = 3$, true for every $x$, so there are infinitely many solutions.

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

$10x + 25 = 35, \qquad 10x = 10, \qquad x = 1$

Cross-multiply.

$x + 7 = 3(x + 1), \qquad x + 7 = 3x + 3, \qquad 4 = 2x, \qquad x = 2$

Since $2 \neq -1$, the solution is $x = 2$.