Solving for a Variable: Additional Practice (Set 1)

Both terms on the left contain $x$, so factor it out, then divide.

$x(m - n) = p, \qquad x = \frac{p}{m - n}$

Only the first term holds $h$, so subtract $2\pi r^2$ from both sides, then divide the whole result by $2\pi r$.

$2\pi rh = A - 2\pi r^2, \qquad h = \frac{A - 2\pi r^2}{2\pi r}$

Factor $x$ from the left, combine the two fractions, then divide by the result.

$x \cdot \frac{b - a}{ab} = 1, \qquad x = \frac{ab}{b - a}$

Distribute both sides, gather the $x$ terms on the left, then factor and divide.

$kx + ka = mx + mb, \qquad x(k - m) = mb - ka, \qquad x = \frac{mb - ka}{k - m}$

Multiply both sides by $t$ to clear the denominator, then add $u$.

$v - u = at, \qquad v = at + u$

Add $k$ to both sides to isolate $y$.

$y = m(x - h) + k$

Divide both sides by $P$ to free the parenthesis, then subtract $1$.

$1 + i = \frac{A}{P}, \qquad i = \frac{A}{P} - 1$

Add $cx$ to both sides and subtract $d$, then divide by $c$.

$cx = ab - d, \qquad x = \frac{ab - d}{c}$

Multiply both sides by $x$, gather the $x$ terms, then factor and divide.

$x + a = bx, \qquad a = x(b - 1), \qquad x = \frac{a}{b - 1}$

Multiply both sides by $\frac{2}{n}$ to free the parenthesis, then subtract $a$.

$a + L = \frac{2S}{n}, \qquad L = \frac{2S}{n} - a$

Combine the fractions on the right over the common denominator $r_1 r_2$, then take the reciprocal of both sides.

$\frac{1}{R} = \frac{r_1 + r_2}{r_1 r_2}, \qquad R = \frac{r_1 r_2}{r_1 + r_2}$

Add $cx$ to both sides to gather the $x$ terms, then factor and divide.

$ax + cx = b, \qquad x(a + c) = b, \qquad x = \frac{b}{a + c}$