Solving for a Variable: Additional Practice (Set 2)

The target $P$ is in both terms on the right, so factor it out and divide.

$A = P(1 + rt), \qquad P = \frac{A}{1 + rt}$

The lone $P$ is $P \cdot 1$, so a $1$ stays inside the factor; dropping it gives the wrong $\dfrac{A}{rt}$.

Both terms on the left contain $y$, so factor it out, then divide by the sum.

$y(a + b) = c - d, \qquad y = \frac{c - d}{a + b}$

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

$x(y + 1) = z, \qquad x = \frac{z}{y + 1}$

The second term $x$ is $x \cdot 1$, so a $1$ joins $y$ inside the factor.

Subtract $a$, divide by $d$, then add $1$ to undo the $n - 1$.

$\frac{L - a}{d} = n - 1, \qquad n = \frac{L - a}{d} + 1$

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

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

The right side is a sum, so subtract $b_1$ from both sides.

$b_2 = \frac{2A}{h} - b_1$

Cross-multiply the proportion, then divide by $P$.

$Px = QR, \qquad x = \frac{QR}{P}$

Add $t$ to both sides, then multiply by $s$ to clear the denominator.

$\frac{r}{s} = q + t, \qquad r = s(q + t)$

Gather the $x$ terms on the left and the constants on the right, then factor and divide.

$px - rx = s - q, \qquad x(p - r) = s - q, \qquad x = \frac{s - q}{p - r}$

Subtract $c$ from both sides, then factor $x$ from the two $x$ terms and divide.

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

Subtract $k$, then clear the denominator by multiplying by $V$ and dividing by $T - k$.

$\frac{D}{V} = T - k, \qquad D = V(T - k), \qquad V = \frac{D}{T - k}$

Multiply both sides by $k - x$ to clear the denominator, then isolate $x$.

$g(k - x) = h, \qquad k - x = \frac{h}{g}, \qquad x = k - \frac{h}{g}$

The value $x = k$ is excluded, since it makes the original denominator zero.