In certain situations we may need to rearrange an equation or formula to make a different variable the subject. As an example, consider the formula for finding the circumference $C$C of any circle with radius $r$r:
$C=2\pi r$C=2πr
In this formula, $C$C is currently the subject. By rearranging the formula, however, we can make $r$r the subject. This can be done with the same methods we used for solving equations. In this case however, we don't have any values to substitute, so we are working mainly with variables. Although we use equation-solving techniques, we are not actually solving an equation. Instead, we are rearranging the formula or equation into a different form. In this case, the rearranged formula becomes:
$r=\frac{C}{2\pi}$r=C2π
The subject of an equation or formula is the single variable by itself on one side of the equals sign, usually the left-hand side.
For an object that travels a distance $d$d in time $t$t, the average speed of the object is given by the formula $S=\frac{d}{t}$S=dt. Rearrange the formula to change the subject to:
(a) $d$d
Solution
$S$S | $=$= | $\frac{d}{t}$dt | ||
$S\times t$S×t | $=$= | $\frac{d}{t}\times t$dt×t | (Multiply both sides by $t$t) | |
$St$St | $=$= | $d$d | ||
$d$d | $=$= | $St$St | (Swap so that the subject is on the left-hand side) |
(b) $t$t
Solution
$S$S | $=$= | $\frac{d}{t}$dt | ||
$S\times t$S×t | $=$= | $\frac{d}{t}\times t$dt×t | (Multiply both sides by $t$t) | |
$St$St | $=$= | $d$d | ||
$\frac{St}{S}$StS | $=$= | $\frac{d}{S}$dS | (Divide both sides by $S$S) | |
$t$t | $=$= | $\frac{d}{S}$dS |
Rearrange $y=mx+c$y=mx+c to change the subject of the equation to the following:
(a) $c$c
Solution
$y$y | $=$= | $mx+c$mx+c | ||
$y-mx$y−mx | $=$= | $mx+c-mx$mx+c−mx | (Subtract $mx$mx from both sides) | |
$y-mx$y−mx | $=$= | $c$c | ||
$c$c | $=$= | $y-mx$y−mx | (Swap so that the subject is on the left-hand side) |
(b) $m$m
Solution
$y$y | $=$= | $mx+c$mx+c | ||
$y-c$y−c | $=$= | $mx+c-c$mx+c−c | (Subtract $c$c from both sides) | |
$y-c$y−c | $=$= | $mx$mx | ||
$\frac{y-c}{m}$y−cm | $=$= | $\frac{mx}{m}$mxm | (Divide both sides by $m$m) | |
$\frac{y-c}{m}$y−cm | $=$= | $x$x | ||
$x$x | $=$= | $\frac{y-c}{m}$y−cm | (Swap so that the subject is on the left-hand side) |
The kinetic energy of an object is given by the formula $E=\frac{1}{2}mv^2$E=12mv2. Rearrange to make $v^2$v2 the subject of the formula.
Solution
We begin here by multiplying both sides by $2$2 in order to remove the fraction.
$E$E | $=$= | $\frac{1}{2}mv^2$12mv2 | ||
$E\times2$E×2 | $=$= | $\frac{1}{2}mv^2\times2$12mv2×2 | (Multiply both sides by $2$2) | |
$2E$2E | $=$= | $mv^2$mv2 | ||
$\frac{2E}{m}$2Em | $=$= | $\frac{mv^2}{m}$mv2m | (Divide both sides by $m$m) | |
$\frac{2E}{m}$2Em | $=$= | $v^2$v2 | ||
$v^2$v2 | $=$= | $\frac{2E}{m}$2Em | (Swap so that the subject is on the left-hand side) |
Rearrange $ax=b-cx$ax=b−cx to make $x$x the subject.
Solution
In this question we have $x$x variables on both sides. We want to get all the $x$x variables on the left, so we begin by adding $cx$cx to both sides. This will give us the expression $ax+cx$ax+cx on the left. We can factorise this expression by taking $x$x out as a common factor. To isolate the $x$x, our final step is to divide both sides by $a+c$a+c.
$ax$ax | $=$= | $b-cx$b−cx | ||
$ax+cx$ax+cx | $=$= | $b-cx+cx$b−cx+cx | (Add $cx$cx to both sides) | |
$ax+cx$ax+cx | $=$= | $b$b | ||
$x\left(a+c\right)$x(a+c) | $=$= | $b$b | (Take out the common factor $x$x) | |
$\frac{x\left(a+c\right)}{a+c}$x(a+c)a+c | $=$= | $\frac{b}{a+c}$ba+c | (Divide both sides by $a+c$a+c) | |
$x$x | $=$= | $\frac{b}{a+c}$ba+c |
Make $m$m the subject of the following equation:
$\frac{m}{y}=gh$my=gh
Make $m$m the subject of the following equation:
$y=6mx-9$y=6mx−9
Rearrange the formula $r=\frac{k}{k-9}$r=kk−9 to make $k$k the subject.
uses algebraic and graphical techniques to compare alternative solutions to contextual problems