Example 1 · Basic method
Make x the subject of y = mx
This equation describes a direct proportional relationship: multiplying the input by a constant produces the output.
y is the dependent output; m is a known constant multiplier or gradient; x is the input to isolate.
y = mxStart
\frac{y}{m}=\frac{mx}{m}Divide both sides by m
\frac{y}{m}=\frac{\cancel{m}x}{\cancel{m}}Cancel the common m in the numerator and denominator
\frac{y}{m}=xBecause \frac{m}{m}=1, for m\ne0
x=\frac{y}{m}Write the subject on the left
Example 2 · Engineering formula
Make resistance the subject of V = IR
Ohm's law relates the voltage across an ohmic component to the current through it and its resistance.
V is voltage in volts; I is current in amperes; R is resistance in ohms.
V = IROhm's law
\frac{V}{I}=\frac{IR}{I}Divide both sides by current
\frac{V}{I}=\frac{\cancel{I}R}{\cancel{I}}Cancel I from the numerator and denominator
\frac{V}{I}=RBecause \frac{I}{I}=1, for I\ne0
R=\frac{V}{I}Write resistance as the subject
Example 3 · Two operations
Make acceleration the subject of v = u + at
This constant-acceleration equation gives the final velocity after acceleration acts for a specified time.
v is final velocity in \mathrm{m\,s^{-1}}; u is initial velocity in \mathrm{m\,s^{-1}}; a is acceleration in \mathrm{m\,s^{-2}}; t is time in seconds.
v - u = atSubtract u from both sides
\frac{v-u}{t}=\frac{at}{t}Divide both sides by t
\frac{v-u}{t}=\frac{a\cancel{t}}{\cancel{t}}Cancel t from the numerator and denominator
\frac{v-u}{t}=aBecause \frac{t}{t}=1, for t\ne0
a=\frac{v-u}{t}Write acceleration as the subject
Example 4 · Variable in a denominator
Make area the subject of \sigma=\frac{F}{A}
Normal stress measures how a force acting perpendicular to a surface is distributed over its cross-sectional area.
\sigma is normal stress in pascals; F is normal force in newtons; A is cross-sectional area in \mathrm{m^2}.
\sigma A=FMultiply both sides by A
\frac{\sigma A}{\sigma}=\frac{F}{\sigma}Divide both sides by \sigma
\frac{\cancel{\sigma}A}{\cancel{\sigma}}=\frac{F}{\sigma}Cancel \sigma from the numerator and denominator
A=\frac{F}{\sigma}Because \frac{\sigma}{\sigma}=1, for \sigma\ne0
Example 5 · Square and square root
Make velocity the subject of E_k=\frac{1}{2}mv^2
Kinetic energy is the energy an object possesses because it is moving; it increases with mass and with the square of speed.
E_k is kinetic energy in joules; m is mass in kilograms; v is speed in \mathrm{m\,s^{-1}}.
2E_k=mv^2Multiply both sides by 2
\frac{2E_k}{m}=\frac{mv^2}{m}Divide both sides by m
\frac{2E_k}{m}=\frac{\cancel{m}v^2}{\cancel{m}}Cancel m from the numerator and denominator
\frac{2E_k}{m}=v^2Because \frac{m}{m}=1, for m\ne0
v=\sqrt{\frac{2E_k}{m}}Take the physically relevant positive square root
Example 6 · Examination bridge
Make acceleration the subject of s=ut+\frac{1}{2}at^2
This equation gives displacement during motion with constant acceleration over an elapsed time.
s is displacement in metres; u is initial velocity in \mathrm{m\,s^{-1}}; a is acceleration in \mathrm{m\,s^{-2}}; t is elapsed time in seconds.
s-ut=\frac{1}{2}at^2Subtract ut from both sides
2(s-ut)=at^2Multiply both sides by 2
\frac{2(s-ut)}{t^2}=\frac{at^2}{t^2}Divide both sides by t^2
\frac{2(s-ut)}{t^2}=\frac{a\cancel{t^2}}{\cancel{t^2}}Cancel t^2 from the numerator and denominator
\frac{2(s-ut)}{t^2}=aBecause \frac{t^2}{t^2}=1, for t\ne0
a=\frac{2(s-ut)}{t^2}Write acceleration as the subject