For example, you might be told to solve the area of a triangle formula for h{\displaystyle h}. Or, you might know that you have the area and base of the triangle, so you need to solve for the height. So, you need to rearrange the formula and isolate the h{\displaystyle h} variable.
Multiplication and division. Addition and subtraction. Squaring and taking a square root.
For example, to solve the area of a triangle formula (A=12bh{\displaystyle A={\frac {1}{2}}bh}) for h{\displaystyle h}: Cancel the fraction by multiplying each side by 2:A×2=2×12bh{\displaystyle A\times 2=2\times {\frac {1}{2}}bh}2A=bh{\displaystyle 2A=bh} Isolate h{\displaystyle h} by dividing each side by b{\displaystyle b}:2Ab=bhb{\displaystyle {\frac {2A}{b}}={\frac {bh}{b}}}2Ab=h{\displaystyle {\frac {2A}{b}}=h} Rearrange the formula, if desired: h=2Ab{\displaystyle h={\frac {2A}{b}}}
For example, you might have the equation of a line 3x+2y=4{\displaystyle 3x+2y=4}. This is in standard form. If you need to find the y-intercept of the line, you need to rearrange the formula to slope-intercept form by isolating the y{\displaystyle y} variable:[6] X Research source Subtract 3x{\displaystyle 3x} from both sides of the equation:3x+2y−3x=4−3x{\displaystyle 3x+2y-3x=4-3x}2y=4−3x{\displaystyle 2y=4-3x}. Divide both sides by 2{\displaystyle 2}:2y2=4−3x2{\displaystyle {\frac {2y}{2}}={\frac {4-3x}{2}}}y=4−3x2{\displaystyle y={\frac {4-3x}{2}}}
For example, to change y=4−3x2{\displaystyle y={\frac {4-3x}{2}}} to the correct slope-intercept formula, you need to switch the order of the number in the numerator, then simplify:y=−3x+42{\displaystyle y={\frac {-3x+4}{2}}}y=−32x+2{\displaystyle y={\frac {-3}{2}}x+2}Now, since the formula is in proper slope-intercept form, it is easy to identify the y-intercept as 2.
Factor out the b{\displaystyle b}: R=b(5d−6a){\displaystyle R=b(5d-6a)}. Isolate the b{\displaystyle b} by dividing each side by the expression in parentheses:R5d−6a=b(5d−6a)5d−6a{\displaystyle {\frac {R}{5d-6a}}={\frac {b(5d-6a)}{5d-6a}}}R5d−6a=b{\displaystyle {\frac {R}{5d-6a}}=b}
Understand what each variable stands for. In this formula, C{\displaystyle C} is the circumference, and r{\displaystyle r} is the radius. So you need to isolate the r{\displaystyle r} to solve for the radius. Isolate the r{\displaystyle r} by dividing both sides of the equation by 2π{\displaystyle 2\pi }:C2π=2πr2π{\displaystyle {\frac {C}{2\pi }}={\frac {2\pi {r}}{2\pi }}}C2π=r{\displaystyle {\frac {C}{2\pi }}=r} If desired, reverse the order of the equation for standard form: r=C2π{\displaystyle r={\frac {C}{2\pi }}}.
Recall that standard form is Ax+By=C{\displaystyle Ax+By=C}. Cancel the fraction by multiplying each side of the equation by 2:2y=2(12x+5){\displaystyle 2y=2({\frac {1}{2}}x+5)}2y=x+10{\displaystyle 2y=x+10} Subtract x{\displaystyle x} from both sides of the equation:2y−x=x+10−x{\displaystyle 2y-x=x+10-x}2y−x=10{\displaystyle 2y-x=10} Rearrange the y{\displaystyle y} and x{\displaystyle x} variables so that they are in the standard form: −x+2y=10{\displaystyle -x+2y=10}. Multiply both sides by −1{\displaystyle -1}, since A{\displaystyle A} should be a positive integer for standard form:[9] X Research source −1(−x+2y)=−1(10){\displaystyle -1(-x+2y)=-1(10)}x−2y=−10{\displaystyle x-2y=-10}
