Solve for a Variable

Isolate any letter — in a plain equation or in a formula full of other symbols
Solve 5x - 7 = 3x + 9 for x
Solve A = (1/2)bh for h
Solve F = (9/5)C + 32 for C
Solve 2(y - 4) = 3y + 1 for y

What a Variable Problem Asks

A variable is a letter standing for a number you do not yet know. "Solve for the variable" means rewrite the equation until that letter sits alone on one side and everything else sits on the other.

Two situations look different but obey the same rules:

  • Numeric equations such as 5x7=3x+95x - 7 = 3x + 9, where the answer is a number.
  • Literal equations (formulas) such as A=12bhA = \tfrac{1}{2}bh, where the answer is another expression: h=2Abh = \dfrac{2A}{b}.

The second kind is the one students find alien, but nothing changes. The letters AA and bb are simply numbers whose values have not been announced. Rearranging a formula once and substituting afterwards beats substituting first and rearranging every single time.

The skill transfers directly into science and finance, where almost every formula gets used in more than one direction. The same relationship between distance, speed and time is three different formulas depending on which quantity is missing, and rearranging is what turns one memorised statement into all three.

Isolate by Undoing, in Reverse Order

Whatever you do to one side, do to the other. That single rule keeps the equation balanced.

The procedure

  1. Clear brackets and fractions. Distribute; multiply through by the common denominator.
  2. Gather the target variable on one side. Add or subtract terms so every copy of the variable is together, usually on the side where its coefficient is larger.
  3. Undo addition and subtraction attached to it.
  4. Undo multiplication and division last, dividing by the coefficient.
  5. Substitute back into the original equation to verify.

The order matters: you unwrap an expression in the opposite order to how it was built, the way you take off a coat before a sweater.

Attached operationUndo with
+7+7subtract 77
×3\times 3divide by 33
÷5\div 5multiply by 55
squaredtake the square root (±\pm)

Common Mistakes to Avoid

  • Dividing before collecting. Divide by 22 while a +7+7 is still glued on and you must divide the 77 too. Undo the addition first.
  • Multiplying only one term by the denominator. Clearing a fraction means multiplying every term on both sides.
  • Panicking at letters. In A=12bhA = \tfrac{1}{2}bh, treat AA and bb exactly as you would 2424 and 66.
  • Losing the negative coefficient. From x=5-x = 5 the answer is x=5x = -5, not x=5x = 5.
  • Skipping the check. Substituting your value back into the original equation catches nearly every sign slip in under ten seconds.

Examples

Step 1: Subtract 3x3x from both sides to gather the variable: 2x7=92x - 7 = 9
Step 2: Add 77 to both sides: 2x=162x = 16
Step 3: Divide both sides by 22: x=8x = 8
Step 4: Check: left =5(8)7=33= 5(8) - 7 = 33; right =3(8)+9=33= 3(8) + 9 = 33
Answer: x=8x = 8

Step 1: Multiply both sides by 22 to clear the fraction: 2A=bh2A = bh
Step 2: Divide both sides by bb: h=2Abh = \dfrac{2A}{b}, valid for b0b \neq 0
Step 3: Substitute A=24A = 24, b=6b = 6: h=2(24)6=486=8h = \dfrac{2(24)}{6} = \dfrac{48}{6} = 8
Step 4: Check in the original formula: 12(6)(8)=24\tfrac{1}{2}(6)(8) = 24
Answer: h=2Abh = \dfrac{2A}{b}, and h=8h = 8 when A=24A = 24, b=6b = 6

Step 1: Subtract 3232 from both sides: F32=95CF - 32 = \dfrac{9}{5}C
Step 2: Multiply both sides by 59\dfrac{5}{9}: C=5(F32)9C = \dfrac{5(F - 32)}{9}
Step 3: Substitute F=68F = 68: C=5(36)9=1809=20C = \dfrac{5(36)}{9} = \dfrac{180}{9} = 20
Step 4: Check: 95(20)+32=36+32=68\dfrac{9}{5}(20) + 32 = 36 + 32 = 68
Answer: C=5(F32)9C = \dfrac{5(F - 32)}{9}, so 68 F=20 C68\ ^\circ\text{F} = 20\ ^\circ\text{C}

Frequently Asked Questions

It means rearranging the equation so the named letter stands alone on one side. The other side then holds either a number or an expression built from the remaining symbols.

Exactly as with numbers: undo the operations attached to the target letter, applying each step to both sides. Treat the other letters as ordinary constants. A = (1/2)bh becomes h = 2A/b in two steps.

Add or subtract a term so all copies land on one side. From 5x - 7 = 3x + 9, subtract 3x to get 2x - 7 = 9, then finish as usual.

Because you only do the algebra once. If you must apply the same formula to twenty data points, a rearranged version turns each case into one substitution instead of a fresh rearrangement, and there are far fewer places to make an arithmetic slip.

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