Algebra

Algebra

7 min read Updated Mar 26, 2026

Think of algebra as a balance scale. Whatever you do to one side, you must do to the other. Add 5 kg to the left pan → add 5 kg to the right. Cut the left side in half → cut the right side in half. That single principle — keep the scale balanced — is all algebra really is.

On the MCAT, you’ll rarely solve algebra for its own sake. Instead you’ll rearrange physics and chemistry formulas to isolate an unknown variable, then plug in values. The faster you can manipulate equations, the more time you have for the actual reasoning — which is what the MCAT is really testing. This section drills the techniques: isolating variables, cross-multiplication, the quadratic formula, and common MCAT rearrangements.

Isolating Variables

The core MCAT algebra skill: given a formula, solve for any variable in it.

Strategy: Undo operations in reverse order of operations (PEMDAS backward). Undo addition/subtraction first, then multiplication/division, then exponents.

Example: Solve KE = (1/2)m$v^2$ for v.

  1. Multiply both sides by 2: 2KE = m$v^2$
  2. Divide both sides by m: 2KE/m = $v^2$
  3. Take the square root: v = sqrt(2KE/m)

Example: Solve PV = nRT for T.

  1. Divide both sides by nR: T = PV/(nR)

That is it. One step. Know which variable you are solving for and systematically peel away everything else.

Cross-Multiplication

When you have a proportion (two fractions set equal), cross-multiplication is the fastest tool.

If a/b = c/d, then ad = bc.

Example: If 5 mL of solution contains 20 mg of drug, how many mg are in 12 mL?

Set up the proportion: 20/5 = x/12

Cross-multiply: 5x = 240

Solve: x = 48 mg

The Quadratic Formula

Most MCAT algebra involves linear equations (no exponents on the variable) or simple rearrangement of squared terms. Occasionally, however, you will encounter a quadratic equation of the form a$x^2$ + bx + c = 0.

In practice, the MCAT usually sets up quadratics where one root is physically meaningless (negative concentration, negative time), so you take the positive root and move on. Also, many MCAT quadratics simplify because one term is negligible - for example, when Ka is very small in a weak acid problem, you can often ignore the x in the denominator.

Substitution

When you have two equations and two unknowns, solve one equation for one variable and substitute into the other.

Example: A passage states that v = v_0 + at and x = v_0 t + (1/2)at$^2$. If you know x and v0v_{0} and a but not t, you use the second equation directly. If you know v and v0v_{0} and need a, you rearrange the first: a = (v - v_0)/t.

The key is choosing which equation to start with. Pick the one that has fewer unknowns or is simpler to rearrange.

Common Algebraic Moves on the MCAT

Starting FormGoalOperation
y = kx + bSolve for xSubtract b, divide by k
y = k/xSolve for xMultiply both sides by x, divide by y: x = k/y
y = k$x^2$Solve for xDivide by k, take square root
1/f = 1/d_o + 1/d_iSolve for did_{i}Subtract 1/dod_{o}, take reciprocal
a/b = c/dSolve for aMultiply both sides by b: a = bc/d

”What Happens When…” Questions

These are the MCAT’s favorite algebra questions. They do not ask you to calculate - they ask you to reason.

Template: “If variable X is doubled/tripled/halved, what happens to variable Y?”

Strategy:

  1. Write the formula relating X and Y
  2. Replace X with the new value (2X, 3X, X/2)
  3. See what multiplier ends up in front of Y

Example: Kinetic energy is KE = (12\frac{1}{2})mv2v^2. If velocity triples, what happens to KE?

Replace v with 3v: KE_new = (12\frac{1}{2})m(3v)² = (12\frac{1}{2})m(9v2v^2) = 9 x (12\frac{1}{2})mv2v^2 = 9 KE

Kinetic energy increases by a factor of 9.

Solve the ideal gas law PV = nRT for the pressure P when n = 2 mol, R = 8.3 J/(mol K), T = 300 K, and V = 0.05 m³.
Click to reveal answer
P = nRT/V = (2)(8.3)(300) / 0.05. Numerator: 2 x 8 x 300 = 4800 (rounding 8.3 to 8). Denominator: 0.05. P = 48000.05\frac{4800}{0.05} = 96,000 Pa. More precisely: 2 x 8.3 x 300 = 4980, so P = 49800.05\frac{4980}{0.05} = 99,600 Pa, roughly 10510^5 Pa (1 atm). Either estimate works.
In the equation F = kx (Hooke's Law), the spring constant k is tripled and the displacement x is halved. What happens to the force F?
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FnewF_{\text{new}} = (3k)(x/2) = (32\frac{3}{2})kx = 1.5 F. The force increases by a factor of 32\frac{3}{2}, or 50%. You do not need numbers - just substitute the changes directly into the formula.