SAT · Math · Systems of Two Linear Equations

Solving Systems: Substitution and Elimination

9 min readPreviewBy Uzair Khan

What you'll be able to do

Substitution, elimination, and structural shortcuts such as adding or subtracting the equations to get a requested expression like x + y directly; checking the intersection with Desmos.

Introduction

A system of two linear equations in two variables appears throughout the Algebra domain, which accounts for roughly 35% of SAT Math questions. The test doesn't just ask you to solve the system for xx and yy — it sometimes asks for a combined expression like x+yx + y or 3x−y3x - y, where a structural shortcut gets the answer in one line. This note covers three fluent methods: substitution, elimination, and structural shortcuts, plus how to confirm any answer with Desmos.

Related notes: For systems with no solution or infinitely many solutions, see Systems with No Solution or Infinitely Many Solutions. For word problems that require setting up the system from context, see Systems of Equations in Context.


Core Concept

Every system of two linear equations in two variables has a solution (x,y)(x, y) that satisfies both equations simultaneously. Geometrically, it is the intersection point of two lines. Three algebraic strategies reach that point:

Strategy 1 — Substitution

Best when: One equation is already solved for one variable (or is easy to solve).

Steps:

  1. Isolate one variable in one equation.
  2. Substitute that expression into the other equation.
  3. Solve the resulting single-variable equation.
  4. Back-substitute to find the second variable.

Quick illustration:

y=2x+1,3x+y=11y = 2x + 1, \quad 3x + y = 11

Substitute y=2x+1y = 2x+1: 3x+(2x+1)=11⇒5x=10⇒x=2,  y=53x + (2x+1) = 11 \Rightarrow 5x = 10 \Rightarrow x = 2, \; y = 5.

Strategy 2 — Elimination (Addition/Subtraction)

Best when: The same variable has equal (or opposite) coefficients, or can be made equal by multiplying one equation by a constant.

Steps:

  1. Multiply one (or both) equation(s) so one variable has matching coefficients.
  2. Add or subtract the equations to eliminate that variable.
  3. Solve for the remaining variable; back-substitute.

Quick illustration:

3x+2y=13,3x−y=43x + 2y = 13, \quad 3x - y = 4

Subtract: 3y=9⇒y=33y = 9 \Rightarrow y = 3; back-sub: x=13−63=73x = \frac{13-6}{3} = \frac{7}{3}.

Strategy 3 — Structural Shortcut

Best when: The question asks for a combined expression (e.g., x+yx + y, 2x−3y2x - 3y) rather than individual values.

Key idea

Key idea: Adding or subtracting the two equations often directly produces the target expression. No need to find xx and yy separately.

Quick illustration:

4x+y=17,2x−y=7→add6x=244x + y = 17, \quad 2x - y = 7 \quad \xrightarrow{\text{add}} \quad 6x = 24

If the question asked for 6x6x, the answer is 2424 — one step.


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