DWG NO. 7.1 — Lesson 1 of 5

Solving Systems by Substitution and Elimination

Unit 7: Systems of Equations and Matrices · ~20 min

Objective Solve systems of two linear equations in two variables using substitution and elimination, and interpret the solution as a point of intersection.

Unit 1 taught you to graph a linear function and read off its key features. Put two linear equations side by side and ask where their graphs agree, and you have a system of equations. Reading a solution off a graph is quick but imprecise — the two algebraic methods in this lesson, substitution and elimination, find that intersection point exactly, without ever drawing a line.

Substitution

Solve one equation for one variable, then substitute that expression into the other equation. This collapses two equations in two unknowns into one equation in one unknown.

Step 1 Solve one equation for one variable (pick whichever is already isolated or easiest to isolate).
Step 2 Substitute that expression into the other equation, giving one equation in one variable.
Step 3 Solve for that variable, then substitute back to find the other.

Elimination

Add or subtract multiples of the two equations so that one variable cancels out entirely.

Step 1 Multiply one or both equations by a constant so that the coefficients of one variable are opposites.
Step 2 Add the equations — that variable disappears, leaving one equation in one unknown.
Step 3 Solve for that variable, then substitute back into either original equation to find the other.

The solution of a system is the single point where both lines' graphs cross

How many solutions?

Two linear equations in two variables graph as two lines, and two lines relate to each other in exactly three ways:

One solution — the lines cross at a single point (different slopes). Elimination or substitution produces one value for each variable.
No solution — the lines are parallel and distinct (same slope, different intercepts). Elimination produces a false statement like 0 = 6.
Infinitely many solutions — the two equations describe the same line. Elimination produces a true statement like 0 = 0.

Worked Example 1 · Substitution

ProblemSolve the system x + y = 5 and 2x − y = 4.

1Solve the first equation for y: y = 5 − x.

2Substitute into the second equation: 2x − (5 − x) = 4.

3Simplify: 2x − 5 + x = 4 → 3x − 5 = 4 → 3x = 9 → x = 3.

4Back-substitute: y = 5 − 3 = 2.

Solution: (3, 2)

Worked Example 2 · Elimination

ProblemSolve the system 4x + 3y = 11 and 2x − 3y = 1.

1The y-coefficients (+3 and −3) are already opposites, so add the equations directly: 6x = 12.

2Solve: x = 2.

3Substitute into 2x − 3y = 1: 4 − 3y = 1 → −3y = −3 → y = 1.

Solution: (2, 1)

Guided practice

1.Solve by substitution: x − y = 2 and 2x + y = 13.
Show answer

From the first equation, y = x − 2. Substitute: 2x + (x − 2) = 13 → 3x − 2 = 13 → 3x = 15 → x = 5, so y = 3. Solution: (5, 3).
2.Solve by elimination: 3x + y = 9 and x − y = 3.
Show answer

Add the equations: 4x = 12 → x = 3. Substitute into x − y = 3: 3 − y = 3 → y = 0. Solution: (3, 0).
3.Without fully solving, how many solutions does 2x + 4y = 8 and x + 2y = 4 have?
Show answer

Divide the first equation by 2: x + 2y = 4, which is identical to the second equation. The two equations describe the same line, so there are infinitely many solutions.
4.Without fully solving, how many solutions does x − y = 1 and x − y = 5 have?
Show answer

Both equations have the same slope (same left side) but different constants, so they graph as parallel, distinct lines — no solution. (Subtracting the equations gives 0 = −4, a false statement, confirming this.)
5.Two numbers have a sum of 20 and a difference of 4. Find the numbers.
Show answer

Let the numbers be x and y: x + y = 20 and x − y = 4. Adding gives 2x = 24 → x = 12, so y = 8. The numbers are 12 and 8.

← Unit Overview Unit Overview 7.2 Systems of Inequalities →