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Algebra I to AP Precalculus · Module 5 of 14

Systems of Linear Equations

Solving systems by graphing, substitution, and elimination, plus systems of inequalities.

In this module
  1. Solving by Graphing
  2. Solving by Substitution
  3. Solving by Elimination
  4. Systems of Linear Inequalities
  5. Check yourself

Solving by Graphing

  • The solution to a system is the point where the two lines intersect — the (x, y) that satisfies both equations.
  • Parallel lines (same slope, different intercept) give no solution; identical lines give infinitely many solutions.
  • Graphing is best for estimating or confirming a solution found algebraically, since reading exact intersections can be imprecise.
Solving a system by graphing
xyy = x + 1y = −x + 5solution (2, 3)
y = x + 1 and y = −x + 5 cross at (2, 3). That point satisfies both equations, so it is the solution of the system.
Worked example

How many solutions does the system y = 2x + 1 and y = 2x − 3 have?

Solution. Same slope (2), different y-intercepts → parallel lines → no solution.

Solving by Substitution

  • Solve one equation for one variable, then substitute that expression into the other equation.
  • Best when one equation is already solved for a variable, or is easy to solve for one.
Worked example

Solve: y = x + 2 and 3x + y = 14.

Solution. Substitute: 3x + (x + 2) = 14 → 4x + 2 = 14 → 4x = 12 → x = 3. Then y = 3 + 2 = 5. Solution: (3, 5).

Solving by Elimination

  • Add or subtract the two equations to eliminate one variable — multiply one or both equations first if needed so a variable's coefficients match or are opposites.
  • After eliminating one variable and solving for the other, substitute back into either original equation.
Worked example

Solve: 2x + 3y = 12 and 2x − y = 4.

Solution. Subtract to eliminate x: 3y − (−y) = 12 − 4 → 4y = 8 → y = 2. Substitute into 2x − y = 4: 2x − 2 = 4 → x = 3. Solution: (3, 2).

Systems of Linear Inequalities

  • Graph each inequality's boundary line — dashed for < or >, solid for ≤ or ≥ — then shade each solution region.
  • The solution to the system is the overlapping (double-shaded) region.
  • Test a point not on either line (often the origin) in each inequality to confirm which side to shade.
Worked example

Describe the solution region for y > x and y ≤ −x + 4.

Solution. Shade above the dashed line y = x, and on/below the solid line y = −x + 4; the solution is the overlapping wedge between them.

Check yourself

2 of this module's 5 quiz questions. Pick your answers, then check them in the app, with an explanation for every question.

1. Solve by substitution: y = 2x − 1 and x + y = 11.

2. Solve by elimination: x + y = 10 and x − y = 2.

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