Systems of linear equations

College Algebra. Also known as: math 101, math 1314, math 1111, math 110, mat 140, intermediate algebra, precalculus i, algebra for college students, math 1010, college algebra and trigonometry

Solve two equations in two unknowns by graphing, substitution or elimination, and recognise no-solution and infinite-solution cases.

What is a system?

A system is two or more equations that must be true at the same time. For two lines, the solution is the point (x, y) that lies on both. Graphically, that is where the lines cross.

There are three possible outcomes:

  • One solution: the lines cross once (different slopes).
  • No solution: the lines are parallel (same slope, different intercepts). The system is inconsistent.
  • Infinitely many solutions: both equations describe the same line. The system is dependent.

Substitution

Solve one equation for one variable, then substitute into the other. It is best when one variable already has coefficient 1.

Elimination

Add or subtract the equations (after scaling) so that one variable cancels. It is best when the equations are in standard form.

Worked example (substitution)

Solve 2x + 3y = 12 and x − y = 1.

  1. From the second equation, x = y + 1.
  2. Substitute: 2(y + 1) + 3y = 12, so 5y + 2 = 12 and y = 2.
  3. Then x = 2 + 1 = 3.
  4. Check in both: 2(3) + 3(2) = 12 ✓; 3 − 2 = 1 ✓.

The solution is (3, 2).

Worked example (elimination)

Solve 3x + 2y = 7 and 5x − 2y = 9.

  • The y-terms are +2y and −2y, so adding the equations eliminates y: 8x = 16, so x = 2.
  • Put x = 2 in the first: 6 + 2y = 7, so y = 1/2.
  • Solution: (2, 1/2).

Spotting special cases

When you eliminate and get a false statement such as 0 = 7, there is no solution. When you get a true statement such as 0 = 0, there are infinitely many. For instance 2x + y = 5 and 4x + 2y = 10: doubling the first gives the second exactly, so every point on the line works.

Word problems

Define two unknowns, write two equations, solve.

Adult tickets cost $8 and child tickets $5. Ten tickets cost $62 in all. Let a be adults, c be children.

  • a + c = 10
  • 8a + 5c = 62

Substituting c = 10 − a gives 8a + 50 − 5a = 62, so 3a = 12 and a = 4. There are 4 adults and 6 children: 32 + 30 = 62. ✓

Common mistakes

  • Solving for only one variable and stopping.
  • Not multiplying every term when scaling an equation.
  • Not checking the answer in both original equations.

Practice questions

  1. Solve x + y = 10 and x − y = 4.

  2. Solve y = 2x and x + y = 9.

  3. The system y = 3x + 1 and y = 3x − 4 has…

  4. The system 2x + y = 5 and 4x + 2y = 10 has…

  5. Adding 3x + 2y = 7 and 5x − 2y = 9 gives what value of x?

  6. Adult tickets cost $8, child tickets $5. Ten tickets cost $62. How many adult tickets?