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How to solve a system of equations

A system of equations is two or more equations that must be true at the same time, and its solution is the pair of values that works in every equation. For two linear equations, you can graph both lines and find where they cross, substitute one equation into the other, or add the equations to eliminate a variable. A system can have one solution, no solution, or infinitely many.

What a system of equations is

One equation with two unknowns, such as x + y = 10, has endless solutions: 3 and 7, 4 and 6, 2.5 and 7.5, and so on. Add a second condition, such as x − y = 4, and only one pair fits both: x = 7 and y = 3. That pair is the solution of the system.

Every solution of a linear equation in x and y is a point on its line. A solution of the system has to lie on both lines at once, so it is the point where they meet. This picture explains why there are only three possible outcomes for two lines: they cross once, they never meet, or they are the same line.

You can always check a solution by substituting it into every original equation. If any one of them fails, it is not a solution.

Method 1: graphing

Write each equation in slope-intercept form, y = mx + b, graph both lines, and read the point where they cross. For y = x + 1 and y = −x + 5, the first line starts at 1 on the y-axis and climbs 1 per step; the second starts at 5 and drops 1 per step. They cross at (2, 3). Checking: 2 + 1 = 3 and −2 + 5 = 3.

Graphing shows what a solution means, and it is a good first look. Its weakness is precision: if the lines cross at (1.4, 2.7), a hand-drawn graph will not tell you that exactly. Use graphing to estimate, and algebra to get the exact answer.

Method 2: substitution

Substitution works best when one equation already has a variable on its own, such as y = 2x − 1, or when a variable has a coefficient of 1 and is easy to isolate. Replace that variable in the other equation with its expression, solve the resulting one-variable equation, then substitute back.

ProblemSolve {y=2x−13x+y=14

  1. 3x+(2x−1)=14

    The first equation says y equals 2x − 1, so put that expression in place of y in the second equation. Use parentheses.

  2. 5x−1=14

    Combine like terms. Now there is only one variable.

  3. 5x=15,x=3

    Add 1 to both sides, then divide by 5.

  4. y=2(3)−1=5

    Substitute x = 3 back into the simpler equation to find y.

  5. 3(3)+5=14

    Check in the other original equation: 9 + 5 is 14, so (3, 5) works in both.

Answer(3,5)

Method 3: elimination

Elimination, also called the addition method, works best when both equations are in the form ax + by = c. Multiply one or both equations so that one variable has opposite coefficients, such as 4y and −4y, then add the equations so that variable disappears. Multiplying a whole equation by a number does not change its solutions, for the same reason it keeps a single equation balanced.

ProblemSolve {3x+2y=135x−4y=7

  1. 6x+4y=26

    Multiply every term of the first equation by 2, so its y-term becomes 4y, the opposite of −4y.

  2. (6x+4y)+(5x−4y)=26+7

    Add the new first equation to the second, left side to left side and right side to right side.

  3. 11x=33,x=3

    The y-terms cancel. Divide by 11.

  4. 3(3)+2y=13,y=2

    Substitute into the first original equation: 9 + 2y = 13, so 2y = 4.

  5. 5(3)−4(2)=7

    Check in the second original equation: 15 − 8 is 7.

Answer(3,2)

No solution and infinitely many solutions

Sometimes both variables disappear at once. What is left tells you which special case you have:

Worked example: a word problem

Word problems with two unknowns and two facts are natural systems. Name each unknown, write one equation per fact, and say what the answer means at the end.

ProblemA school sold 50 tickets for $480. Adult tickets cost $12 and student tickets $8. How many of each were sold?

  1. {a+s=5012a+8s=480

    Let a be adult tickets and s be student tickets. One equation counts tickets, the other counts dollars.

  2. s=50−a

    Solve the first equation for s, ready to substitute.

  3. 12a+8(50−a)=480

    Substitute into the money equation.

  4. 12a+400−8a=480,4a=80,a=20

    Distribute, combine like terms, subtract 400, and divide by 4.

  5. s=50−20=30

    Find the student tickets.

  6. 12(20)+8(30)=240+240=480

    Check: 20 + 30 is 50 tickets, and the money adds to $480. So 20 adult and 30 student tickets were sold.

Answera=20,s=30

Common mistakes

Watch for these when solving systems:

Choosing a method and checking your work

Choose substitution when a variable is already alone or has a coefficient of 1 or −1. Choose elimination when both equations are in standard form, especially when coefficients already match or are easy to match. Use graphing to see the situation, to estimate, or when a graphing tool is allowed.

Always check the final pair in both original equations, not in a rearranged version, which may already contain a mistake. For extra practice, solve the same system with two different methods and confirm that they agree. Systems with three equations and three unknowns, which come up in Algebra II, use the same elimination idea twice.

Questions

What are the three ways to solve a system of equations?

Graphing, substitution and elimination. Graphing shows the crossing point, substitution replaces one variable with an expression, and elimination adds equations to cancel a variable.

How do you know if a system has no solution?

When solving leads to a false statement such as 0 = 5, there is no solution. On a graph, the lines are parallel: they have the same slope but different y-intercepts.

What does infinitely many solutions mean?

Both equations describe the same line, so every point on it satisfies both. Solving leads to a statement that is always true, such as 0 = 0.

Which method is best, substitution or elimination?

Both always work for linear systems. Substitution is quicker when a variable is already isolated; elimination is quicker when the equations are in standard form, ax + by = c.

What grade learns systems of equations?

Grade 8 introduces systems of two linear equations solved by graphing, by inspection and algebraically. Algebra I covers substitution and elimination in depth, and Algebra II extends them.