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G12 · Solve linear systems by substitution or elimination

Learn to solve linear systems by substitution or elimination through clear examples and targeted practice.

Ontario Grade 10 Mathematics

Modelling Linear Relations

Find the values that make two linear equations true at the same time.

A linear equation can have many solutions. For example, different pairs of numbers can make x+y=8x+y=8 true. A linear system is a pair of equations that use the same variables. Its solution is a pair of values that makes both equations true. In this lesson, you will solve systems using substitution or elimination. These methods build on familiar skills: solving an equation, keeping both sides balanced, and combining like terms.

What you will learn

Grade 9 bridge: what does a solution mean?

Start with the equation x+y=8x+y=8. The pair (3,5)(3,5) is a solution because 3+5=83+5=8. The pair (2,4)(2,4) is not, because 2+42+4 is not 88.
Now consider two equations together: x+y=8x+y=8 and x−y=2x-y=2. A solution must work in both. The pair (5,3)(5,3) works: 5+3=85+3=8 and 5−3=25-3=2.
The variables are the unknown values. An ordered pair, such as (5,3)(5,3), lists the value of xx first and the value of yy second. A linear system is a set of linear equations with the same variables. Solving the system means finding the values that satisfy every equation in it.

Two methods for solving a system

Substitution means replacing a variable with an equal expression. It is useful when one equation already gives a variable by itself, or when that is easy to arrange. For example, if y=2x+1y=2x+1, you can replace yy with 2x+12x+1 in the other equation. This leaves one equation with one variable.
Elimination means combining equations so one variable disappears. You may add or subtract the equations. If a variable has matching coefficients with opposite signs, adding can remove it. If the coefficients match with the same sign, subtracting can remove it. A coefficient is the number multiplying a variable; in 3x3x, the coefficient is 33.
Whichever method you use, keep equations balanced. If you multiply an equation by a number, multiply every term on both sides. Once you find one variable, put its value into an original equation to find the other. Then check both values in both original equations.
\begin{aligned} ax+by&=c\\ dx+ey&=f \end{aligned}

A table view: what does the solution represent?

Each equation describes pairs of values that make it true. A system asks for a pair that belongs to both sets of pairs. The table shows this idea for a simple system. The shared pair is the solution.
The methods in this lesson find the shared pair using equations. They do not depend on guessing values from a table.

Choosing a method and interpreting the result

After solving, you may find one pair that works in both equations. Some systems instead lead to a statement that is never true, such as 0=50=5. That means there is no solution: no pair can satisfy both equations.
A different system may reduce to a statement that is always true, such as 0=00=0. This means the equations describe the same relationship, so there are many solutions. In that case, every pair that works in one equation also works in the other.
These outcomes come from simplifying the given equations. They do not change the main test: a solution must make both original equations true.

Checking candidate pairs for the system $x+y=8$ and $x-y=2$

Candidate pairFirst equationSecond equationWorks in both?
(5,3)(5,3)5+3=85+3=85−3=25-3=2Yes
(4,4)(4,4)4+4=84+4=84−4=04-4=0No

Worked example

Solve by elimination

Solve the system 2x+y=112x+y=11 and x−y=1x-y=1.
  1. Add to remove a variable
    The yy terms have opposite signs. Add the equations vertically. The yy terms cancel, leaving an equation in xx. \begin{aligned}2x+y&=11\\x-y&=1\\\hline 3x&=12\end{aligned}
  2. Find the value of x
    Divide both sides by 33 to keep the equation balanced. This gives the value of xx.
    x=4x=4
  3. Find the value of y
    Substitute x=4x=4 into the original equation x−y=1x-y=1. Solve for yy.
    4−y=1⇒y=34-y=1 \Rightarrow y=3
  4. Check both equations
    Replace xx with 44 and yy with 33 in each original equation. Both statements are true, so the pair is the solution.
    2(4)+3=114−3=12(4)+3=11 4-3=1
Answer: The solution is (4,3)(4,3).
Check: The pair makes both original equations true.

Common mistakes and how to avoid them

Stopping after finding one variable.
Correction: Substitute that value into an original equation to find the second variable.
Changing only one term when multiplying an equation.
Correction: Multiply every term on both sides by the same number so the equation stays balanced.
Checking the answer in only one equation.
Correction: Substitute the ordered pair into both original equations. Both must be true.
Thinking 0=00=0 means there is no solution.
Correction: It is a true statement and can show that the equations describe the same relationship, giving many solutions.

Lesson summary

Check your understanding

Question 1

For the system x+y=9x+y=9 and x−y=3x-y=3, what is the solution?
  1. (6,3)(6,3)
  2. (3,6)(3,6)
  3. (5,4)(5,4)
  4. (6,6)(6,6)
Show answer and explanation
(6,3)(6,3)
For (6,3)(6,3), the equations give 6+3=96+3=9 and 6−3=36-3=3. The other pairs do not satisfy both equations.

Question 2

In the system y=2x+1y=2x+1 and x+y=10x+y=10, which substitution gives an equation with one variable?
  1. Replace yy in the second equation with 2x+12x+1.
  2. Replace xx in the first equation with 2x+12x+1.
  3. Add yy to both equations.
  4. Replace 2x+12x+1 with 1010.
Show answer and explanation
Replace yy in the second equation with 2x+12x+1.
Since the first equation states that yy equals 2x+12x+1, replace yy in the other equation. This gives x+(2x+1)=10x+(2x+1)=10.

Question 3

A system simplifies to 0=50=5. What does this tell you?
  1. The system has no solution.
  2. The system has one solution, (0,5)(0,5).
  3. The system has many solutions.
  4. The value of each variable is 55.
Show answer and explanation
The system has no solution.
The statement 0=50=5 is false. The original equations cannot both be true for any pair of values.

Key terms

Linear system
A group of linear equations that use the same variables and must be true at the same time.
Solution
The value or values that make every equation in a system true.
Substitution
A method that replaces a variable with an equal expression.
Elimination
A method that combines equations to cancel one variable.
Coefficient
The number multiplying a variable in a term.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 10 Mathematics (MFM2P), study topic G12. It is a study resource, not an official curriculum publication.

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