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G4 · Solve two-variable linear systems by substitution or elimination

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

Ontario Grade 10 Mathematics

Analytic Geometry

Use substitution or elimination to find the pair that makes both equations true.

A linear equation with two variables can have many solutions. For example, different pairs of numbers may make an equation true. A system is a group of equations that must be true at the same time. In this lesson, you will find the ordered pair that works in both equations. You will use substitution or elimination, then check your answer in the original equations.

What you will learn

1. Grade 9 bridge: equations and ordered pairs

An equation says that two expressions have the same value. You can keep an equation balanced by doing the same operation to both sides. For example, if you add the same number to each side, the equality stays true.
A variable is a letter that stands for a number. In a two-variable equation, such as x+y=10x+y=10, both xx and yy stand for numbers. An ordered pair, written (x,y)(x,y), gives a value for each variable in that order.
A solution to a system is one ordered pair that makes every equation in the system true. The two equations might describe two number relationships. The solution is the pair that satisfies both relationships.

2. What substitution and elimination do

Substitution means replacing a variable with an equal expression. First, isolate one variable in one equation. Isolate means to get that variable by itself. Then replace that variable in the other equation with the expression that equals it. This leaves an equation with one variable.
For example, if one equation tells you y=8−xy=8-x, then every place you see yy in the other equation can be replaced with 8−x8-x. The replacement is valid because the two expressions have the same value.
Elimination means combining equations so that one variable disappears. To eliminate a variable, make its coefficients opposites, then add the equations. A coefficient is the number multiplying a variable. For example, in 3x+2y=143x+2y=14, the coefficient of yy is 22.
If the coefficients of a variable are already opposites, add the equations directly. If they are not, multiply one or both entire equations by a number first. Multiply every term on both sides so the equation remains balanced. After one variable disappears, solve the remaining equation, then substitute the value back to find the other variable.
Choose substitution when a variable is already alone or can be isolated easily. Choose elimination when a variable’s coefficients are equal or can be made opposites without much work. Both methods aim to reduce the system to one equation with one variable.

3. Guided example: use both methods and check

A school club sells adult tickets for CAD 8 and student tickets for CAD 5. One order contains 7 tickets and costs CAD 47. Let aa be the number of adult tickets and ss the number of student tickets. The total number of tickets gives one equation. The total cost gives another.
We can solve this system by substitution. The ticket-count equation makes it easy to express one variable in terms of the other. We can also use elimination because the ss terms have opposite coefficients after we write the equations in standard order.
In the steps below, both methods lead to the same pair. The final check puts the pair into the original equations, not just the rearranged equations.
a+s=7,8a+5s=47a+s=7,\quad 8a+5s=47

4. Independent practice and a reliable check

Try these systems on your own. For the first one, use substitution because one variable is already isolated. For the second, use elimination because the yy coefficients are opposites. Write each solution as an ordered pair and check it in both equations.
System A: y=2x+1y=2x+1 and x+y=10x+y=10. System B: 3x+2y=163x+2y=16 and x−2y=0x-2y=0. These are practice prompts, not additional worked examples.
A check is simple: replace each variable in each original equation with your proposed value, then calculate both sides. The left side and right side must be equal in both equations. If either equation fails, review the signs, arithmetic, and substitution.

What each equation tells us

ConditionEquationMeaning
Ticket counta+s=7a+s=7The two ticket types total 7.
Ticket cost8a+5s=478a+5s=47The total cost is CAD 47.
Solution(a,s)=(4,3)(a,s)=(4,3)Both conditions are true.

Worked example

Ticket sales system

Solve a+s=7a+s=7 and 8a+5s=478a+5s=47. Here, aa is the number of adult tickets and ss is the number of student tickets.
  1. Isolate a variable
    The first equation is easy to rearrange. Subtract aa from both sides to express ss by itself.
    s=7−as=7-a
  2. Substitute
    Replace ss in the cost equation with 7−a7-a. This is valid because the first equation says these expressions have the same value.
    8a+5(7−a)=478a+5(7-a)=47
  3. Solve for a
    Distribute 55, combine like terms, and then isolate aa using the same operation on both sides.
    8a+35−5a=47⇒3a=12⇒a=48a+35-5a=47\quad\Rightarrow\quad 3a=12\quad\Rightarrow\quad a=4
  4. Find s
    Replace aa with 44 in the equation s=7−as=7-a. This gives the number of student tickets.
    s=7−4=3s=7-4=3
  5. Verify by elimination
    The same system can be written with the terms in matching order. Add the equations to cancel ss. The result agrees with a=4a=4.
    a+s=78a+5s=478a+5s=47\begin{aligned}a+s&=7\\8a+5s&=47\\8a+5s&=47\end{aligned}
Answer: There are 4 adult tickets and 3 student tickets, so the solution is (a,s)=(4,3)(a,s)=(4,3).
Check: The count is 4+3=74+3=7. The cost is 8(4)+5(3)=32+15=478(4)+5(3)=32+15=47. Both original equations are true.

Common mistakes and how to avoid them

Changing a sign while rearranging an equation.
Correction: Use the same operation on both sides. For example, subtract aa from both sides of a+s=7a+s=7 to get s=7−as=7-a.
Multiplying only one term when preparing for elimination.
Correction: Multiply every term on both sides of the equation by the same number.
Stopping after finding only one variable.
Correction: Substitute that value into an original equation to find the second variable.
Checking the answer in just one equation.
Correction: Substitute the ordered pair into both original equations. A solution must make both true.

Lesson summary

Check your understanding

Question 1

Which ordered pair solves x+y=9x+y=9 and x−y=3x-y=3?
  1. (6,3)(6,3)
  2. (3,6)(3,6)
  3. (9,3)(9,3)
  4. (5,4)(5,4)
Show answer and explanation
(6,3)(6,3)
For (6,3)(6,3), the sum is 99 and the difference is 33. The other pairs do not make both equations true.

Question 2

In the system y=4−xy=4-x and 2x+y=112x+y=11, what should replace yy in the second equation when using substitution?
  1. 4−x4-x
  2. 4+x4+x
  3. 11−2x11-2x
  4. x−4x-4
Show answer and explanation
4−x4-x
The first equation states that yy equals 4−x4-x, so that expression replaces yy.

Question 3

What is the main purpose of adding two equations in elimination?
  1. To make one variable cancel
  2. To make both variables disappear
  3. To change the solution
  4. To turn the equations into inequalities
Show answer and explanation
To make one variable cancel
When the coefficients of one variable are opposites, adding the equations makes that variable’s total coefficient zero.

Key terms

System
A group of equations that must be true at the same time.
Ordered pair
A pair of values written in order, usually (x,y)(x,y), that gives values for two variables.
Substitution
Replacing a variable with an expression that has the same value.
Elimination
Combining equations so that one variable cancels.
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 (MPM2D), study topic G4. It is a study resource, not an official curriculum publication.

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