Grade by Grade
Analytic Geometry · Analytic Geometry

Solving Linear Systems by Substitution

1.1About 70 minutes
Learning goals
  • I can solve a linear system by substitution.
  • I can clear fractions before solving a system.
  • I can verify a solution algebraically and graphically.
  • I can recognize when substitution shows no solution or infinitely many solutions.

Words to know

substitution
Replacing a variable with an equal expression or value.
isolate
Rearrange an equation so one variable is alone on one side.
back-substitute
Put a value you have found into an equation to find the other variable.
equivalent equations
Equations that have exactly the same solutions, e.g. x/2 + y/3 = 4 and 3x + 2y = 24.
verify
Show that a solution is correct by substituting it into every original equation.

Why we need algebra

In the last lesson you solved systems by graphing. That works well when the lines cross on a grid point. But what if they cross at (2.4, −1.75)? You would only get an estimate. Substitution is an algebraic method that gives the exact solution every time.

The idea is simple: if one equation tells you what y equals, you can replace y with that expression in the other equation. Then you have one equation with one variable, which you already know how to solve.

The method of substitution

  1. Isolate one variable in one equation. Pick a variable with a coefficient of 1 or −1 if you can, so you avoid fractions.
  2. Substitute that expression into the other equation. Use brackets!
  3. Solve the new equation, which has only one variable.
  4. Back-substitute: put that value into the isolated equation to find the other variable.
  5. Check the ordered pair in both original equations, and write the solution as an ordered pair.
1. Isolateone variable2. Substituteinto the other3. Solveone variable4. Back-substitutefind the other5. Checkin bothy = ½x − 53x + 2(½x − 5) = −24x − 10 = −2, x = 2y = ½(2) − 5 = −46 − 8 = −2 ✓
The five steps of substitution, with Example 1 written under each step. Step 2 uses brackets so the 2 multiplies the whole expression ½x − 5.
Worked example

Example 1: One equation is already solved for y

Solve algebraically: y = ½x − 5 and 3x + 2y = −2. Verify algebraically and graphically.

  1. Isolate: the first equation already gives y = ½x − 5.
  2. Substitute into the second equation: 3x + 2(½x − 5) = −2.
  3. Solve: expand: 3x + x − 10 = −2. Collect like terms: 4x − 10 = −2. Add 10 to both sides: 4x = 8. Divide by 4: x = 2.
  4. Back-substitute into y = ½x − 5: y = ½(2) − 5 = 1 − 5 = −4.
  5. Check algebraically. First equation: ½(2) − 5 = −4 ✓. Second: 3(2) + 2(−4) = 6 − 8 = −2 ✓.
  6. Check graphically: graph both lines (see the graph below). They cross at (2, −4).

The solution is (2, −4).

−2−11234567−7−6−5−4−3−2−1120xy(2, −4)y = ½x − 53x + 2y = −2
A graph verifies Example 1: the solid line y = ½x − 5 and the dashed line 3x + 2y = −2 cross at (2, −4).
Worked example

Example 2: Isolate first

Solve: x + 3y = 7 and 2x − 5y = −8.

  1. Isolate: x has a coefficient of 1 in the first equation. Subtract 3y: x = 7 − 3y.
  2. Substitute into the second equation: 2(7 − 3y) − 5y = −8.
  3. Solve: 14 − 6y − 5y = −8, so 14 − 11y = −8. Subtract 14: −11y = −22. Divide by −11: y = 2.
  4. Back-substitute: x = 7 − 3(2) = 7 − 6 = 1.
  5. Check: 1 + 3(2) = 7 ✓ and 2(1) − 5(2) = 2 − 10 = −8 ✓.

The solution is (1, 2).

Brackets save marks

The most common substitution error is forgetting brackets. In Example 2, 2(7 − 3y) means 14 − 6y. Writing 2 × 7 − 3y = 14 − 3y is wrong, because the 2 must multiply both terms. Also watch the signs when you subtract: −(4 − 2y) = −4 + 2y.

Worked example

Example 3: Equations with fractions

Solve: x/2 + y/3 = 4 and x − y = 3.

  1. Clear the fractions first. The lowest common denominator of 2 and 3 is 6. Multiply every term of the first equation by 6: 3x + 2y = 24.
  2. Isolate x in the second equation: x = y + 3.
  3. Substitute: 3(y + 3) + 2y = 24, so 3y + 9 + 2y = 24, so 5y = 15 and y = 3.
  4. Back-substitute: x = 3 + 3 = 6.
  5. Check in the original equations: 6/2 + 3/3 = 3 + 1 = 4 ✓ and 6 − 3 = 3 ✓.

The solution is (6, 3).

When substitution gives a strange result

Sometimes the variable disappears completely.

  • If you get a false statement such as 0 = 8 or −2 = 6, the system has no solution (the lines are parallel).
  • If you get a true statement such as 0 = 0 or −4 = −4, the system has infinitely many solutions (the equations describe the same line).
Worked example

Example 4: No solution or infinitely many?

Solve each system. (a) y = 2x + 1 and 4x − 2y = 6. (b) y = 3x + 2 and 6x − 2y = −4.

  1. (a) Substitute: 4x − 2(2x + 1) = 6. Expand: 4x − 4x − 2 = 6, which gives −2 = 6. This is false for every x.
  2. So (a) has no solution. Rearranging 4x − 2y = 6 gives y = 2x − 3: same slope as y = 2x + 1, different intercept, so the lines are parallel.
  3. (b) Substitute: 6x − 2(3x + 2) = −4. Expand: 6x − 6x − 4 = −4, which gives −4 = −4. This is true for every x.
  4. So (b) has infinitely many solutions. Rearranging 6x − 2y = −4 gives y = 3x + 2, the same line.

(a) No solution (parallel lines). (b) Infinitely many solutions: every point on y = 3x + 2.

Just for fun

Why are substitute teachers so good at linear systems?

Because they know exactly how to take someone else's place.

Hands-on activity

Mystery coin envelope

You need: a handful of loonies ($1 coins) and toonies ($2 coins), an envelope or a cup, paper and a pencil, and a grown-up.

  1. The grown-up secretly puts some loonies and toonies in the envelope (try for 6 to 12 coins in total).
  2. The grown-up tells you only two things: the number of coins and the total value. For example: "9 coins worth $14."
  3. Let x be the number of loonies and y the number of toonies. Write the system: x + y = 9 (coins) and x + 2y = 14 (dollars).
  4. Solve by substitution: x = 9 − y, so 9 − y + 2y = 14, so y = 5 and x = 4.
  5. Open the envelope and count. Were you right? Swap roles and make a puzzle for the grown-up.

What to look for: the two equations count different things (coins and dollars). If the coins do not match your answer, check the equations first, then the algebra.

Your turn

1

Solve by substitution: y = x + 4 and 2x + y = 13.

2

Solve by substitution: x = 2y − 1 and 3x − 4y = 5.

3

Solve: x + y = 10 and 3x − 2y = 5.

4

Solve: 2x − y = 7 and 5x + 3y = 1.

5

Solve: y = −2x + 3 and 6x + 3y = 9. What does your result tell you?

6

Solve: x/4 + y/2 = 3 and x − y = 3.

More practice: warm-up

1

Isolate y in 3x + y = 8.

2

Isolate x in x − 4y = 6.

3

Substitute y = 2x into x + y = 12 and solve for x.

4

Expand 3(5 − 2y).

5

Solving a system gives 0 = 7. How many solutions are there?

More practice: on your own

1

Solve: y = 3x − 5 and y = −x + 7.

2

Solve: 4x + y = 2 and 3x − 2y = 18.

3

Solve: x − 3y = −1 and 2x + 5y = 20.

4

Solve: y = 0.5x + 2 and 3x − 4y = −6.

5

Ali says the solution of y = x − 2 and 2x + 3y = 9 is (1, −1). Check his answer. If it is wrong, find the correct solution.

Challenge

1

Solve: (x + 1)/3 − y/2 = 1 and 2x − y = 5.

2

For what value of k does the system y = kx + 3 and 4x − 2y = 1 have no solution? Explain using substitution.

3

The sum of two numbers is 42. The larger number is 6 less than twice the smaller. Write a system and solve it by substitution.

Check your understanding

Quiz

1.

In the substitution method, which variable is usually best to isolate first?

2.

Substitute y = x − 3 into 2x + y = 9. Which equation results?

3.

What is the solution of y = 2x and x + y = 15?

4.

Which expansion of 3(4 − 2y) is correct?

5.

Substitution produces the statement 5 = 5. What does this mean?

6.

Solve x = 3y and 2x − y = 10.

For parents and teachers

Big idea: substitution replaces one variable with an equivalent expression, turning two equations in two variables into one equation in one variable. This lesson teaches expectation 1.1 (solving systems by substitution). The worked Example 1 is the sample problem from the Ontario curriculum.

How to help: Ask your teen to say each step out loud: "I isolated..., I substituted..., I solved..., I back-substituted..., I checked...". Have them check every answer in both original equations. The coin-envelope activity is a fun way to practise.

Common difficulties:

  • Forgetting brackets when substituting, e.g. writing 2 × 7 − 3y instead of 2(7 − 3y).
  • Sign errors when subtracting a bracket.
  • Substituting back into the same equation that was used to isolate the variable (this always gives a true statement and does not find a value).
  • Stopping after finding one variable. A solution is an ordered pair.
  • Being confused when the variable disappears: a false statement means no solution; a true statement means infinitely many.