MPM2D Solving by Substitution: Notes and Practice

Solving by Substitution

Substitution turns two equations in two variables into one equation in one variable: isolate a variable in one equation, replace it in the other, solve, then back-substitute. It gives exact answers, even fractions, and shows when a system has no solution or infinitely many.

Key things to remember

1Substitution: isolate a variable in one equation, substitute into the other equation, solve, back-substitute, check.

2Isolate a variable with coefficient

1

- 1

to avoid fractions.

3Put the substituted expression in brackets:

2 x - 3 y = - 9

with

y = 14 - 3 x

becomes

2 x - 3 (14 - 3 x) = - 9

4If the variables cancel and leave a true statement (such as

4 = 4

), there are infinitely many solutions; a false statement (such as

12 = 5

) means no solution.

5The solution is an ordered pair

(x, y)

: finding one variable is only halfway.

A smoothie stand sold 60 smoothies on Saturday, and it sold twice as many small smoothies as large ones. How many of each? Let $s$ be the number of small smoothies and $l$ the number of large ones. The facts give two equations: $s + l = 60$ and $s = 2 l$ . You could graph both lines, but look at the second equation: it tells you that $s$ is $2 l$ . So in the first equation you can cross out $s$ and write $2 l$ in its place. That leaves $2 l + l = 60$ , an equation with only one variable, which you already know how to solve. This move, replacing a variable with an expression that is equal to it, is called substitution, and it is one of the two main algebra methods for solving linear systems.

s + l = 60 s = 2 l 2 l + l = 60 \Rightarrow 3 l = 60 \Rightarrow l = 20, s = 2 (20) = 40

So the stand sold 40 small and 20 large smoothies. Check both facts: $40 + 20 = 60$ , and $40$ is twice $20$ . Why is it allowed? The two equations must be true at the same time, so at the solution $s$ really does equal $2 l$ . Replacing one by the other cannot change whether the first equation is true. The payoff is that the second variable disappears, and a two-variable problem becomes a one-variable equation.

The steps

Isolate one variable in one of the equations, so it reads $y = \dots$ or $x = \dots$ . Choose a variable whose coefficient is $1$ or $- 1$ if you can.
Substitute that expression into the other equation, inside brackets.
Solve the new equation, which has only one variable.
Back-substitute: put the value you found into the isolated equation from step 1 to find the other variable.
Check the ordered pair in both original equations, then state the solution as $(x, y)$ .

Fig 1Solving 3x + y = 14 and 2x − 3y = −9 by substitution: y = 14 − 3x is placed in brackets in the second equation, giving x = 3, then y = 5, and the check in both equations.

Which variable should you isolate

Look for a coefficient of

1

- 1

. In

3 x + y = 14

the

y

has coefficient

1

, so subtracting

3 x

from both sides gives

y = 14 - 3 x

with no fractions. In

4 x - y = 7

, isolate

y

- y = 7 - 4 x

, then multiply every term by

- 1

to get

y = 4 x - 7

. If no coefficient is

1

- 1

, substitution still works but often creates fractions, and elimination (next lesson) is usually easier.

VideoAn Ontario Grade 10 lesson on isolating a variable, substituting it in brackets into the other equation, and checking the answer.Watch on YouTube

Why the brackets matter

In the figure, $y$ in $2 x - 3 y = - 9$ is replaced by $14 - 3 x$ . The $- 3$ in front of $y$ multiplies the whole expression, so you must write $2 x - 3 (14 - 3 x) = - 9$ . Distributing gives $- 3 \times 14 = - 42$ and $- 3 \times (- 3 x) = + 9 x$ , so the equation becomes $2 x - 42 + 9 x = - 9$ . Without brackets, it is easy to write $2 x - 42 - 3 x$ , and that one sign error makes every later step wrong. Brackets also keep a product clear when you back-substitute: $y = 14 - 3 (3) = 14 - 9 = 5$ , not $14 - 33$ .

When the variable disappears

Sometimes, after you substitute and simplify, both variables cancel and you are left with a statement containing only numbers. That is not an error: it is the algebra telling you which special case you have. Try $y = 3 x - 2$ and $6 x - 2 y = 4$ . Substituting gives $6 x - 2 (3 x - 2) = 4$ , so $6 x - 6 x + 4 = 4$ , which is $4 = 4$ . That is always true, whatever $x$ is, so every point on the line is a solution: the two equations describe the same line and there are infinitely many solutions. Now try $x - 2 y = 4$ and $3 x - 6 y = 5$ . Isolating gives $x = 4 + 2 y$ , and substituting gives $3 (4 + 2 y) - 6 y = 5$ , so $12 + 6 y - 6 y = 5$ , which is $12 = 5$ . That is never true, so no point works: the lines are parallel and there is no solution.

AVariables cancel, true statement

Example: $4 = 4$ or $0 = 0$
Every point on the line works
The equations describe the same line
Infinitely many solutions

BVariables cancel, false statement

Example: $12 = 5$ or $0 = 7$
No point works
The lines are parallel and never meet
No solution

Substitute into the other equation

If you isolate

y = 14 - 3 x

from

3 x + y = 14

and then substitute it back into

3 x + y = 14

, you get

3 x + 14 - 3 x = 14

, so

14 = 14

. That true statement only says the equation agrees with itself; it does not mean infinitely many solutions. The expression must go into the other equation.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Solve by substitution:

y = 2 x - 1

and

3 x + y = 14

Example 2

Solve by substitution:

x + 3 y = 7

and

2 x - 5 y = - 8

Example 3

Solve

y = \frac{1}{2} x - 5

and

3 x + 2 y = 2

algebraically, then verify the solution algebraically and graphically.

Example 4

Solve by substitution:

x - 2 y = 4

and

3 x - 6 y = 5

. Explain the result.

Where marks get lost

Substituting the isolated expression back into the same equation it came from. You get something like

14 = 14

and wrongly conclude there are infinitely many solutions. It must go into the other equation.

Dropping the brackets: replacing

y

2 x - 3 y = - 9

14 - 3 x

and writing

2 x - 42 - 3 x

instead of

2 x - 3 (14 - 3 x) = 2 x - 42 + 9 x

Isolating with a sign error: from

4 x - y = 7

, writing

y = 7 - 4 x

. Subtracting

4 x

gives

- y = 7 - 4 x

, and multiplying every term by

- 1

gives

y = 4 x - 7

Stopping after one variable. "

x = 3

" is not the solution of a system; back-substitute to find

y

and give the ordered pair

(3, 5)

Reading

0 = 0

as "the solution is

(0, 0)

". A true statement with no variables left means infinitely many solutions.

Key things to remember

Worked examples

Where marks get lost

Flashcards 7

Practice