VertexMath · MPM2D · Free worksheet

Systems of Linear Equations Review

Try each problem first. You don’t need to finish everything.

Core path ~55 min Full sheet ~1 h 45 min

Recommended: K1–K5, then one A and one T problem

K = Knowledge & Understanding · A = Application · T = Thinking · Boss = Challenge

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Core Skills (K)

Graphing · Substitution · Elimination · Number of Solutions · Modelling

K1 Is It a Solution?

Consider the system $x+y=5$ and $2x-y=-2$ .

a)
Determine whether $(1,4)$ is a solution of the system.
b)
Determine whether $(4,1)$ is a solution of the system.
c)
A different system is graphed below. Determine its solution.

Stuck? Peek at a hint

Substitute the pair into each equation; a solution must make both true.

K2 Substitution

Solve each system by substitution.

a)
$\begin{cases} y=3x-5 \\ 4x+2y=20 \end{cases}$
b)
$\begin{cases} x=2y+1 \\ 3x-4y=7 \end{cases}$
c)
$\begin{cases} 2x+y=1 \\ 5x-3y=19 \end{cases}$

Stuck? Peek at a hint

Put the substituted expression in brackets so the coefficient multiplies every term. In (c), isolate $y$ first.

K3 Elimination

Solve each system by elimination.

a)
$\begin{cases} 3x+2y=11 \\ 5x-2y=13 \end{cases}$
b)
$\begin{cases} 2x+3y=4 \\ x-2y=9 \end{cases}$
c)
$\begin{cases} 3x+4y=2 \\ 2x+3y=1 \end{cases}$

Stuck? Peek at a hint

Multiply every term of an equation, constant included, so one variable has matching coefficients. Add if their signs are opposite, subtract if they are the same.

K4 How Many Solutions?

Determine the number of solutions of each system without solving it.

a)
$\begin{cases} y=2x+5 \\ y=3x+5 \end{cases}$
b)
$\begin{cases} 6x-4y=8 \\ 3x=2y+4 \end{cases}$
c)
The system graphed below. Tick marks are 1 unit apart.

Stuck? Peek at a hint

Write each equation in the form $y=mx+b$ . Compare slopes first, then $y$ -intercepts.

K5 Words to Equations

Define two variables and write a linear system for each situation. Do not solve.

a)
Two numbers have a sum of $42$ . The larger number is $6$ more than twice the smaller.
b)
A theatre sold $220$ tickets. Student tickets cost $8, adult tickets cost $12, and sales totalled $2160.
c)
A class trip has $3$ times as many students as teachers, and $52$ people in all.

Stuck? Peek at a hint

Say what each variable counts. Test each equation with small numbers: with $2$ teachers, how many students are there?

Real World (A)

Modelling with Systems · Interpreting Solutions

A1 Food Truck Sales

A food truck in Ottawa sells poutine for $9 and hot dogs for $5. On Saturday it sold $64$ items for a total of $440.

a)
Write a linear system that models Saturday's sales.
b)
Solve the system. How many of each item were sold?
c)
On Sunday the owner records $50$ items sold for $500. Solve this new system. What does the result tell you about the record?

Stuck? Peek at a hint

Write one equation that counts items and one that counts dollars.

A2 Climbing Gym Plans

Two climbing gyms in Toronto charge for a season as shown. $A(v)$ and $B(v)$ are the season costs, in dollars, for $v$ visits at Gym A and Gym B.

$v$	$0$	$5$	$10$
$A(v)$	$60$	$100$	$140$
$B(v)$	$0$	$75$	$150$

a)
Write equations for $A(v)$ and $B(v)$ .
b)
Solve the system $C=A(v)$ , $C=B(v)$ . Give $v$ to one decimal place and $C$ to the nearest cent.
c)
Visits come in whole numbers. Which gym is cheaper for $8$ visits, and which for $9$ ? Will the two gyms ever cost exactly the same?

Stuck? Peek at a hint

From the table, find each gym's cost for one more visit and its cost at $0$ visits.

A3 Against the Current

Leah paddles her canoe $12$ km downstream on the Ottawa River in $1.5$ h, then paddles back upstream in $3$ h. Let $c$ be her paddling speed in still water and $r$ the speed of the current, both in km/h.

a)
Write a linear system for $c$ and $r$ .
b)
Solve the system and state both speeds.
c)
In spring the current reaches $7$ km/h. Find Leah's upstream speed at the same paddling speed. Can she paddle the $12$ km upstream?

Stuck? Peek at a hint

Her speed downstream is $c+r$ and upstream is $c-r$ ; each one equals distance divided by time.

Think It Through (T)

Error Analysis · Building Systems

T1 Spot the Flaw

A student solves the system $4x+3y=10$ and $2x-y=10$ :
"Multiply the second equation by $2$ : $4x-2y=10$ . Subtract it from the first: $5y=0$ , so $y=0$ . Then $4x=10$ , so $x=2.5$ . Check: $4(2.5)+3(0)=10$ . The solution is $(2.5,0)$ ."

a)
Show that $(2.5,0)$ is not a solution of the system.
b)
Identify the student's error, then solve the system correctly.
c)
Explain why the student's check worked even though the answer is wrong. What should the check have been?

Stuck? Peek at a hint

Substitute the student's answer into the equation they did not check.

T2 Tune the Constant

Consider the system $2x+ky=6$ and $3x-6y=9$ , where $k$ is a constant.

a)
Find the value of $k$ for which the system has infinitely many solutions.
b)
Keep $3x-6y=9$ . Write a new first equation so that the system has no solution.
c)
A student claims some value of $k$ makes the original system have no solution. Explain why the student is wrong.

Stuck? Peek at a hint

Write $3x-6y=9$ in the form $y=mx+b$ , then match the first equation's slope to it.

Boss Level

Optional challenge

Systems in Disguise

These systems are not linear in $x$ and $y$ , but each becomes linear after a substitution.

a)
Solve $\begin{cases} \dfrac{3}{x}+\dfrac{2}{y}=2 \\[4pt] \dfrac{6}{x}-\dfrac{4}{y}=0 \end{cases}$
b)
Solve $\begin{cases} \dfrac{4}{x-1}+\dfrac{3}{y+2}=5 \\[4pt] \dfrac{2}{x-1}-\dfrac{9}{y+2}=-1 \end{cases}$
c)
The system $\dfrac{1}{x}+\dfrac{1}{y}=2$ and $\dfrac{2}{x}+\dfrac{2}{y}=4$ has infinitely many solutions. Find two of them, and the one nonzero value that $x$ can never take.

Stuck? Peek at a hint

Let $u=\frac{1}{x}$ and $v=\frac{1}{y}$ , solve the linear system in $u$ and $v$ , then convert back.

Answer key

Final answers only. Try first, then check.

Core Skills

a) Yes: $1+4=5$ and $2(1)-4=-2$
b) No: $4+1=5$ , but $2(4)-1=7\ne-2$
c) $(6,4)$

a) $(3,4)$
b) $(5,2)$
c) $(2,-3)$

a) $(3,1)$
b) $(5,-2)$
c) $(2,-1)$

a) One solution (slopes $2$ and $3$ differ)
b) Infinitely many (both are $y=\frac{3}{2}x-2$ )
c) One solution (slopes $\frac{1}{2}$ and $\frac{2}{3}$ differ)

a) $s$ = smaller, $L$ = larger: $s+L=42$ , $L=2s+6$
b) $s$ = student tickets, $a$ = adult tickets: $s+a=220$ , $8s+12a=2160$
c) $s$ = students, $t$ = teachers: $s=3t$ , $s+t=52$

Real World

a) $p+h=64$ , $9p+5h=440$ ( $p$ poutines, $h$ hot dogs)
b) $30$ poutines and $34$ hot dogs
c) $p=62.5$ , $h=-12.5$ : a negative, fractional number of hot dogs is impossible, so the record has an error (even $50$ poutines bring in only $450).

a) $A(v)=60+8v$ , $B(v)=15v$
b) $v=\frac{60}{7}\approx8.6$ , $C\approx\$128.57$
c) $8$ visits: Gym B ($120 vs $124). $9$ visits: Gym A ($132 vs $135). Never: the costs are equal only at $v=\frac{60}{7}$ , which is not a whole number.

a) $1.5(c+r)=12$ and $3(c-r)=12$ , i.e. $c+r=8$ , $c-r=4$
b) $c=6$ km/h, $r=2$ km/h
c) $c-r=6-7=-1$ km/h: she drifts downstream at $1$ km/h, so she never makes it. The time $\frac{12}{-1}=-12$ h has no meaning.

Think It Through

a) $2(2.5)-0=5\ne10$ , so it fails the second equation.
b) The constant was not doubled: $2x-y=10$ becomes $4x-2y=20$ . Then $5y=-10$ , so the solution is $(4,-2)$ .
c) $x$ was found from the first equation, so the point lies on that line by construction. A check must use both equations, especially the one not used to find $x$ .

a) $k=-4$
b) e.g. $x-2y=5$ (any $x-2y=c$ with $c\ne3$ )
c) $(3,0)$ satisfies $2x+ky=6$ for every $k$ , and also $3x-6y=9$ , so the lines always share a point. When the slopes match ( $k=-4$ ), they are the same line.

Boss Level

Boss

a) $(3,2)$
b) $(2,1)$
c) e.g. $(1,1)$ and $\left(2,\frac{2}{3}\right)$ ; $x=\frac{1}{2}$ (it would force $\frac{1}{y}=0$ )

Free cheat sheet

Linear Systems Cheat Sheet

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Unit Summary

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Most testedA school sold $50$ tickets for $480. Adult tickets cost $12 and student tickets $8. How many of each were sold? (card 6)

1Solution of a system

UseA point that satisfies both equations: the point of intersection of the lines. To graph, write each line as $y=mx+b$ .

Example $y=x+1$ and $y=-x+5$ cross at $(2,3)$ , since $3=2+1$ and $3=-2+5$

Trap $(1,6)$ works in $x+y=7$ , but $2(1)-6=-4\ne5$ , so it does not solve $x+y=7$ , $2x-y=5$ .

2Substitution

UseIsolate a variable with coefficient $1$ or $-1$ . Put it into the other equation, in brackets. Solve, then substitute back for the second variable.

Example $y=2x-1$ , $3x+2y=12$ : $3x+2(2x-1)=12$ , $7x=14$ , so $x=2$ , $y=3$ .

Trap $x=y+3$ into $2x+y=12$ is $2(y+3)+y=12$ , so $y=2$ . Without brackets, $2y+3+y=12$ gives $y=3$ .

3Elimination

UseMultiply every term, constants too, until one variable has matching coefficients. Opposite signs: add. Same signs: subtract.

Example $3x+2y=16$ , $x-y=2$ : $\times2$ gives $2x-2y=4$ ; add: $5x=20$ , so $x=4$ , $y=2$ .

TipCheck the answer in both original equations.

Trap $\times2$ on $x-y=2$ gives $2x-2y=4$ , not $2x-2y=2$ : the constant doubles too.

Trap $+3y$ and $-3y$ cancel when you add; subtracting gives $6y$ .

4Number of solutions

UseNone: an inconsistent system. Infinitely many: a dependent system. A false result like $0=6$ : none.

Both lines in $y=mx+b$ form:

different slopes	one: they cross
same slope, different $b$	none: parallel
same slope, same $b$	infinitely many: same line

Trap $y=2x+3$ and $y=-x+3$ share $b=3$ , yet they cross at $(0,3)$ .

TrapBoth variables cancel and you get $0=0$ : infinitely many solutions, not none.

5Equivalent systems

UseMultiplying an equation by a non-zero number, or adding the two equations, gives equivalent systems: they have the same solution.

Example $x+y=5$ , $x-y=1$ and $2x+2y=10$ , $2x=6$ both have solution $(3,2)$ .

6Setting up a problem

UseSay what each variable counts. Write one equation per fact, often one for items and one for value. Answer in a sentence.

Example $50$ tickets, $480, adults $12, students $8: $a+s=50$ , $12a+8s=480$ , so $a=20$ , $s=30$ .

Trap"4 times as many cars as buses" is $c=4b$ , not $4c=b$ : try $b=10$ .

TrapA count like $-2.5$ tickets means a wrong setup or an impossible situation.

7Distance and mixture

$\text{distance}=\text{speed}\times\text{time}$

UseWith a current $c$ : boat speed $b+c$ downstream, $b-c$ upstream. Mixtures: one equation for the volumes, one for the pure part (percent $\times$ volume).

Example $36$ km downstream in $2$ h, back in $3$ h: $b+c=18$ , $b-c=12$ , so $b=15$ km/h, $c=3$ km/h.

TrapSpeed is distance $\div$ time: $36\div2=18$ km/h, not $2\div36$ .

Which method?

A variable is alone or has coefficient

\pm1

Substitution

Coefficients match or are small multiples Elimination

You need only how many solutions Compare slopes

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