LemonmathSelf-study sheet

Grade 9 · Lesson 1.2

Solving simultaneous linear equations

Name: Date: Time spent: min

Goals

Solve simultaneous linear equations by the substitution method.
Solve them by the elimination method, including when you first multiply both sides by a number.
Recognise equations with no solution or infinitely many solutions, and check answers on a calculator.

Remember

The common idea: transform until you have an equation in one unknown.
Substitution: write one unknown in terms of the other and substitute into the other equation.
Elimination: make the coefficients of one unknown equal (then subtract) or opposite (then add); multiply both sides when needed.
If you reach $0x=c$ with $c\neq0$ : no solution. If you reach $0x=0$ : infinitely many solutions.

Exercises

Exercise 1

Solve $\begin{cases} x=2y\\ x+y=9 \end{cases}$ by substitution.

Exercise 2

Solve $\begin{cases} 2x-y=3\\ x+2y=4 \end{cases}$ by substitution.

Exercise 3

Solve $\begin{cases} x+y=7\\ x-y=1 \end{cases}$ by elimination.

Exercise 4

Solve $\begin{cases} 3x+4y=5\\ 2x-3y=-8 \end{cases}$

Exercise 5

How many solutions does each pair of equations have? a) $\begin{cases} 2x-4y=6\\ x-2y=3 \end{cases}$ b) $\begin{cases} x-2y=1\\ -2x+4y=5 \end{cases}$

Exercise 6

Let $x$ and $y$ (thousand đồng) be the price of an exercise book and of a pen. $3$ books and $2$ pens cost $34$ thousand đồng; $2$ books and $3$ pens cost $31$ thousand đồng. Solve $\begin{cases} 3x+2y=34\\ 2x+3y=31 \end{cases}$ to find each price.

Exercise 7 Review

Find $a$ and $b$ so that the graph of $y=ax+b$ passes through the points $A(1, 3)$ and $B(2, 5)$ .

Exercise 8

Consider $\begin{cases} x+2y=5\\ 3x-y=1 \end{cases}$

a) Solve the equations by substitution.

b) Solve them by elimination and compare the results.

c) Show that the line $y=x+1$ passes through the point where the lines $x+2y=5$ and $3x-y=1$ meet.

Exercise 9 Challenge

Solve $\begin{cases} 2(x+y)+3(x-y)=13\\ 3(x+y)-2(x-y)=13 \end{cases}$

Mini test

15 minutes, without looking back at the lesson. Part I: circle one letter; Part II: write T or F for each statement; Part III: write the answer in the box.

Part I. Multiple choice

Question 1. The solution of

\begin{cases} x+y=5\\ x-y=3 \end{cases}

(4, 1)

.B.

(1, 4)

.C.

(3, 2)

.D.

(8, -3)

Question 2. To eliminate

y

from

\begin{cases} 2x+3y=7\\ 5x-2y=8 \end{cases}

, you should

A. multiply the first equation by

2

and the second by

3

, then add.B. multiply the first equation by

3

and the second by

2

, then add.C. multiply the first equation by

2

and the second by

3

, then subtract.D. multiply the first equation by

5

and the second by

2

, then subtract.

Question 3. The simultaneous equations

\begin{cases} x-2y=3\\ -2x+4y=1 \end{cases}

A. have exactly one solution.B. have no solution.C. have infinitely many solutions.D. have the solution

(3, 0)

Question 4.

2

bánh mì (Vietnamese baguettes) and

1

carton of milk cost

31

thousand đồng;

1

bánh mì and

2

cartons of milk cost

26

thousand đồng. The price of one bánh mì is

12

thousand đồng.B.

7

thousand đồng.C.

19

thousand đồng.D.

5

thousand đồng.

Part II. True or false

Question 1. Consider

\begin{cases} 3x+y=7\\ x-y=1 \end{cases}

a) Adding the two equations gives $4x=8$ .	T / F
b) The solution is $(2, 1)$ .	T / F
c) From the second equation, $y=x+1$ .	T / F
d) The solution satisfies $x+y=3$ .	T / F

Part III. Short answer

Question 1. Solve

\begin{cases} 2x+y=1\\ 3x-y=9 \end{cases}

and find

x+y

Question 2. The graph of

y=ax+b

passes through

(1, 4)

and

(3, 0)

. Find

b

Score: ........ / 10

Hints (only when stuck)

Exercise 1.

What you need: the substitution method.

Steps:

Substitute $x=2y$ into the second equation.
Find $y$ , then $x$ .

Exercise 2.

What you need: it is easiest to make an unknown with coefficient $1$ or $-1$ the subject.

Steps:

From the first equation: $y=2x-3$ .
Substitute into the second equation and solve.

Exercise 3.

What you need: the coefficients of $y$ are opposite, so add the equations.

Steps:

Add to find $x$ .
Substitute into $x+y=7$ to find $y$ .

Exercise 4.

What you need: multiply both sides so that the coefficients of one unknown are equal.

Steps:

Multiply the first equation by $2$ and the second by $3$ .
Subtract to remove $x$ .

Exercise 5.

What you need: after substituting, $0x=0$ means infinitely many solutions and $0x=c$ ( $c\neq0$ ) means no solution.

Steps:

Write $x=2y+\dots$ from the equation in which $x$ has coefficient $1$ .
Substitute into the other equation.

Exercise 6.

What you need: the elimination method (or a trick: add the equations, then subtract them).

Steps:

Add: $5x+5y=65$ , so $x+y=13$ .
Subtract: $x-y=3$ .
Solve the new pair.

Exercise 7.

What you need: the point $(x_0, y_0)$ is on the graph when $y_0=ax_0+b$ (Grade 8). Each point gives one equation in the unknowns $a$ , $b$ .

Steps:

Substitute the coordinates of $A$ and $B$ to get two equations.
Subtract them to find $a$ .

Exercise 8.

What you need: the solution of the simultaneous equations gives the coordinates of the point where the two lines meet.

Steps:

a) Write $x=5-2y$ .
b) Multiply the second equation by $2$ and add.
c) Substitute the coordinates of that point into $y=x+1$ .

Exercise 9.

What you need: you can put $u=x+y$ , $v=x-y$ to get simpler equations, or expand the brackets and simplify.

Steps:

Solve $2u+3v=13$ , $3u-2v=13$ .
From $x+y=u$ , $x-y=v$ find $x$ and $y$ .

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