Linear equations and simultaneous linear equations in two unknowns

Lan has $60$ thousand đồng to spend on school supplies. Each pen costs $5$ thousand đồng, each exercise book costs $8$ thousand đồng, and Lan wants to spend all of it, with nothing left over. How many pens and how many books can she buy? And what if she wants to buy exactly $9$ items?

The problem has two numbers we do not know: the number of pens and the number of books. We need a new kind of equation, one with two unknowns.

Linear equations in two unknowns

Explore: ways to spend all the money

Let $x$ be the number of pens and $y$ the number of books. The total cost is $5x+8y$ (thousand đồng), so "spending all the money" means $5x+8y=60$ . We try a few values of $x$ and compute $y=\dfrac{60-5x}{8}$ :

$x$	$0$	$4$	$8$	$12$
$y$	$7.5$	$5$	$2.5$	$0$

Each value of $x$ gives one value of $y$ , so a great many pairs fit. But the numbers of pens and books must be whole numbers, so there are only two ways to shop: $4$ pens and $5$ books, or $12$ pens.

Showing the solution set as a line

A linear equation in two unknowns always has infinitely many solutions. Each solution $(x_0, y_0)$ is a point in the coordinate plane, and all these points lie on one straight line.

If $b\neq 0$ : we write $y=-\dfrac{a}{b}x+\dfrac{c}{b}$ . This is the graph of a linear function (or a constant function when $a=0$ ), which you met in Grade 8.
If $b=0$ (then $a\neq0$ ): the equation becomes $x=\dfrac{c}{a}$ , a line parallel to the $y$ -axis (or the $y$ -axis itself when $c=0$ ).

Example Drawing the line of solutions

Show the solution set of each equation on the same coordinate plane: a) $x+2y=4$ ; b) $0x+y=3$ ; c) $2x+0y=-2$ .

Solution. We rewrite each equation in the form $y=\dots$ or $x=\dots$ :

$\begin{aligned} \text{a) }& x+2y=4\iff y=-\tfrac12x+2;\\ \text{b) }& 0x+y=3\iff y=3;\\ \text{c) }& 2x+0y=-2\iff x=-1. \end{aligned}$

So a) is the line through $(0, 2)$ and $(4, 0)$ ; b) is the line parallel to the $x$ -axis that crosses the $y$ -axis at $(0, 3)$ ; c) is the line parallel to the $y$ -axis that crosses the $x$ -axis at $(-1, 0)$ .

Three lines showing solution sets

Simultaneous linear equations

Back to Lan's problem. If she wants to buy exactly $9$ items, then $x$ and $y$ must satisfy two equations at the same time: $5x+8y=60$ and $x+y=9$ .

Geometrically, the solution gives the coordinates of the point where the two lines meet.

Now we answer the second question from the start.

Key terms

English	Tiếng Việt
linear equation in two unknowns	phương trình bậc nhất hai ẩn
unknown	ẩn
coefficient	hệ số
solution	nghiệm
solution set	tập nghiệm
simultaneous linear equations in two unknowns	hệ hai phương trình bậc nhất hai ẩn
solution of the simultaneous equations	nghiệm của hệ
coordinate plane	mặt phẳng toạ độ

Summary

Linear equation in two unknowns: $ax+by=c$ with $a\neq0$ or $b\neq0$ .
$(x_0, y_0)$ is a solution when $ax_0+by_0=c$ ; the equation has infinitely many solutions.
The solution set is a straight line: $y=-\dfrac{a}{b}x+\dfrac{c}{b}$ if $b\neq0$ ; $x=\dfrac{c}{a}$ if $b=0$ .
A solution of two simultaneous equations is a pair that satisfies both equations, that is, the point where the two lines meet.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Which of these are linear equations in two unknowns? a) $3x-y=2$ ; b) $0x+5y=-10$ ; c) $x+y^2=4$ ; d) $0x+0y=7$ .

What you need and how to start

What you need: $ax+by=c$ is a linear equation in two unknowns when $a\neq0$ or $b\neq0$ , and $x$ , $y$ appear with no power other than $1$ .

Steps:

Read off the coefficients of $x$ and $y$ in each equation.
Rule out the equation with $y^2$ and the one where both coefficients are $0$ .

Show solution

a) $a=3$ , $b=-1$ : the right form. b) $a=0$ , $b=5\neq0$ : the right form. c) contains $y^2$ : no. d) $a=b=0$ : no.

Answer. a) and b).

Exercise 2

Which of the pairs $(2, 1)$ , $(0, 4)$ , $(4, -2)$ , $(1, 3)$ are solutions of the equation $3x+2y=8$ ?

What you need and how to start

What you need: $(x_0, y_0)$ is a solution when $3x_0+2y_0=8$ .

Steps:

Substitute each pair into $3x+2y$ .
Compare the result with $8$ .

Show solution

$3\cdot2+2\cdot1=8$ ; $3\cdot0+2\cdot4=8$ ; $3\cdot4+2\cdot(-2)=8$ ; $3\cdot1+2\cdot3=9\neq8$ .

Answer. $(2, 1)$ , $(0, 4)$ and $(4, -2)$ .

Exercise 3

Find $x$ so that $(x, -1)$ is a solution of the equation $4x+3y=9$ .

What you need and how to start

What you need: substituting the known value turns the equation into a linear equation in one unknown.

Steps:

Substitute $y=-1$ .
Solve the equation in one unknown that you get.

Show solution

$4x-3=9\iff 4x=12\iff x=3$ .

Answer. $x=3$ .

Exercise 4

Show the solution set of each equation: a) $2x-y=1$ ; b) $0x+2y=6$ .

What you need and how to start

What you need: when $b\neq0$ , write $y$ in terms of $x$ and draw the line through two points.

Steps:

Rewrite in the form $y=\dots$ .
Find two points on the line (for example with $x=0$ and $x=1$ ).

Show solution

a) $y=2x-1$ : the line through $(0, -1)$ and $(1, 1)$ . b) $2y=6\iff y=3$ : the line parallel to the $x$ -axis that crosses the $y$ -axis at $(0, 3)$ .

Answer. a) The line $y=2x-1$ ; b) the line $y=3$ .

Exercise 5

A group of $7$ people go to the cinema. A student ticket costs $45$ thousand đồng, an adult ticket costs $70$ thousand đồng, and the group pays $365$ thousand đồng. Let $x$ be the number of students and $y$ the number of adults. Form the simultaneous equations, then say which of $(5, 2)$ and $(4, 3)$ is a solution.

What you need and how to start

What you need: each condition in the problem gives one equation; a solution satisfies both.

Steps:

The number of people gives $x+y=7$ .
The cost gives $45x+70y=365$ .
Substitute each pair into both equations.

Show solution

The equations: $x+y=7$ and $45x+70y=365$ . For $(5, 2)$ : $5+2=7$ and $225+140=365$ . For $(4, 3)$ : $4+3=7$ but $180+210=390\neq365$ .

Answer. $(5, 2)$ : the group has $5$ students and $2$ adults.

Exercise 6 Review

Write the equation $3x+y=5$ in the form $y=ax+b$ . What is the gradient of the line of solutions, and where does it cross the $y$ -axis?

What you need and how to start

What you need: the line $y=ax+b$ has gradient $a$ and crosses the $y$ -axis at $(0, b)$ (Grade 8).

Steps:

Move $3x$ to the right-hand side.
Read off $a$ and $b$ .

Show solution

$3x+y=5\iff y=-3x+5$ , so $a=-3$ and $b=5$ .

Answer. Gradient $-3$ ; it crosses the $y$ -axis at $(0, 5)$ .

Exercise 7

Consider the equation $x-2y=4$ .

a) Find the solution with $x=0$ and the solution with $y=0$ .

b) Draw the line showing the solution set of the equation.

c) Of the two solutions $(6, 1)$ and $(2, -1)$ of the equation, which one is also a solution of $2x+y=3$ ?

What you need and how to start

What you need: two solutions give two points of the line; a common solution of two equations solves the simultaneous equations.

Steps:

Put $x=0$ , then $y=0$ .
Draw the line through the two points you found.
Substitute each pair into $2x+y$ .

Show solution

a) $x=0$ : $-2y=4\iff y=-2$ . $y=0$ : $x=4$ . The two solutions are $(0, -2)$ and $(4, 0)$ .

b) The line through $(0, -2)$ and $(4, 0)$ , that is $y=\dfrac12x-2$ .

c) $2\cdot6+1=13\neq3$ ; $2\cdot2+(-1)=3$ .

Answer. a) $(0, -2)$ , $(4, 0)$ ; b) the line through those two points; c) $(2, -1)$ .

Exercise 8 Challenge

A shop sells stickers at $3$ thousand đồng a sheet and at $5$ thousand đồng a sheet. Hùng wants at least one sheet of each kind and wants to spend exactly $40$ thousand đồng. Find all the ways he can do it.

What you need and how to start

What you need: a way of buying is a solution $(x, y)$ of $3x+5y=40$ with $x$ , $y$ positive whole numbers.

Steps:

Write $5y=40-3x$ ; since $40$ and $5y$ are multiples of $5$ , so is $3x$ .
So $x$ is a multiple of $5$ ; try $x=5, 10, 15$ .

Show solution

$x$ must be a multiple of $5$ with $3x<40$ , so $x\in\{5, 10\}$ . For $x=5$ : $5y=25$ , $y=5$ . For $x=10$ : $5y=10$ , $y=2$ . (For $x=15$ , $3x=45>40$ .)

Answer. Two ways: $5$ sheets at $3$ thousand and $5$ sheets at $5$ thousand, or $10$ sheets at $3$ thousand and $2$ sheets at $5$ thousand.

a) $(3, 1)$ is a solution of the equation.	True	False
b) The equation has exactly one solution.	True	False
c) The line showing the solution set passes through the point $(0, 7)$ .	True	False
d) $(3, 1)$ is also a solution of the simultaneous equations $\begin{cases} 2x+y=7\\ x-y=1 \end{cases}$ .	True	False

Linear equations in two unknowns

Explore: ways to spend all the money

Showing the solution set as a line

Simultaneous linear equations

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer