Linear inequalities in two variables | Grade 10

You have $300$ units of pocket money for the cinema. A ticket for an adult costs $60$ and a ticket for a child costs $30$ . How many tickets of each kind can you buy? There are many answers, and each answer is a pair of numbers. A single inequality with two unknowns describes all these pairs at once, and a picture on the coordinate plane shows them nicely.

What is a linear inequality in two variables?

"Ordered" means that the order matters: $(1, 2)$ and $(2, 1)$ are different pairs.

There are infinitely many solutions, so we cannot list them. We draw them.

The solution region

Take the inequality $x+y\le 3$ . The equation $x+y=3$ is a straight line. It cuts the plane into two half-planes. On one side we always have $x+y<3$ , on the other side $x+y>3$ . So the solutions of $x+y\le 3$ form one half-plane together with the line.

The region

x+y\le 3

How to draw it: the test-point method

The region

x+2y<4

The region

3x-y\ge 3

Modelling real situations

Example Cinema tickets

Tickets cost $60$ units for an adult and $30$ units for a child. We have at most $300$ units. Let $x$ be the number of adult tickets and $y$ the number of child tickets. (a) Write an inequality. (b) Can we buy $3$ adult and $4$ child tickets? And $4$ adult and $3$ child?

Solution.

(a) The total cost is $60x+30y$ , which must be at most $300$ : $60x+30y\le 300$ . Dividing by $30$ gives $2x+y\le 10$ . Also $x\ge 0$ , $y\ge 0$ are whole numbers.
(b) For $(3, 4)$ : $2\cdot 3+4=10\le 10$ is true, so we can buy them (the cost is exactly $300$ ).
For $(4, 3)$ : $2\cdot 4+3=11\le 10$ is false, so we cannot (the cost would be $330$ ).

Possible ticket numbers

The real solutions are only the points of the region with whole-number coordinates, because we cannot buy half a ticket.

Key terms

English	Tiếng Việt
linear inequality in two variables	bất phương trình bậc nhất hai ẩn
solution (ordered pair)	nghiệm (cặp số)
boundary line	đường thẳng biên (bờ)
half-plane	nửa mặt phẳng
solution region	miền nghiệm
test point	điểm kiểm tra
solid line	nét liền
dashed line	nét đứt

Summary

A solution of $ax+by\le c$ is an ordered pair $(x_0, y_0)$ that makes it true; there are infinitely many.
The line $ax+by=c$ splits the plane into two half-planes; the solution region is one of them (plus the line when the sign is $\le$ or $\ge$ ).
Use a solid line for $\le,\ge$ and a dashed line for $<,>$ .
Test a point not on the line (usually the origin) to choose the side to shade.
In real problems, translate the limit into an inequality and keep only the points that make sense (whole numbers, not negative).

Practice

Which of $(2, -1)$ , $(0, 3)$ , $(6, 1)$ are solutions of $x-2y\ge 4$ ?
Which of the following are linear inequalities in two variables? (a) $2x-y<1$ , (b) $x^2+y\ge 0$ , (c) $3x>5$ , (d) $x+y+z\le 1$ .
For each inequality, say whether the boundary line is solid or dashed: (a) $y>2x-1$ , (b) $x+y\le 3$ , (c) $2x-3y\ge 6$ , (d) $x<-1$ .
Describe the solution region of $2x-y\le 4$ : give two points of the boundary line, its type, the test point and the side to shade.
Describe the solution region of $x-y\ge 0$ . (The line passes through the origin, so choose another test point.)
Which of the points $A(1, 1)$ , $B(-2, 3)$ , $C(4, -1)$ satisfy $3x+2y>5$ ?
A phone plan costs $2000$ per minute of calls and $500$ per text message. Mai has at most $40000$ . Let $x$ be the minutes and $y$ the messages. Write an inequality, then decide whether she can use $15$ minutes and $20$ messages, and whether $18$ minutes and $10$ messages.
For which values of $m$ is the pair $(1, m)$ a solution of $2x-y>3$ ?
Challenge. Student tickets cost $20$ units and adult tickets cost $50$ units. A group has at most $200$ units and must buy at least $2$ adult tickets. Write the inequalities, and find the largest total number of people in the group.
Challenge. Find an inequality of the form $ax+by\le c$ whose boundary line passes through $(0, 2)$ and $(3, 0)$ and whose solution region contains the origin.

Show solutions

Compute $x-2y$ . $(2, -1)$ : $2+2=4\ge 4$ , yes. $(0, 3)$ : $0-6=-6\ge 4$ , no. $(6, 1)$ : $6-2=4\ge 4$ , yes.
(a) Yes. (b) No, $x^2$ is not linear. (c) Yes, it has the form $ax+by>c$ with $a=3$ , $b=0$ . (d) No, it has three variables.
(a) Dashed ( $>$ ). (b) Solid ( $\le$ ). (c) Solid ( $\ge$ ). (d) Dashed ( $<$ ); the line is the vertical line $x=-1$ .
Boundary $2x-y=4$ passes through $(2, 0)$ and $(0, -4)$ . The sign is $\le$ , so it is solid. Test $(0, 0)$ : $0\le 4$ is true, so shade the side containing the origin.
Boundary $x-y=0$ (the line $y=x$ ), solid. Test point $(1, 0)$ : $1-0=1\ge 0$ is true. Shade the side containing $(1, 0)$ , that is, the half-plane below the line $y=x$ .
$A$ : $3+2=5>5$ false. $B$ : $-6+6=0>5$ false. $C$ : $12-2=10>5$ true. Only $C$ .
The cost is $2000x+500y\le 40000$ ; dividing by $500$ : $4x+y\le 80$ . For $(15, 20)$ : $60+20=80\le 80$ , true, so she can. For $(18, 10)$ : $72+10=82\le 80$ , false, so she cannot.
Substitute: $2\cdot 1-m>3$ , so $-m>1$ , so $m<-1$ .
Let $x$ be the student tickets and $y$ the adult tickets. Then $20x+50y\le 200$ , that is $2x+5y\le 20$ , with $y\ge 2$ and $x\ge 0$ whole numbers. For $y=2$ : $2x\le 10$ , so $x\le 5$ and the total is at most $7$ . For $y=3$ : $2x\le 5$ , so $x\le 2$ , total $5$ . For $y=4$ : $x=0$ , total $4$ . The largest total is $7$ people ( $5$ students, $2$ adults).
The line through $(0, 2)$ and $(3, 0)$ is $\dfrac{x}{3}+\dfrac{y}{2}=1$ , that is $2x+3y=6$ . At the origin $2\cdot 0+3\cdot 0=0\le 6$ , so the origin lies on the " $\le$ " side. The inequality is $2x+3y\le 6$ .