Linear inequalities in one unknown

Minh has $200$ thousand đồng. He buys a box of pens for $50$ thousand đồng and wants to buy some exercise books at $12$ thousand đồng each. What is the greatest number of books he can buy?

Let $x$ be the number of books. He pays $50+12x$ , which must not exceed $200$ , so $50+12x\le200$ . This is not an equation but an inequality.

Explore: trying a few values

$x$	$10$	$12$	$13$
$50+12x$	$170$	$194$	$206$
$\le200$ ?	yes	yes	no

With $12$ books he has enough money; with $13$ he does not. But trying numbers one by one is slow; we need a method like the one for solving equations.

How to solve

We transform an inequality using the properties of inequalities from the last lesson.

These rules work because adding the same number to both sides keeps the direction, while multiplying by a negative number reverses it (for example $2<3$ but $-2>-3$ ).

When the coefficient of $x$ is negative, the division step needs the most care.

The solutions

x>2

(top) and

x\le3

(bottom) on number lines

A hollow point means that number is not a solution (sign $>$ or $<$ ); a filled point means it is a solution (sign $\ge$ or $\le$ ).

Inequalities that reduce to linear form

Many inequalities have brackets or fractions. We expand the brackets, clear the fractions, move the terms in $x$ to one side and the numbers to the other.

Now we answer Minh's question.

Key terms

English	Tiếng Việt
inequality (to be solved)	bất phương trình
linear inequality in one unknown	bất phương trình bậc nhất một ẩn
solution of the inequality	nghiệm của bất phương trình
rule for moving a term to the other side	quy tắc chuyển vế
rule for multiplying both sides	quy tắc nhân
reverse the inequality sign	đổi chiều
number line	trục số
solve an inequality	giải bất phương trình

Summary

Linear inequality in one unknown: $ax+b<0$ (or $>$ , $\le$ , $\ge$ ), $a\neq0$ .
Moving a term to the other side changes its sign.
Multiplying or dividing by a positive number: keep the direction; by a negative number: reverse it.
Brackets or fractions: expand, multiply by a positive common denominator, reach $ax<c$ , then divide by $a$ .
Real problems: check the conditions (whole numbers…) before choosing the answer.

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 the numbers $-1$ , $0$ , $3$ are solutions of the inequality $2x+1>0$ ?

What you need and how to start

What you need: $x_0$ is a solution if substituting it gives a true statement.

Steps:

Substitute each number into $2x+1$ .
Compare with $0$ .

Show solution

$x=-1$ : $-1>0$ false. $x=0$ : $1>0$ true. $x=3$ : $7>0$ true.

Answer. $0$ and $3$ .

Exercise 2

Solve the inequality $4x-8<0$ .

What you need and how to start

What you need: move a term, then divide by a positive number (keep the direction).

Steps:

Move $-8$ to the right-hand side.
Divide by $4$ .

Show solution

$4x<8\iff x<2$ .

Answer. $x<2$ .

Exercise 3

Solve the inequality $-3x+9\le0$ .

What you need and how to start

What you need: dividing by a negative number reverses the sign.

Steps:

Move $9$ to the right-hand side.
Divide both sides by $-3$ and reverse the sign.

Show solution

$-3x\le-9\iff x\ge3$ .

Answer. $x\ge3$ .

Exercise 4

Solve the inequality $2(x+3)>5x-6$ .

What you need and how to start

What you need: expand, move terms, and watch the sign of the coefficient of $x$ .

Steps:

Expand: $2x+6>5x-6$ .
Moving terms gives $-3x>-12$ .
Divide by $-3$ and reverse the sign.

Show solution

$2x+6>5x-6\iff -3x>-12\iff x<4$ .

Answer. $x<4$ .

Exercise 5

Solve the inequality $\dfrac{2x-1}{3}\ge\dfrac{x+2}{4}$ .

What you need and how to start

What you need: multiply both sides by the positive common denominator $12$ ; the direction stays the same.

Steps:

Multiply both sides by $12$ .
Expand, move terms, divide by the coefficient of $x$ .

Show solution

$4(2x-1)\ge3(x+2)\iff 8x-4\ge3x+6\iff 5x\ge10\iff x\ge2$ .

Answer. $x\ge2$ .

Exercise 6

A taxi firm charges a starting fee of $10$ thousand đồng plus $14$ thousand đồng per kilometre. With $150$ thousand đồng, what is the greatest distance you can travel (in whole kilometres)?

What you need and how to start

What you need: the fare must not exceed the money you have.

Steps:

Let $x$ be the number of kilometres; form $10+14x\le150$ .
Solve and choose the greatest whole number.

Show solution

$14x\le140\iff x\le10$ .

Answer. At most $10$ km.

Exercise 7

Nam has scored $7$ , $8$ and $6$ in three tests. What is the lowest mark he needs in the fourth test so that his mean for the four tests is not below $7.5$ ?

What you need and how to start

What you need: "not below" is $\ge$ ; mean $=$ total $\div$ number of tests.

Steps:

Let $x$ be the fourth mark ( $0\le x\le10$ ).
Form $\dfrac{7+8+6+x}{4}\ge7.5$ and multiply both sides by $4$ .

Show solution

$21+x\ge30\iff x\ge9$ .

Answer. At least $9$ .

Exercise 8 Review

Find the solutions of the equation $(x-4)(2x+1)=0$ that satisfy the inequality $3x-2>0$ .

What you need and how to start

What you need: equations in factorised form (lesson 4) and checking solutions of an inequality.

Steps:

Solve the equation in factorised form.
Substitute each solution into $3x-2$ .

Show solution

The equation has solutions $x=4$ and $x=-\dfrac12$ . For $x=4$ : $10>0$ , it works. For $x=-\dfrac12$ : $-\dfrac72<0$ , it does not.

Answer. $x=4$ .

Exercise 9

Consider the expressions $A=3(x-1)$ and $B=x+5$ .

a) Find $x$ such that $A=B$ .

b) Find $x$ such that $A>B$ .

c) Find the least integer $x$ such that $A>B$ .

What you need and how to start

What you need: solving linear equations and linear inequalities in one unknown.

Steps:

a) Solve $3x-3=x+5$ .
b) Solve $3x-3>x+5$ .
c) Choose the least integer that satisfies part b).

Show solution

a) $2x=8\iff x=4$ . b) $2x>8\iff x>4$ . c) The least integer greater than $4$ is $5$ .

Answer. a) $x=4$ ; b) $x>4$ ; c) $x=5$ .

Exercise 10 Challenge

Phone plan A charges $3$ thousand đồng per minute of calls. Plan B charges $45$ thousand đồng a month plus $1$ thousand đồng per minute. In a month, what is the least number of whole minutes of calls for which plan B is cheaper than plan A?

What you need and how to start

What you need: "B is cheaper than A" means the cost of plan B $<$ the cost of plan A.

Steps:

Let $x$ be the number of minutes; plan A costs $3x$ and plan B costs $45+x$ .
Solve $45+x<3x$ .
Choose the least whole number.

Show solution

$45+x<3x\iff -2x<-45\iff x>22.5$ . The least whole number greater than $22.5$ is $23$ . Check: for $23$ minutes plan A costs $69$ and plan B $68$ thousand đồng.

Answer. At least $23$ minutes.

a) $x=3$ is a solution.	True	False
b) Moving terms gives $-3x\ge-9$ .	True	False
c) The solution of the inequality is $x\ge3$ .	True	False
d) Exactly $4$ non-negative integers satisfy the inequality.	True	False

Linear inequalities in one unknown