Inequalities

An inequality is like an equation, but in place of the $\displaystyle {}={}$ sign it has one of the signs $\displaystyle <$, $\displaystyle >$, $\displaystyle \leq$ or $\displaystyle \geq$. A first-degree equation usually has one solution; an inequality has infinitely many.

A solution of an inequality is a number that satisfies it: put it in place of $\displaystyle x$, and the statement is true.
We check a few numbers in the inequality $\displaystyle 2x+1>7$:

• $\displaystyle {x}={5}$: $\displaystyle {2\cdot5+1}={11}$, and $\displaystyle 11>7$ is true – 5 is a solution.
• $\displaystyle {x}={3}$: $\displaystyle {2\cdot3+1}={7}$, and $\displaystyle 7>7$ is false – 3 is not a solution.
• $\displaystyle {x}={0}$: $\displaystyle {2\cdot0+1}={1}$, and $\displaystyle 1>7$ is false – 0 is not a solution.

Every number greater than 3 is a solution, and no other number is. All the solutions together are the inequality’s solution set, written in short as $\displaystyle x>3$. The signs:

• $\displaystyle x>3$ – $\displaystyle x$ is greater than 3;
• $\displaystyle x<3$ – $\displaystyle x$ is less than 3;
• $\displaystyle x\geq3$ – $\displaystyle x$ is greater than or equal to 3;
• $\displaystyle x\leq3$ – $\displaystyle x$ is less than or equal to 3.

Here are the numbers $\displaystyle -2$, $\displaystyle 0$, $\displaystyle 2.5$, $\displaystyle 4$, $\displaystyle 7$.
Which of them are solutions of the inequality $\displaystyle 3x-2\leq10$?
Substitute each one and check.

A solution set is marked on the number line – here, on the $\displaystyle x$-axis. Each coloured strip is all the numbers that satisfy one inequality:

• $\displaystyle x>3$ – every number to the right of 3. The line at the strip’s end is dashed: 3 itself is not a solution.
• $\displaystyle x\leq-1$ – every number to the left of $\displaystyle -1$, and $\displaystyle -1$ itself. The line at the strip’s end is solid: $\displaystyle -1$ is a solution.

Drawn by hand, the end of the interval is a circle: an empty circle – the number is not a solution ($\displaystyle <$, $\displaystyle >$); a filled circle – it is ($\displaystyle \leq$, $\displaystyle \geq$).

An inequality is solved like an equation: do the same to both sides until $\displaystyle x$ is alone. These steps keep the sign:

• add or subtract the same number on both sides: from $\displaystyle 5>2$, after $\displaystyle +3$, we get $\displaystyle 8>5$;
• multiply or divide both sides by the same positive number: from $\displaystyle 5>2$, after multiplying by 2, we get $\displaystyle 10>4$.

Multiplying or dividing both sides by a negative number turns the sign: $\displaystyle <$ becomes $\displaystyle >$, and $\displaystyle \leq$ becomes $\displaystyle \geq$.

• The reason: multiplying by $\displaystyle -1$ takes every number to the other side of 0 on the number line, as in a mirror – so the order turns round.
• For example: 2 is to the left of 5, so $\displaystyle 2<5$. But $\displaystyle -2$ is to the right of $\displaystyle -5$, so $\displaystyle -2>-5$.

We divided by a positive number, 3, so the sign stays. The solution set: $\displaystyle x<4$.
A check: 0 is a solution, since $\displaystyle {3\cdot0-5}={-5}$ and $\displaystyle -5<7$; 10 is not, since $\displaystyle {3\cdot10-5}={25}$ and $\displaystyle 25<7$ is false.

We divided by $\displaystyle -2$, so the sign turned: $\displaystyle \geq$ became $\displaystyle \leq$.
A check: $\displaystyle -5$ is a solution, since $\displaystyle {4-2\cdot(-5)}={14}$ and $\displaystyle 14\geq10$.

Solve the inequalities.
Show each one’s solution set on the $\displaystyle x$-axis below.

As with an equation: open the brackets, bring the terms with $\displaystyle x$ to one side and the numbers to the other.

Solve.
In an inequality with fractions, first multiply both sides by a common denominator.

Sometimes $\displaystyle x$ cancels out, and a statement with no $\displaystyle x$ is left – as in the special equations.

• $\displaystyle 2(x+1)<2x-3$, that is $\displaystyle 2x+2<2x-3$. After $\displaystyle -2x$, $\displaystyle 2<-3$ is left – a false statement. The inequality has no solution.
• $\displaystyle 3x+4>3(x-1)$, that is $\displaystyle 3x+4>3x-3$. After $\displaystyle -3x$, $\displaystyle 4>-3$ is left – a true statement. Every number is a solution.

Solve.

In word problems, these words become signs:

• at least 10 – 10 or more: $\displaystyle \geq10$;
• at most 10 – 10 or less: $\displaystyle \leq10$;
• more than 10: $\displaystyle >10$;
• less than 10: $\displaystyle <10$.

Noa has 120 ₪, and every week she saves another 35 ₪.
After how many weeks will she have at least 400 ₪?

Let $\displaystyle x$ be the number of weeks. After $\displaystyle x$ weeks Noa will have $\displaystyle 120+35x$ shekels, and that must be at least 400:

After 8 weeks or more, Noa will have at least 400 ₪.

A lift can carry at most 600 kg.
A worker who weighs 80 kg takes boxes of 25 kg each up in the lift.
At most how many boxes can he take up with him at one time?
Write an inequality and solve it.

A gym has two memberships.
Membership A: a joining fee of 200 ₪, and 50 ₪ a month.
Membership B: no joining fee, 90 ₪ a month.
From how many months is membership A cheaper?

After $\displaystyle x$ months membership A costs $\displaystyle 200+50x$ shekels, and membership B $\displaystyle 90x$ shekels. A is cheaper when:

Membership A is cheaper for more than 5 months, that is from 6 months on. The graph shows it too: to the left of $\displaystyle {x}={5}$ the line $\displaystyle {y}={200+50x}$ is above the line $\displaystyle {y}={90x}$, and to the right of $\displaystyle {x}={5}$ it is below it.

Hiring a bike.
At shop A you pay 30 ₪, and 6 ₪ for every hour.
At shop B you pay 12 ₪ for every hour.
Mark on the graph the point where the two lines meet.
Write an inequality and solve it: from how many hours of riding is shop A cheaper?