2.07 Properties of inequality

When we manipulate with equalities, we can apply the same operation to both sides and the equality statement remains true. Take the following equality:

$x+7$x+7

$=$=

$12$12

We can subtract both sides of the equation in order to find the value of $x$x. This is because both sides of the equation are identical, so what we do to one side, we should do to the other side.

$x+7$x+7	$=$=	$12$12	(rewriting the equation)
$x+7-7$x+7−7	$=$=	$12-7$12−7	(subtracting $7$7 from both sides)
$x$x	$=$=	$5$5	(simplifying both sides)

When working with inequalities, this is not necessarily always the case.

Exploration

Consider the inequality $9<15$9<15. If we add or subtract both sides by any number, say $3$3, we can see that the resulting inequality remains true. More specifically we can write $9+3<15+3$9+3<15+3 and $9-3<15-3$9−3<15−3.

Adding $3$3 to $9$9 and $15$15.

Subtracting $3$3 from $9$9 and $15$15.

 

Now consider if we multiply or divide both sides of the inequality by $3$3. We get $9\times3<15\times3$9×3<15×3 and $\frac{9}{3}<\frac{15}{3}$93​<153​. These statements are true, since we increase (or decrease) $9$9 and $15$15 by the same positive factor, so the signs of each side are unchanged.

However, if we had chosen a negative number, like $-3$−3, the signs of each side are changed and we must swap the inequality sign around. So the correct statements are $9\times\left(-3\right)>15\times\left(-3\right)$9×(−3)>15×(−3) and $-\frac{9}{3}>-\frac{15}{3}$−93​>−153​.

 

Practice questions

Question 1

question 2

question 3