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Equivalent Inequalities

Equivalent inequalities are inequalities that communicate the same information.

There are a few different ways to produce equivalent inequalities.

First, an inequality is equivalent to its mirror image. Let's see what that looks like:

$$x > 4$$ $$4 < x$$

Notice how one inequality is essentially a backwards version of the other. The pointy part of the inequality sign is always pointing towards the $4$! When this is the case, the inequalities are equivalent. However, it is important that you are VERY mindful of the direction of the inequality sign.

Example 1: These two inequalities are NOT equivalent because the pointy part of the inequality sign is pointing to $4$ in the first inequality, and $x$ in the second inequality:

$$x > 4$$ $$4 > x$$

Example 2: These two inequalities are equivalent because they are mirror images....the pointy part of the inequality sign is facing the $15x$ in both cases: $$15x<2$$ $$2>15x$$

Example 3: These two inequalities are not equivalent because the orientation of the inequality signs are opposites....these two inequalities convey opposite information! $$-7x \ge 3 + x$$ $$-7x \le 3 + x$$

It helps to read the inequalities aloud. For our two equivalent inequalities, the first one tells us that "$x$ is greater than $4$" and the second says that "$4$ is less than $x$." That is equivalent information.

If that's still confusing, there are a few more ways to determine if inequalities are equivalent.

Method One: Plug in the same value for the variable(s) in both of the following inequalities: $$x > 4$$ $$4 < x$$

Method Two: Create a number line that represents each inequality. Then compare.

Method Three: Write the solutions to each inequality in interval notation and then compare.

Method Four: Another way to produce equivalent inequalities is to perform the same operation on both sides of the inequality.

NOTE that we FLIPPED the inequality sign in the last example. Whenever you multiply or divide by a negative number, you need to reverse the inequality sign.

(Video) Equivalent Inequalities

by Anusha Rahman

In this lesson, Anusha Rahman covers what equivalent inequalities are and how to determine whether inequalities are equivalent.

Summary

Equivalent inequalities: inequalities that have the same range of solutions.

Let's walk through some examples to figure out whether inequalities are equivalent.

Example 1: Say we have the two inequalities, $$2x>3+x$$ $$2x>6$$

Let's solve the first inequality.

$$2x>3+x$$ $$x>3$$

Now, let's solve the second inequality.

$$2x>6$$ $$x>3$$

Both of these inequalities have the same range of solutions, $x>3$, so these two inequalities are equivalent!

Now, try the next example on your own!

To figure out which of the inequalities are equivalent to $2x+4>12$, we want to find the range of solutions for all of the inequalities.

Let's walk through the first inequality (in blue). $$-2x+5<-3$$ $$2x>8$$ $$x>4$$

Let's walk through the second inequality (in yellow). $$x+3>7$$ $$x>4$$

Let's walk through the third inequality (in purple). $$3x+9<-4$$ $$3x<-13$$ $$x<\frac{-13}{3}$$

Lastly, let's walk through the fourth inequality (in orange). $$\frac{7}{4} < 7x$$ $$\frac{1}{4} < x$$ $$x>\frac{1}{4}$$

Equivalent Inequalities

If a set of inequalities are equivalent, it means that they all have exactly the same solutions. Say you have a certain inequality and you want to produce an equivalent inequality. You can do this by taking the inequality you have and performing the same operation on each side of the inequality sign, just like you learned when solving inequalities with addition or subtraction or multiplying or dividing by a positive or negative number. Then, you will be left with an equivalent inequality!

Take a look at the graphic above. All of the inequalities on there are equivalent, because they all have the same solution of $x<2!$

$x<2$ is equivalent to $x<2,$ clearly.

If you flip the numbers and the sign of $2>x,$ you get $x<2.$

If you divide both sides of $-x>-2$ by $-1$ (a negative number!), you get $x<2.$

If you subtract $1$ from both sides of $x+1<3,$ you get $x<2.$

If you subtract $1$ from both sides of $7>3x+1,$ you get $6>3x.$ If you divide both sides of $6>3x$ by $3,$ you get $2>x.$ If you flip the numbers and the sign of $2>x,$ you get $x<2.$

If you try multiplying both sides of $\dfrac{x}{2} < 1$ by $2,$ you get $x<2.$

If you divide both sides of $6>3x$ by $3,$ you get $2>x.$ If you flip the numbers and the sign of $2>x,$ you get $x<2.$

If you square both sides of $\sqrt{x} < \sqrt{2},$ you get $x<2.$ (Here is a refresher on square roots!)

All of these inequalities are equivalent because they all have the same solution of $x<2!$

Solution

Let's determine which of the following inequalities are equivalent to $2x+1 \le 4.$

$2x+1 \le 4$ is clearly equivalent to $2x+1 \le 4.$

$2x+1 \ge 4$ is not equivalent to $2x+1 \le 4$ because the sign is $\ge$ instead of $\le.$

If you add $1$ to both sides of $2x \le 3,$ you get $2x+1 \le 4,$ so $2x \le 3$ is equivalent to $2x+1 \le 4.$

If you multiply both sides of $x \le \dfrac{3}{2}$ by $2,$ you get $2x \le 3.$ If you add $1$ to both sides of $2x \le 3,$ you get $2x+1 \le 4.$ So $x \le \dfrac{3}{2}$ is equivalent to $2x+1 \le 4.$

We just found that $x \le \dfrac{3}{2}$ is equivalent to $2x+1 \le 4.$ So, $x \ge \dfrac{3}{2}$ cannot also be equivalent, because the sign is flipped. Not convinced? If you multiply both sides of $x \ge \dfrac{3}{2}$ by $2,$ you get $2x \ge 3.$ If you add $1$ to both sides of $2x \ge 3,$ you get $2x+1 \ge 4.$ We've seen earlier that $2x+1 \ge 4$ is not equivalent to $2x+1 \le 4$ because the sign is $\ge$ instead of $\le.$

If you subtract $3y$ and $6z$ from both sides of $3y+6z+ 2x+1 \le 3y+6z+4,$ you get you get $2x+1 \le 4,$ so $3y+6z+ 2x+1 \le 3y+6z+4$ is equivalent to $2x+1 \le 4.$

If you subtract $17z$ and $6y$ from $17z+6y+ 2x+1 \ge 17z+6y+4,$ you get $2x+1 \ge 4.$ We've seen earlier that $2x+1 \ge 4$ is not equivalent to $2x+1 \le 4$ because the sign is $\ge$ instead of $\le.$

If you multiply both sides of $-x \le -\dfrac{3}{2}$ by $-1,$ you get $x \ge \dfrac{3}{2}.$ We have already shown that $x \ge \dfrac{3}{2}$ is not equivalent to $2x+1 \le 4$ because of the wrong inequality sign.