How do you solve compound inequalities 5a-4>16 or 3a + 2 <17?

Jul 21, 2018

$a > 4$ or $a < 5$

Explanation:

We have the following:

$\textcolor{\lim e g r e e n}{5 a - 4 > 16}$ and $\textcolor{b l u e}{3 a + 2 < 17}$

Let's start with our green inequality. We can add $4$ to both sides to get

$\textcolor{\lim e g r e e n}{5 a > 20}$

Next, divide both sides by $5$ to get

$\textcolor{\lim e g r e e n}{a > 4}$

Next, let's look at our blue inequality. Let's subtract $2$ from both sides to get

$\textcolor{b l u e}{3 a < 15}$

Lastly, let's divide bot hsides by $3$ to get

$\textcolor{b l u e}{a < 5}$

Our solutions are

$a > 4$ or $a < 5$

Hope this helps!

$a > 4$ or $a < 5$

Explanation:

1) Solving first inequality:

$5 a - 4 > 16$

$5 a - 4 + 4 > 16 + 4$

$5 a > 20$

$\frac{5 a}{5} > \frac{20}{5}$

$a > 4$

2) Solving second inequality:

$3 a + 2 < 17$

$3 a + 2 - 2 < 17 - 2$

$3 a < 15$

$\frac{3 a}{3} < \frac{15}{3}$

$a < 5$