# How do you solve compound Inequality 2/5x<6 and  -1/2x <= -10?

Jul 8, 2018

No solution

#### Explanation:

At first we will solve

$\frac{2}{5} x < 6$
multiplying by $\frac{5}{2}$
we get
$x < \frac{30}{2} = 15$
No we solve the second inequality

$- \frac{1}{2} x \le - 10$
multiplying by $- 2$

$x \ge 20$ ($\le$ must be reversed to $\ge$)
putting Things to gether

$20 \le x < 15$
which is impossible.

Jul 8, 2018

No solution satisfies the conditions.

#### Explanation:

The first inequality is:

$\frac{2}{5} x < 6$

Multiplying both sides by $\frac{5}{2}$ yields:

$x < 6 \cdot \frac{5}{2}$

$x < 15$

The second inequality is:

$- \frac{1}{2} x \le - 10$

Multiplying both sides by $- 2$ yields:

$x \ge - 10 \cdot - 2$

$x \ge 20$

Since both inequalities of $x$ cannot exist at the same time, we say the solution is $x < 15$ or $x \ge 20$.

But since the question asks for $\boldsymbol{\mathmr{and}}$ and not $\boldsymbol{\mathmr{and}}$, we say that there is no solution that satisfies the given conditions.