How do you solve compound Inequality $\frac{2}{5} x < 6$ $2/5x<6$ and $- \frac{1}{2} x \leq - 10$ $-1/2x <= -10$ ?

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Answer 1 · 2018-07-08T12:50:20

No solution

At first we will solve

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

$- \frac{1}{2} x \leq - 10$ $-1/2x<=-10$
multiplying by $- 2$ $-2$

$x \geq 20$ $x>=20$ ( $\leq$ $<=$ must be reversed to $\geq$ $>=$ )
putting Things to gether

$20 \leq x < 15$ $20<=x<15$
which is impossible.

Answer 2 · 2018-07-08T12:52:10

No solution satisfies the conditions.

The first inequality is:

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

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

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

$x < 15$ $x<15$

The second inequality is:

$- \frac{1}{2} x \leq - 10$ $-1/2x<=-10$

Multiplying both sides by $- 2$ $-2$ yields:

$x \geq - 10 \cdot - 2$ $x>=-10*-2$

$x \geq 20$ $x>=20$

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

But since the question asks for $and$ $bb(and)$ and not $or$ $bb(or)$ , we say that there is no solution that satisfies the given conditions.

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