How do you solve $8 (a - 2) \leq 10 (a + 2)$ $8(a-2)<=10(a+2)$ ?

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Answer 1 · 2018-04-16T01:03:58

$a \geq - 18$ $a >= -18$

We solve inequalities similarly to equations.

$8 (a - 2) \leq 10 (a + 2)$ $8(a-2) <= 10(a+2)$

First, let's distribute:
$8 a - 16 \leq 10 a + 20$ $8a - 16 <= 10a + 20$

Now subtract $8 a$ $8a$ from both sides of the inequality:
$8a - 16 quadcolor(red)(-quad8a) <= 10a + 20 quadcolor(red)(-quad8a)$

$-16 <= 2a + 20$

Subtract $20$ from both sides of the inequality:
$-16 quadcolor(red)(-quad20) <= 2a + 20 quadcolor(red)(-quad20)$

$-36 <= 2a$

Divide both sides by $2$ :
$(-36)/color(red)(2) <= (2a)/color(red)(2)$

$-18 <= a$

Put $a$ on the left side of the inequality:
$a >= -18$

This means that $a$ must be more than or equal to $-18$ .

Hope this helps!

How do you solve 8(a-2)<=10(a+2)8(a−2)≤10(a+2)8(a-2)<=10(a+2)?