Question

Solve for x.

Answer 1

Answer:

$x\leq6\text{ or } x>30$

Step-by-step explanation:

So we have the compound inequality:

$9x+4\leq 58\text{ or } \frac{1}{6}x-2>3$

This is an "or" inequality. So, we just have to solve each inequality individually and then combine the answers!

1)

We have:

$9x+4\leq58$

Subtract 4 from both sides:

$9x\leq54$

Divide both sides by 9:

$x\leq6$

So, our first solution is all numbers less than or equal to 6.

2)

We have:

$\frac{1}{6}x-2>3$

Add 2 to both sides:

$\frac{1}{6}x>5$

To cancel out the fraction, let's multiply both sides by 6. So:

$6(\frac{1}{6}x)>6(5)$

The left side will cancel. Therefore:

$x>30$

So, our second solution is all numbers greater than 30.

Therefore, our full solution will be:

$x\leq6\text{ or } x>30$

And we're done!

Answer 2

Answer:

−8x−7y=96 (* 5)

−3x+5y=−25  (* 7)

-40x - 35y = 480

-21x + 35y = -175

=> -61x = 305

x = -305/61

x = -5

-3x + 5y = -25  

-3 * (-5) + 5y = -25

5y = -40

y = -8 =>

x = -5

y = -8

9x+4≤58 or 16x−2>3

9x ≤ 54 or 16x > 5

x ≤ 6 or x > 5/16