Question #4c249

Nov 19, 2017

Selection B. $y < \frac{5}{3} x - 2$

Explanation:

Given:

$5 x - 3 y > 6$

Subtract 5x from both sides of the inequality:

$- 3 y > - 5 x + 6$

When we divide both sides by negative number (in this case -3), we must change the direction of the inequality:

$y < \frac{- 5}{-} 3 x + \frac{6}{-} 3$

Simplify:

$y < \frac{5}{3} x - 2$

This matches selection B.

Nov 19, 2017

B
$y < \frac{5}{3} x - 2$

Explanation:

$5 x - 3 y > 6$

All of these inequalities have y as the subject. So lets manipulate our given inequality, almost like it is an equation, to make y the subject.

$5 x - 3 y > 6$
$- 3 y > 6 - 5 x$
$- y > 2 - \frac{5}{3} x$

Now, this part is where we need to remember we are dealing with an inequality, not an equation. With an inequality, when dividing by a negative number, you need to flip the sign. I'll show why later on.
Keeping this in mind:

$y < - 2 + \frac{5}{3} x$
$y < \frac{5}{3} x - 2$

Now, to show why you can't divide by -1 normally

$4 = 4$ (I hope you'll agree)
$4 + 1 = 4 + 1$
$\frac{4 + 1}{7} = \frac{4 + 1}{7}$
$12 \left(\frac{4 + 1}{7}\right) = 12 \left(\frac{4 + 1}{7}\right)$
$- 12 \left(\frac{4 + 1}{7}\right) = - 12 \left(\frac{4 + 1}{7}\right)$

Everything we do with an equation is fine.

Now, for an inequality:

$3 < 4$
$3 + 1 < 4 + 1$
$\frac{3 + 1}{7} < \frac{4 + 1}{7}$
$12 \left(\frac{3 + 1}{7}\right) < 12 \left(\frac{4 + 1}{7}\right)$
So far, so good. Now dividing by a negative.
$- 12 \left(\frac{3 + 1}{7}\right) < - 12 \left(\frac{4 + 1}{7}\right)$
Oops, this is wrong. Evaluating both sides we get:
$- \frac{48}{7} < - \frac{60}{7}$
Which is wrong. Therefore, flipping the sign:
$- \frac{48}{7} > - \frac{60}{7}$ is correct, and all is good.