Question

Need help w 3 and 4 pls

Answer 1

Since you are solving for x, your objective is to get x by itself

3a).
$ax + b \geqslant cd$
first subtract b on both sides
$ax \geqslant cd - b$
then divide by a
$x \geqslant \frac{cd - b}{a}$
3b).
$\frac{a(x + 2)}{b} > c$
first multiply by b on both sides
$a(x + 2) > bc$
then divide by a
$x + 2 > \frac{bc}{a}$
subtract by 2
$x = \frac{bc}{a} - 2$
4).
Since a is negative, when dividing the sign of the inequality would change such that > would be < and vice versa. The same applies to greater than or equal to and less than or equal to.

ax+b>d

first subtract b

ax>d-b

then divide by a to get x bu itself, but since a is negative the > sign would change to <

$x < \frac{d - b}{a}$
So the answer choice 2 is the solution for question 4