Since you are solving for x, your objective is to get x by itself
3a).
[tex]ax + b \geqslant cd[/tex]
first subtract b on both sides
[tex]ax \geqslant cd - b[/tex]
then divide by a
[tex]x \geqslant \frac{cd - b}{a} [/tex]
3b).
[tex] \frac{a(x + 2)}{b} > c[/tex]
first multiply by b on both sides
[tex]a(x + 2) > bc[/tex]
then divide by a
[tex]x + 2 > \frac{bc}{a} [/tex]
subtract by 2
[tex]x = \frac{bc}{a} - 2[/tex]
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 <
[tex]x < \frac{d - b}{a} [/tex]
So the answer choice 2 is the solution for question 4
