Question

Solve for x, assuming a, b, and c are negative constants.

(ax + b)/c ≤ b

Answer 1

(ax + b)/c ≤ b

ax+b ≤ cb

ax ≤ cb - b

x ≤ (cb-b)/a

Answer 2

Solving for some variable means finding its value for which the given statement is true. Thus, the solution to given inequality is $x \leq \dfrac{b(c-1)}{a}$ (if a/c is non-negative)or $x \geq \dfrac{b(c-1)}{a}$ (if a/c is negative)

What type of operations are allowed in inequality?

All such operations such that inequality doesn't gets modified( for example, if left expression is bigger than right one, then that expression cannot go bigger than the later expression).

For the given case, the inequality given is:

$(ax+b)/c \leq b$

Its solution can be found by doing operations which doesn't modify the inequality and takes all the terms such that one side is 'x' and other side are constants.
$(ax+b)/c \leq b\\\\\dfrac{a}{c}x + \dfrac{b}{c} \leq b\\\\\text{Subtracting b/c from both the sides}\\\\\dfrac{a}{c}x \leq b - \dfrac{b}{c}\\\\\text{Supposing that a/c is non-negative, multiplying both sides by c/a}\\\\x \leq \dfrac{c}{a} \times \dfrac{bc - b}{c} = \dfrac{b(c-1)}{a}\\\\x \leq \dfrac{b(c-1)}{a}$

(i assumed a/c non-negative, since multiplying with negative quantity can modify the sign of inequality, so ≤ would have become ≥ )

If a/c is negative, then:

$x \geq \dfrac{c}{a} \times \dfrac{bc - b}{c} = \dfrac{b(c-1)}{a}\\\\x \geq \dfrac{b(c-1)}{a}$

Thus, the solution to given inequality is $x \leq \dfrac{b(c-1)}{a}$ (if a/c is non-negative)

or $x \geq \dfrac{b(c-1)}{a}$ (if a/c is negative)

Learn more about inequality here:

https://brainly.com/question/11901702