If the 0 is in the numerator, the value is 0
if the 0 is in the denominator, the value is undefined
so 0/8 = 0
and 3/0 = undefined
So basically you just want to replace all the x's in the h(x) equation with the f(x) equation.
h(x-7) = 2(x-7) + 3
h(x) = 2x -14 + 3
h(x) = 2x - 11
Break down the problem into these two equations;
x = -31
-x = -31
Solve for the 1st equation; x = -31
x = -31
Solve for the 2nd equation; -x = -31
x = 31
Collect all solutions
x = ±31
Check the solution
When x = -31, the original equation; |x| = -31 does not hold true. Thus, we will drop x = -31 from the solution set.
Check the solution again;
When x = 31, the original equation |x| = -31 does not hold true as well. Thus we will drop x = 31 from the solution set.
Therefore,
<u>No solution exists to this equation. </u>
