Answer:
C. The equation has two valid solutions
Step-by-step explanation:
A rational equation like this is often solved by "cross-multiplying" to eliminate the fractions. When we do that, we get ...
4(2x+10) = x(3x+1) . . . . . . . multiply both sides by (3x+1)(2x+10)
3x² -7x -40 = 0 . . . . . . . . . . rearrange to standard form
3x² -15x +8x -40 = 0 . . . . . . separate -7 into factors of 3(-40)
3x(x -5) +8(x -5) = 0 . . . . . factor pairs of terms
(3x +8)(x -5) = 0 . . . . . . . finish factoring
The zeros of these factors are not zeros of the original denominators, so neither of the solutions these give is extraneous.
3x +8 = 0 ⇒ x = -8/3
x -5 = 0 ⇒ x = 5
