Another Simplifying Question?

Question

(2x^2-x-6)/(x^2-3x+2) divided by (2x^2+5x+3)/(3x^2-3)

Answer 1

[(2x²-x-6)/(x²-3x+2)] / [(2x²+5x+3)/(3x²-3)] =

Factor the numerator and denominator of each fraction.

[((x-2)(2x+3))/((x-2)(x-1))] / [((x+1)(2x+3))/(3(x-1)(x+1))] =

Cancel any term that appears in both the numerator and denominator.

[(2x+3)/(x-1)] / [(2x+3)/(3(x-1))] =

When you divide a fraction by another fraction you multiply the first fraction by the flip of the second fraction, so (a/b) / (c/d) = (a/b)(d/c)

[(2x+3)/(x-1)] * [(3(x-1))/(2x+3)] =

((2x+3)3(x-1)) / ((2x+3)(x-1)) =

Cancel (x-1)/(x-1) = 1.

(3(2x+3)) / (2x+3) =

Cancel (2x+3)/(2x+3) = 1

3

Then (2x^2-x-6)/(x^2-3x+2) divided by (2x^2+5x+3)/(3x^2-3) = 3.

Hope this helps you!