MATH SOLVE

4 months ago

Q:
# quantity x squared plus 4 x minus 5 over quantity 5 x squared minus 8 x plus 3 times quantity 20 x minus 12 over x squared minus 6 x minus 55.

Accepted Solution

A:

The answer is [tex]\frac{4}{x-11}[/tex].

We want to factor each numerator and denominator:

[tex]\frac{x^2+4x-5}{5x^2-8x+3} \times \frac{20x-12}{x^2-6x-55}[/tex]

For the numerator of the ratio on the left, we want factors of -5 that sum to 4; 5(-1) = -5 and 5+(-1) = 4:

[tex]\frac{(x+5)(x-1)}{5x^2-8x+3} \times \frac{20x-12}{x^2-6x-55}[/tex]

Factoring the denominator of the ratio on the left, we want factors of 5*3 that sum to -8; -5(-3) = 15 and -5+(-3) = -8:

[tex]\frac{(x+5)(x-1)}{5x^2-5x-3x+3} \times \frac{20x-12}{x^2-6x-55}[/tex]

We factor by grouping on the denominator on the left now:

[tex]\frac{(x+5)(x-1)}{(5x^2-5x)+(-3x+3)} \times \frac{20x-12}{x^2-6x-55} \\ \\ \frac{(x+5)(x-1)}{5x(x-1)-3(x-1)} \times \frac{20x-12}{x^2-6x-55} \\ \\ \frac{(x+5)(x-1)}{(5x-3)(x-1)} \times \frac{20x-12}{x^2-6x-55}[/tex]

Factor the GCF out of the numerator on the right:

[tex]\frac{(x+5)(x-1)}{(5x-3)(x-1)} \times \frac{4(5x-3)}{x^2-6x-55}[/tex]

For the denominator on the right, we want factors of -55 that sum to -6; -11(5) = -55 and -11+5 = -6:

[tex]\frac{(x+5)(x-1)}{(5x-3)(x-1)} \times \frac{4(5x-3)}{(x-11)(x+5)}[/tex]

Cancelling everything that is common on the numerators and denominators, we are left with

[tex]\frac{4}{x-11}[/tex]

We want to factor each numerator and denominator:

[tex]\frac{x^2+4x-5}{5x^2-8x+3} \times \frac{20x-12}{x^2-6x-55}[/tex]

For the numerator of the ratio on the left, we want factors of -5 that sum to 4; 5(-1) = -5 and 5+(-1) = 4:

[tex]\frac{(x+5)(x-1)}{5x^2-8x+3} \times \frac{20x-12}{x^2-6x-55}[/tex]

Factoring the denominator of the ratio on the left, we want factors of 5*3 that sum to -8; -5(-3) = 15 and -5+(-3) = -8:

[tex]\frac{(x+5)(x-1)}{5x^2-5x-3x+3} \times \frac{20x-12}{x^2-6x-55}[/tex]

We factor by grouping on the denominator on the left now:

[tex]\frac{(x+5)(x-1)}{(5x^2-5x)+(-3x+3)} \times \frac{20x-12}{x^2-6x-55} \\ \\ \frac{(x+5)(x-1)}{5x(x-1)-3(x-1)} \times \frac{20x-12}{x^2-6x-55} \\ \\ \frac{(x+5)(x-1)}{(5x-3)(x-1)} \times \frac{20x-12}{x^2-6x-55}[/tex]

Factor the GCF out of the numerator on the right:

[tex]\frac{(x+5)(x-1)}{(5x-3)(x-1)} \times \frac{4(5x-3)}{x^2-6x-55}[/tex]

For the denominator on the right, we want factors of -55 that sum to -6; -11(5) = -55 and -11+5 = -6:

[tex]\frac{(x+5)(x-1)}{(5x-3)(x-1)} \times \frac{4(5x-3)}{(x-11)(x+5)}[/tex]

Cancelling everything that is common on the numerators and denominators, we are left with

[tex]\frac{4}{x-11}[/tex]