How do you evaluate #frac { x + y } { 8x - y } - \frac { 7x } { y - 8x }#?

Question

Isaiah Anthony · Accepted Answer

See a solution process below:

First, multiply the second fraction by #-1/-1#. This is equal to #1# and therefore does not change the value of the fraction: #(x + y)/(8x - y) - (-1/-1 xx (7x)/(y - 8x)) =># #(x + y)/(8x - y) - (-1 xx 7x)/(-1(y - 8x)) =># #(x + y)/(8x - y) - (-7x)/((-1 xx y) - (-1 xx 8x)) =># #(x + y)/(8x - y) - (-7x)/(-y - (-8x)) =># #(x + y)/(8x - y) - (-7x)/(-y + 8x) =># #(x + y)/(8x - y) - (-7x)/(8x - y)# Now, with the fractions over common denominators we can subtract the numerators: #((x + y) - (-7x))/(8x - y) =># #(x + y + 7x)/(8x - y) =># #(x + 7x + y)/(8x - y) =># #(1x + 7x + y)/(8x - y) =># #((1 + 7)x + y)/(8x - y) =># #(8x + y)/(8x - y)#

Julia Adams · Answer

undefined To evaluate $ \frac{x + y}{8x - y} - \frac{7x}{y - 8x} $, follow these steps:

1. Find a common denominator for the fractions.
2. Simplify each fraction separately by multiplying both numerator and denominator by the common denominator.
3. Subtract the simplified fractions.

$$ \frac{x + y}{8x - y} - \frac{7x}{y - 8x} $$

$$ \text{Common denominator} = (8x - y)(y - 8x) $$

$$ \frac{(x + y)(y - 8x)}{(8x - y)(y - 8x)} - \frac{7x(8x - y)}{(y - 8x)(8x - y)} $$

$$ \frac{xy - 8x^2 + y^2 - 56x^2 + 7xy}{(8x - y)(y - 8x)} $$

$$ \frac{xy + 7xy - 8x^2 - 56x^2 + y^2}{(8x - y)(y - 8x)} $$

$$ \frac{8xy - 64x^2 + y^2}{(8x - y)(y - 8x)} $$