Question

How do you solve `\frac { x + 5} { 4} = \frac { x ^ { 2} - x } { 8}`?

Answer 1

Eliminate the fractions combine like terms and solve as a quadratic. Gives the answer of x = 5 or x = -2

Explanation:

First eliminate the fractions by multiplying both sides by the least common multiply 8

`8 xx ( x+5)/4 = 8 xx ( x^2-x)/8` This gives

`2 xx ( x + 5) = x^2 -x` removing the parenthesis gives

`2x + 10 = x^2 -x` subtract -2x and -10 from both sides

`2x-2x + 10-10 = x^2 -x -2x -10` combining like terms gives.

`0 = x^2 -3x -10` Since the C term (-10) is negative one binomial must be negative and one binomial must be positive. Since the B term -3x is negative the negative binomial must be greater than the positive binomial when they are algebraically added.

so `0 = ( x -5) xx ( x +2)` solve for x in each binomial

`0 = x -5` add five to both sides

`0 + 5 = x -5 +5` which gives.

`5 = x`

`0 = x +2` subtract 2 from both sides

`0 -2 = x + 2 -2` which gives

`-2 = x`

Answer 2

`x=5,-2`

Explanation:

`8(x+5)=4(x^2-x)`

`2(x+5)=(x^2-x)`

`2x+10=x^2-x`

`x^2-x-2x-10=0`

`x^2-3x-10=0`

`(x-5)(x+2)=0`

`x=5,-2`

Answer 3

`color(blue)(x=5 or x=-2`

Explanation:

`(x+5)/4=(x^2-x)/8`

multily both sides by `32`

`:.32 xx (x+5)/4=32 xx (x^2-x)/8`

`:.(cancel32^8(x+5))/cancel4^1=(cancel32^4(x^2-8))/cancel8^1`

`:.8(x+5)=4(x^2-8)`

`:.8x+40=4x^2-32`

`:.4x^2-32=8x+40`

`:.4x^2-12x-40`

`:.(2x+4)(2x-10)`

`:.2x=-4 or 2x=10`

`:.x=-4/2 or x=10/2`

`:.color(blue)(x=-2 or x= 5`

substitute `color(blue)(x=-2`

`:.((color(blue)(-2))+5)/4=((color(blue)(-2))^2-(color(blue)(-2)))/8`

`:.3/4=(4+2)/8`

`:.3/4=cancel6^color(blue)3/cancel8^color(blue)4`

`:.3/4=3/4`

substitute `color(blue)(x=5`

`:.((color(blue)5)+5)/4=((color(blue)5)^2-(color(blue)5))/8`

`:.cancel10^color(blue)5/cancel4^color(blue)2=(25-5)/8`

`:.5/2=cancel20^color(blue)5/cancel8^color(blue)2`

`:.color(blue)(5/2=5/2`