How do you solve #4- \frac{1}{x + 2} = \frac{7}{x + 5}#?

2 Answers
May 16, 2018

#x=-3/2# or #x=-7/2#

Explanation:

#4-1/(x+2)=7/(x+5)#

#4=7/(x+5)+1/(x+2)#

#4=((7)(x+2)+(x+5))/((x+2)(x+5))#

#4(x+2)(x+5)=7(x+2)+(x+5)#

#4(x^2+7x+10)=7x+14+x+5#

#4x^2+28x+40=8x+19#

#4x^2+20x+21=0#

#(2x+3)(2x+7)=0#

#x=-3/2# or #x=-7/2#

May 16, 2018

#x=-7/2# or #x=-3/2#

Explanation:

Given
#color(white)("XXX")4-1/(x+2)=7/(x+5)#

Since I don't like working with fractions, let's multiply everything by #(x+2)(x+5)# giving:
#color(white)("XXX")4(x+2)(x+5)-(x+5)=7(x+2)#

then spend a few lines expanding and simplifying
#color(white)("XXX")4(x^2+7x+10)-x-5=7x+14#

#color(white)("XXX")4x^2+27x+35=7x+14#

#color(white)("XXX")4x^2+20x+21=0#

If you are really good, you might be able to factor the left side;
but I will just apply the quadratic formula to get
#color(white)("XXX")x=(-20+-sqrt(20^2-4 * 4 * 21))/(2 * 4)#

#color(white)("XXX"x)=(-20+-8)/8#

#color(white)("XXX")rArr x=(-28)/8=-7/2" or "x=(-12)/8=-3/2#