How do you solve #1/(x+3)+1/(x+5)=1#?

2 Answers
Apr 26, 2018

Answer:

There are two values of #x# that make this equation true. They are #x=-3-sqrt(2)# and #x=-3+sqrt(2)#.

Explanation:

#1/(x+3)+1/(x+5)=1#

Lots of ways to solve this. I'm going to try to do it the easiest way I know how.

First subtract #1/(x+5)# from both sides.

#1/(x+3)=1-1/(x+5)#

Now combine the two terms on the right-hand side of the equation.

#1/(x+3)=(x+5-1)/(x+5)#

Simplify the numerator on the right-hand side.

#1/(x+3)=(x+4)/(x+5)#

Multiply both sides by #x-5#.

#(x+5)/(x+3)=x+4#

Multiply both sides by #x+3#.

#x+5=(x+4)(x+3)#

Now expand the right hand side.

#x+5=x^2+7x+12#

Put this quadratic equation in standard form.

#x^2+6x+7=0#

Use the quadratic formula to determine #x#.

#x=(-6pmsqrt(6^2-4(1)(7)))/(2(1))=(-6pmsqrt(8))/2=-3pmsqrt(2)#

So there are two values of #x# that make this equation true. They are #x=-3-sqrt(2)# and #x=-3+sqrt(2)#.

Apr 26, 2018

Answer:

#x = -3+-sqrt(2)#

Explanation:

Given: #1/(x+3) + 1/(x+5) = 1#

One way to solve is to eliminate the denominators by multiplying both sides of the equation by the common denominator.

The common denominator = #(x+3)(x+5)#

#(x+3)(x+5)(1/(x+3) + 1/(x+5) = 1)#

Distribute and cancel:
#(cancel(x+3)(x+5))/cancel(x+3) + ((x+3)cancel(x+5))/cancel(x+5) = (x+3)(x+5)#
:
Add like terms and use FOIL
#x+5 + x + 3 = (x+3)(x+5)#

#2x + 8 = x^2 + 5x + 3x + 15#

#2x + 8 = x^2 + 8x + 15#

Subtract #2x# and #8# from both sides:
#x^2 + 6x + 7 = 0#

Use the quadratic formula to solve for #x#:

#x = (-B +- sqrt(B^2 - 4AC))/(2A)#,

where the equation is in the form #Ax^2 + Bx + C = 0#

#x = (-6 +- sqrt(36 - 4*1*7))/2 #

#x = -3 +-sqrt(8)/2 = -3 +- (sqrt(4) sqrt(2))/2#

#x = =-3 +- (cancel(2) sqrt(2))/cancel(2)#

#x = -3 +- sqrt(2)#