How do you solve #7/(x-4) = 1 + 9/(x+4)#?

2 Answers
Apr 8, 2018

Answer:

#x = -8# or #x = 10#

Explanation:

first, a common denominator for all the fractions needs to be found.

this is the lowest common multiple of #1#, #x - 4#, and #x + 4#.

since #x + 4# and #x - 4# have no common factors, their lowest common multiple is the two expressions multiplied together.

this gives #(x+4)(x-4)#.

#(x+4)(x-4)# can be used as the denominator for both fractions:

#(7)/(x-4) = (7(x+4))/((x-4)(x+4))#

#(9)/(x+4) = (9(x-4))/((x-4)(x+4))#

#1 = ((x-4)(x+4))/((x-4)(x+4))#

#(7)/(x-4) = 1 + (9)/(x+4)#

# (7(x+4))/((x-4)(x+4)) = ((x-4)(x+4))/((x-4)(x+4)) + (9(x-4))/((x-4)(x+4))#

the denominator can then be taken away by multiplying everything by #(x-4)(x+4):#

#7(x+4)= (x-4)(x+4) + 9(x-4)#

then the brackets can be expanded:

#7(x+4) = 7x + 28#

#(x-4)(x+4) = x^2 - 16#

(difference of two squares identity: #(a+b)(a-b) = a^2-b^2#)

#9(x-4) = 9x - 36#

#(7x + 28) = (x^2-16) + (9x-36)#

#7x+28 = x^2-16 + 9x - 36#

collect like terms:

#7x + 28 = x^2 + 9x - 16 - 36#

#7x + 28 = x^2 + 9x - 52#

#x^2 + 9x - 52 = 7x + 28#

then subtract #7x:#

#x^2 + 2x - 52 = 28#

subtract #28# so that #0# is on the right-hand side:

#x^2 - 2x - 80 = 0#

this forms a quadratic eqaution that can be factorised.

to do this, find two numbers that add to make #-2#

and multiply to make #-80#.

#8 + (- 10) = 8 - 10 = -2#
#8 * -10 = -80#

#x^2 - 2x - 80 = (x+8)(x-10)#

#(x+8)(x-10) = 0#

for the number on the right-hand side to be #0#, at least one of the numbers in brackets has to be #0#.

either #x+8 = 0# or #x - 10 = 0#

if #x + 8 = 0#, then #x = -8#.

if #x - 10 = 0#, then #x = 10#.

this means that the two possible values for #x# are #-8# and #10#.

Apr 10, 2018

Answer:

#x# equals #8 and -10#

Explanation:

#7/(x−4)=1/(1)+9/(x+4)#     Solve for #x#

1) Give the addends on the right side a common denominator

#7/(x−4)=(x+4)/(x+4)+9/(x+4)#

2) Add the like fractions on the right by adding the numerators and keeping the common denominator

#7/(x−4)=(x+4+9)/(x+4)#

3) Combine like terms in the numerator on the right

#7/(x−4)=(x+13)/(x+4)#

4) Clear the first fraction by multiplying both sides by #x - 4# and letting the denominator on the left cancel

#7=((x-4)(x+13))/(x+4)#

5) Clear the second fraction by multiplying both sides by #x+4# and letting the right side denominator cancel

#7(x + 4)=(x−4)(x+13)#

6) Clear the parentheses by distributing the #7# and by multiplying the binomials

#7x + 28 = x^2 +9x - 52#

7) Subtract #7x# from both sides to collect all the #x# terms together

#28 = x^2 + 2x - 52#

8) Set the equation to #0# by subtracting #28# from both sides

#x^2 + 2x - 80 = 0#

9) Factor

#(x + 10)(x - 8) = 0#

10) Set the factors equal to #0# and solve for #x#

#x + 10 = 0#
#x = -10#

#x - 8 = 0#
#x = 8#

Answer:
#(8,-10)#

#color(white)(mmmmmmmm)#―――――――――

Check

Sub in #8# in the original equation to be sure it still is an equality
#7/(x−4)=1/(1)+9/(x+4)#

#7/(8−4)# should equal #1/(1)+9/(8+4)#

#7/(4)# should still equal #12/(12)+9/(12)#

#7/(4)# should still equal #21/(12)#

Reduce the fraction to lowest terms

#7/4# does equal #7/4#

#Check#