How do you solve #8/(x+2)+8/2=5#?

1 Answer
May 26, 2017

Answer:

#x = 6#

Explanation:

Given: #8/(x+2) + 8/2 = 5#

One way to solve is by realizing that #8/2 = 4#, so substitute this value into the equation: #" "8/(x+2) + 4 = 5#

Simplify by subtracting #4 # from both sides of the equation:
#8/(x+2) + 4 - 4= 5 - 4#;

#8/(x+2)= 1#

Multiply both sides of the equation by #x+2#:

#cancel(x+2) * 8/(cancel(x+2)) = 1 * (x + 2)#

Simplify: #" "8 = x + 2#

Subtract both sides of the equation by #2#:

#8 - 2 = x + 2 - 2#

#6 = x#

A second way to solve is by finding a common denominator for both sides of the equation #2(x+2)#:

#8/(x+2) * 2/2 + 8/2 * (x+2)/(x+2) = 5 *(2(x+2))/(2(x+2))#

Simplify:

#(16 +8(x+2))/(2(x+2)) = (10(x+2))/(2(x+2))#

Since both denominators are equal, we can set the numerators equal to solve:

#16 +8(x+2) = 10(x+2)#

Distribute:

#16 + 8x + 16 = 10x + 20#

Add like terms on the same side:

#32 + 8x = 10x + 20#

Subtract #20# from both sides:

#32 - 20 + 8x = 10x + 20 - 20#

#12 + 8x = 10x#

Subtract #8x# from both sides: #" "12 + 8x - 8x = 10x - 8x#

Simplify: #" "12 = 2x#

Divide both sides by #2: " "12/2 = (2x)/2#

Simplify: #" "6 = x#