How do you solve #\frac { x + 5} { 2x ^ { 2} } + \frac { 1} { 4x ^ { 2} } = \frac { 4x + 8} { x ^ { 2} }#?

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Answer 1 · 2018-04-10T02:30:46

You'll want to remove the denominators first.

This means you multiply the left and right hand sides of the equation by #4x^2#.

#(x+5)/(2x^2) + 1/(4x^2) = (4x+8)/x^2#

# 2(x+5) + 1 = 4(4x+8)#

Then you expand and move the #x#s to one side and the integers to the other.

#2x+10 + 1 = 16x + 32#

# 10 + 1 -32 = 16x - 2x#

#-21 = 14x#

#x = -3/2#