How do you solve #(6x) /(x+4) + 4 = (2x + 2) /(x - 1)#?

1 Answer
May 28, 2018

Answer:

#x=-1 1/2 vv x=2#

Explanation:

Information

In equations, you are allowed to perform operations on both sides. In order to get rid of denominators, you can multiply by the denominator on both sides. Since a number divided by itself is 1, except for 0, this removes the fraction and leaves the numerator.

The product-sum method can be used to find values of #x# when the equation is shaped like #x^2+bx+c=0#.
If you find two values whose sum is #b# and whose product is #c#, you can find #x# because
#x^2 + ("value1")x + ("value2")x + ("value1")("value2")=0#

#(x+"value1")(x+"value2")=0#

#x+"value1"=0 vv x+"value2"=0#.

Note that you can use the quadratic formula in all cases where the product-sum method is applicable, but it is much faster in a lot of cases to use this.

Calculation

#(6x)/(x+4)+4=(2x+2)/(x-1)#

Multiply both sides by #x+4# to get rid of the denominator on the left side.

#6x+4(x+4)=(2x+2)/(x-1)(x+4)#

#10x+16=(2x+2)/(x-1)(x+4)#

Multiply both sides by #x-1# to do the same for the right side.

#(10x+16)(x-1)=(2x+2)(x+4)#

#10x^2-10x+16x-16=2x^2+8x+2x+8#

#10x^2+6x-16=2x^2+10x+8#

Make the right side equal to 0 by subtracting everything on the right side from both sides.

#8x^2-4x-24=0#

Divide both sides by 8.

#x^2-1/2x-3=0#

Use the product-sum method to find #x#.

The numbers #1 1/2# and #-2# work, because

#1 1/2+(-2)=-1/2=b# and

#1 1/2*-2=-3=c#

So,

#(x+1 1/2)(x-2)=0#

#x+1 1/2=0 vv x-2=0#

#x=-1 1/2 vv x=2#

Neither of these values makes a division by zero when filled back into the equation, so they're both valid.