How do you solve #5/x - 2 = 2/(x+3)#?

Your equation looks like this

#5/x - 2 = 2/(x+3)#

Right from the start, you know that #x# cannot take a value that would make the two denominators equal to zero. This means that you need

#x != 0" "# and #" " x + 3 !=0 implies x != -3#

With this in mind, use the common denominator #x * (x+3)# to get rid of the denominators. Multiply the first fraction by #1 = (x+3)/(x+3)#, multiply #2# by #1 = (x(x+3))/(x(x+3))#, and the second fraction by #1 = x/x# to get

#5/x * (x+3)/(x+3) - 2 * (x(x+3))/(x(x+3)) = 2/(x+3) * x/x#

This is equivalent to

#(5 * (x+3))/(x(x+3)) - (2x(x+3))/(x(x+3)) = (2x)/(x(x+3))#

You can thus say that

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

Expand the parantheses to get

#5x + 15 - 2x^2 - 6x = 2x#

Rearrange by getting all the terms on one side of the equation

#2x^2 + 3x - 15 = 0#

Use the quadratic formula to find the two solutions

#x_(1,2) = (-3 +- sqrt(3^2 - 4 * 2 * (-15)))/(2 * 2)#

#x_(1,2) = (-3 +- sqrt(129))/4#

The two solutions to the original equation will thus be

#x_1 = (-3 - sqrt(129))/4" "# and #" "x_2 = (-3 + sqrt(129))/4#

Do a quick check to make sure that the calculations are correct - I'll do them for #x = (-3 - sqrt(129))/4#

#5/((-3 - sqrt(129))/4) - 2 = 2/( (-3-sqrt(129))/4 + 3)#

#20/(-3-sqrt(129)) = 8/(9-sqrt(129)) + 2#

#-1.39296944... = -3.39296944... + 2color(white)(x)color(green)(sqrt())#