How do you solve #(7x) /(2x+5) + 1 = (10x - 3) /( 3x)#?

1 Answer
May 10, 2018

Answer:

#x=(29+-sqrt(421))/14#

Explanation:

The first thing you do is incorporate the 1 on the left side into the fraction. In order to do that you substitute it by a fraction that is the left denominator divided by itself, like this:

#(7x)/(2x+5)+(2x+5)/(2x+5)=(10x-3)/(3x)#

Now we can add the fractions on the left side. Since they have the same denominator, we only add the numerators.

#(7x+2x+5)/(2x+5)=(10x-3)/(3x)#

#(9x+5)/(2x+5)=(10x-3)/(3x)#

Now we multiply both sides of the equation to remove the #3x# from the bottom of the right side.

#(3x*(9x+5))/(2x+5)=(10x-3)/cancel(3x)*cancel(3x)#

Multiply the top on the left and we get:

#(3x*9x+3x*5)/(2x+5)=10x-3#

#(27x^2+15x)/(2x+5)=(10x-3)#

Do the same thing with the left denominator. Multiply both sides by it in order to remove it.

#cancel(2x+5)(27x^2+15x)/cancel(2x+5)=(10x-3)*(2x+5)#

Now we multiply out the right side:

#27x^2+15x=(10x-3)(2x+5)#

#27x^2+15x=20x^2+50x-6x-15#

#27x^2+15x=20x^2+44x-15#

Finally we can move everything from the right side of the equation to the left:

#27x^2+15x-20x^2-44x+15=0#

#7x^2-29x+15=0#

Now we have a single equation if the form of #ax^2+bx+c=0#

Which means we can use the quadratic formula #x=(-b+-sqrt(b^2-4ac))/(2a)# to solve for #x#

When we plug in the numbers, we get:

#x=(29+-sqrt((-29)^2-4*7*15))/(2*7)#

#x=(29+-sqrt(841-420))/14#

#x=(29+-sqrt(421))/14#