Question

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

Answer 1

`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`