Question

How do you solve `1/(x+3)+1/(x+5)=1`?

Answer 1

There are two values of `x` that make this equation true. They are `x=-3-sqrt(2)` and `x=-3+sqrt(2)`.

Explanation:

`1/(x+3)+1/(x+5)=1`

Lots of ways to solve this. I'm going to try to do it the easiest way I know how.

First subtract `1/(x+5)` from both sides.

`1/(x+3)=1-1/(x+5)`

Now combine the two terms on the right-hand side of the equation.

`1/(x+3)=(x+5-1)/(x+5)`

Simplify the numerator on the right-hand side.

`1/(x+3)=(x+4)/(x+5)`

Multiply both sides by `x-5`.

`(x+5)/(x+3)=x+4`

Multiply both sides by `x+3`.

`x+5=(x+4)(x+3)`

Now expand the right hand side.

`x+5=x^2+7x+12`

Put this quadratic equation in standard form.

`x^2+6x+7=0`

Use the quadratic formula to determine `x`.

`x=(-6pmsqrt(6^2-4(1)(7)))/(2(1))=(-6pmsqrt(8))/2=-3pmsqrt(2)`

So there are two values of `x` that make this equation true. They are `x=-3-sqrt(2)` and `x=-3+sqrt(2)`.

Answer 2

`x = -3+-sqrt(2)`

Explanation:

Given: `1/(x+3) + 1/(x+5) = 1`

One way to solve is to eliminate the denominators by multiplying both sides of the equation by the common denominator.

The common denominator = `(x+3)(x+5)`

`(x+3)(x+5)(1/(x+3) + 1/(x+5) = 1)`

Distribute and cancel:
`(cancel(x+3)(x+5))/cancel(x+3) + ((x+3)cancel(x+5))/cancel(x+5) = (x+3)(x+5)`
:
Add like terms and use FOIL
`x+5 + x + 3 = (x+3)(x+5)`

`2x + 8 = x^2 + 5x + 3x + 15`

`2x + 8 = x^2 + 8x + 15`

Subtract `2x` and `8` from both sides:
`x^2 + 6x + 7 = 0`

Use the quadratic formula to solve for `x`:

`x = (-B +- sqrt(B^2 - 4AC))/(2A)`,

where the equation is in the form `Ax^2 + Bx + C = 0`

`x = (-6 +- sqrt(36 - 4*1*7))/2`

`x = -3 +-sqrt(8)/2 = -3 +- (sqrt(4) sqrt(2))/2`

`x = =-3 +- (cancel(2) sqrt(2))/cancel(2)`

`x = -3 +- sqrt(2)`