Question

How do you solve for `x` in `1/x=1/y+1/z`?

Answer 1

Put on a common denominator:

`(yz)/(xyz) = (xz)/(xyz) + (xy)/(xyz)`

Now that we're on equivalent denominators we can eliminate and just work with the numerators.

`yz = xz + xy`

Factor out an x:

`yz = x(z + y)`

`(yz)/(z + y) = x`

Hopefully this helps!

Answer 2

`x = (yz)/(z+y)`

Explanation:

The biggest problem is that `x` is in the denominator. The fraction cannot be inverted, because there are 2 terms on the right hand side. If the two fractions are added to give one term, then we can Invert both sides.

`1/x = 1/y+ 1/z`

`1/x = (z + y)/(yz)`

`x = (yz)/(z+y)`

OR:

We can get rid of the denominators altogether by multiplying each term by the LCM of the denominators.

`(color(red)(xyz)xx1)/x = (color(red)(xyz)xx1)/y+ (color(red)(xyz)xx1)/z`

`(cancelxyzxx1)/cancelx = (xcancelyzxx1)/cancely+ (xycancelzxx1)/cancelz`

`yz = xz + xy" factorise x"`

`yz = x(z+y)`

`x = (yz)/(y+z)`

Answer 3

Although not the neatest approach, we can take the reciprocal of both sides from the get-go:

`x=1/(1/y+1/z)`

Multiply the fraction by `(yz)/(yz)`:

`x=1/(1/y+1/z)*((yz)/(yz))=(yz)/(yz(1/y+1/z))=(yz)/(z+y)=(yz)/(y+z)`