Question

How do you solve `5/x + 3/y = 4` and `25/x - 2/y = 3`?

Answer 1

`{(x=5), (y = 1) :}`

Explanation:

Right from the start, you know that `x` and `y` cannot be equal to zero, since they represent denominators.

Now, start by rewriting your equations to get rid of the fractions. The common denominator will be `x * y`, so multiply each term accordingly. For the first equation you will get

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

`5y + 3x = 4xy`

For the second equation you will get

`(25y)/(xy) - (2x)/(xy) = (3xy)/(xy)`

`25y - 2x = 3xy`

Multiply the first equation by `-5` to get

`5y + 3x = 4xy | * (-5)`

`-25y -15y = -20xy`

Your two equations are now

`{(-25y - 15x = -20xy), (25y - 2x = 3xy) :}`

Notice that if you were to add these two equations together, the `y`-terms on the left sides of the equations would cancel out,

You will be left with

`-color(red)(cancel(color(black)(25y))) - 15x + color(red)(cancel(color(black)(25y))) - 2x = -20xy + 3xy`

`-17x = -17xy`

You can further simplify this to get the value of `y`

`color(red)(cancel(color(black)(-17x))) = color(red)(cancel(color(black)(-17x))) * y implies y = color(green)(1)`

Take this value of `y` into one of the two equations and find the value of `x`

`5 * 1 + 3x = 4 * x * 1`

`4x - 3x = 5 implies x = color(green)(5)`

Since you have `x!=0` and `y!=0`, the two solutions to your system of equations will be

`{(x=5), (y=1):}`

You can do a quick check to make sure that the calculations are correct

`5/5 + 3/1 = 4`

`1 + 3 = 4" "color(green)(sqrt())`

and

`25/5 - 2/1 = 3`

`5 - 2 = 3" "color(green)(sqrt())`