Question

How to solve for x and y if the equations are: `(5/(x+y))-(2/(x-y))+1=0` and `(15/(x+y))+(7/(x-y))-10=0` ?

Answer 1

`x=3` and `y=2`

Explanation:

I'll use Gabriel Cramer's method to solve the system of equations.

First of all, let `x+y=p` and `x-y=q`

Then the first equation is equal to

`5/p-2/q+1=0`

or, `5/p-2/q=-1`

LET `a_1=5`, `b_1=-2`, `c_1=-1`

Also the second equation is equal to

`15/p+7/q-10=0`

or, `15/p+7/q=10`

LET `a_2=15`, `b_2=7`, `c_2=10`

Now we have,

`Delta=|(a_1,b_1),(a_2,b_2)|=|(5,-2),(15,7)|=7*5-(-2)**15`

`=35+30=65`

Also,

`Delta_1=|(c_1,b_1),(c_2,b_2)|=|(-1,-2),(10,7)|=-1*7-(-2)*10`

`=20-7=13`

Similarly, `Delta_2=|(a_1,c_1),(a_2,c_2)|=|(5,-1),(15,10)|=10*50-(-1)*15`

`=50+15=65`

According to Cramer's rule

`1/p=Delta_1/Delta`

or, `p=Delta/Delta_1=65/13=5`

Similarly , `1/q=Delta_2/Delta`

or `q=Delta/Delta_2=65/65=1`

Putting `p= x+y` and `q= x-y` we have

`x+y=5`....equation(1)

and `x-y=1`.....equation(2)

From equation(1) we have,

`x=5-y`....equation...(3)

Putting the value of `x` from e`qn(3)` to `eqn(2)` we get

`5-y-y=1`

`2y=4`

`y=2`

Putting the value of `y` in `eqn(3)` we get
`x=5-y=5-2=3`

So the value `x` is `3` and `y` is `2`.