Question

How do you solve the simultaneous equations `x^2 + y^2=29` and `y-x=3`?

Answer 1

Use the second equation to provide an expression for `y` in terms of `x` to substitute into the first equation to give a quadratic equation in `x`.

Explanation:

First add `x` to both sides of the second equation to get:

`y = x+3`

Then substitute this expression for `y` into the first equation to get:

`29 = x^2+(x+3)^2 = 2x^2+6x+9`

Subtract `29` from both ends to get:

`0 = 2x^2+6x-20`

Divide both sides by `2` to get:

`0 = x^2+3x-10 = (x+5)(x-2)`

So `x=2` or `x=-5`

If `x=2` then `y = x+3 = 5`.

If `x=-5` then `y = x+3 = -2`

So the two solutions `(x, y)` are `(2, 5)` and `(-5, -2)`

Answer 2

`(x=-5 and y=-2) or (x=2 and y=5)`

Explanation:

Since you have both `x^2+y^2=29` and `y-x=3`,

You want to combine these two equations into one equation with a single variable, solve it and then solve for the other variable. An example on how to do this goes like this:

`y-x=3 rarr y=x+3` and we have `y^2=x^2+6x+9`

Since `x^2+y^2=29`, substitute the expression for `y^2` into this:

`2x^2+6x+9=29`, so `2x^2+6x-20=0`.

We can solve for `x` using the quadratic formula:
`x=(-6pmsqrt(36-4*2*(-20)))/(2*2)=-3/4pm1/4sqrt(196)=(-6pm14)/4`
So `x=-5` or `x=2`.

Since `y=x+3`, this gives `(x=-5 and y=-2) or (x=2 and y=5)`.