Question

Solve the Simultaneous equation
x^2+y^2=58
x+y=10.

Answer 1

That's the intersection of a circle and a line.

`y = 10 - x`

`x^2 + (10 - x)^2 = 58`

`x^2 + 100 -20 x + x^2 = 58`

`2x^2 -20 x + 42 = 0`

`x^2 -10 x + 21 = 0`

`(x-7)(x-3)=0`

`x = 7 \quad \textrm{or} \quad x=3`

By the symmetry of the problem when x is one solution y is the other:

(x,y) = (7,3) or (3,7)

Check:

`7^2+3^2=49+9=58 \quad\checkmark`

Answer 2

From the secon equation, derive

`y = 10-x`

Substitute in the first equation:

`x^2+(10-x)^2 = 58`
Expand and simplify:

`x^2 - 10 x + 21 = 0</span>`
Solve with the quadratic equation to get
`x=3` or
`x=7`
Derive back the values for
`y`:

`x=3 \implies y=7`

`x=7 \implies y=3`
which makes sense, since the problem is symmetric in
`x` and
`y`