How do you solve `x^2+y^2=25` and `y+5=1/2x^2` using substitution?

As `1/2x^2=y+5`, we have `x^2=2*(y+5)=2y+10`.

Putting this in `x^2+y^2=25`, we get `2y+10+y^2=25` or

`y^2+2y-15=0` i.e. `y^2+5y-3y-15=0` or

`y(y+5)-3(+5)=0` or `(y-3)(y+5)=0` i.e. `y=3` or `y=-5`

Hence `x^2=2*3+10=16` i.e. `x=+-4` that is

`x^2=2*(-5)+10=0` i.e. `x=0`

Hence solution set for `(x,y)` is `(0,-5)`, `(4,3)` and `(-4,3)`

Just an observation: `x^2+y^2=25` is the equation of a circle with the centre at the origin.

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Given:
`x^2+y^2=25` ...............................(1)
`y+5=1/2 x^2` ................................(2)

`color(blue)("Solving for "y)`

Rewrite (1) as: `x^2=25-y^2" " ...........(1_a)`
Rewrite (2) as: `x^2=2y+10" " ...........(2_a)`

Equate `(1_a)" to "(2_a)" through "x^2` This is the equivalent of substituting for `x^2`

`25-y^2=2y+10`

This is the same as

`y^2+2y-15=0`

`(y+5)(y-3)=0`

`" "color(blue)(=> y= -5" or " +3)` .........................(3)

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`color(blue)("Solving for "x)`

Substituting into equation `(2_a)`. The y is not squared!

`color(brown)("condition 1 "y=-5)`

`x^2=2y+10" " -> " "x^2=2(-5)+10`

`" "color(green)(x^2=0" " =>" " x=0)`
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`color(brown)("condition 2 "y=+3)`

`x^2=2(3)+10 = 16`

`" "color(green)(x=+-4)`
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`color(blue)("Putting it all together")`

The point of intersection are:

`color(blue)((x,y)->( -4,3)" ; "(+4,3)" ; "(0,-5))`