Lets rewrite our two equestion,

x²+y²=29.....i

y-x=3..........ii

Next we take our equestion ii and equet y in terms of x as follows,

y-x=3

y=x+3... Now we substitute y term in equestion i

:. x²+y²=29 where there's value of y we replace it with x+3

x²+(x+3) (x+3) =29

we rewrite the equestion in quadratic format

x²+x²+3x+3x+9=29

2x²+6x+9=29

2x²+6x+9-29=0

2x²+6x-20=0......now we have formed our quadratic equestion

To solve the equestion we may use factorisation method or completing square method. So in our case here, I prefer to use completing square method because the our equestion doesn't have perfect squeres. Which is [- b±(b²_4ab)¹/²]/2a

Next we substitute the coefficients of x in our equestion above,

where a=2, b=6, and c=-20

[-(6)±(6²-{4×2×-20})¹/²]/2×2

{-6±(36+160)¹/²}/4

{-6±(196)¹/²}/4 where the root of 196=14

(-6±14)/4

x=(-6+14)/4=2

x=(-6-14)/4=-5

Now we have the possible values of x 2&-5, then we substitute these values on our y equestion

y=x+3

y=2+3

y=5

y=-5+3

y=-2

Therefore our values for both x and y will be

x=2 or - 5

y=5 or - 2 This is because our simultaneous equestion had a quadratic equestion.