Solve the following systems of equations
[1] y = x^2 + 1
2x^2 + 3y = 28
Since y =x^2 + 1 let's substitute this into the second equation to solve for 'x'
2 x^2 + 3 (x^2+1) = 28
2 x^2 + 3x^2 + 3 = 28
5 x^2 - 25 = 0 divide by 5
x^2 - 5 = 0
x^2 = 5
x= +- sqrt 5 then y = 5 + 1 = 6 soooooo = sqrt 5 , 6 and -sqrt 5 , 6
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