Solve the system of equations algebraically: y=x^2+4 y=2x+7

We have to solve the system of
equations:

y=x^2+4

y=2x+7

Equate
the value of y

=> x^2+4 =
2x+7

=> x^2 - 2x - 3 =
0

solving the quadratic equation we get two values of
x.

=> x^2 - 3x + x - 3 =
0

=> x(x - 3) + 1(x - 3) =
0

=> (x - 3)(x + 1) =
0

Or x = 3 and x = -1

For x =
3, y = 2x + 7 = 13

For x = -1, y = 2x + 7 =
5

Therefore the required solution is

( 3 , 13) and ( -1 ,
5)