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)**

Given:-

y = (x^2) + 4...........(1)

y = 2x + 7...........(2)

Putting the value of 'y' from (2) into (1) we get

2x + 7 = (x^2) + 4

or, (x^2) - 2x - 3 = 0

or, (x^2) - 3x + x - 3 = 0

or, x(x-3) + (x-3) = 0

or, (x+1)*(x-3) = 0

Thus, either x= -1 which gives y = 5

or, x = 3 which gives y = 13

y=x^2+4

y=2x+7

2x + 7 = x^2 +4

x^2 + 4 - 7 - 2x

x^2 - 2x - 3

x^2 + x - 3x - 3

(x^2 + x) (- 3x - 3)

x (x +1) -3 (x + 1 )

(x - 3)

(x + 1)

x= 3

x= -1

y=x^2+4

y=2x+7

2x + 7 = x^2 +4

combine like terms by moving the terms to the same side:

x^2 + 4 - 7 - 2x

x^2 - 2x - 3

a b c

multiply a by c 1 x -3 = -3 find factors of -3 that subtract to = b

those numbers would be -3 and 1, plug the numbers in as b

x^2 + x - 3x - 3

group

(x^2 + x) (- 3x - 3)

Factor out

x (x +1) -3 (x + 1 )

**(x - 3) (x + 1)**

**x= 3 x= -1**