Question

Y= x^2 + 2x + 1 , y= x+ 3

Answer 1

Y=(x+1)^2
y=x+3

x^2 + 2x + 1 = x+ 3
x^2 + x - 2 = 0

(1;4)
(-2;1)

Answer 2

To solve the system of equations, we can set the expressions for y equal to each other and solve for x:

x^2 + 2x + 1 = x + 3

To simplify the equation, let's move all the terms to one side:

x^2 + 2x - x + 1 - 3 = 0

Combining like terms:

x^2 + x - 2 = 0

Now we can factor the quadratic equation:

(x + 2)(x - 1) = 0

Setting each factor equal to zero and solving for x:

x + 2 = 0 --> x = -2 x - 1 = 0 --> x = 1

So the system of equations is satisfied when x = -2 and x = 1.

Now we can substitute these values of x back into either of the original equations to find the corresponding y-values. Let's use the second equation:

For x = -2: y = -2 + 3 y = 1

For x = 1: y = 1 + 3 y = 4

Therefore, the solution to the system of equations is (x, y) = (-2, 1) and (1, 4).