Question

Solve each system of equations algebraically.
`y = {x}^(2) + 4 \\ y = 2x + 7`

Answer 1

From the problem, we two equations :

`\begin{gathered} y=x^2+4 \\ y=2x+7 \end{gathered}`

Since both equation are defined as y in terms of x, we can equate both equations.

`\begin{gathered} y=y \\ x^2+4=2x+7^{} \end{gathered}`

Simplify and solve for x :

`\begin{gathered} x^2+4=2x+7 \\ x^2-2x+4-7=0 \\ x^2-2x-3=0 \end{gathered}`

Factor completely :

`\begin{gathered} x^2-2x-3=0 \\ (x-3)(x+1)=0 \end{gathered}`

Equate both factors to 0 then solve for x :

x - 3 = 0

x = 3

x + 1 = 0

x = -1

We have two values of x, x = 3 and -1

Substitute x = 3 and -1 to any of the equation, let's say equation 2 :

For x = 3

y = 2x + 7

y = 2(3) + 7

y = 6 + 7

y = 13

One solution is (3, 13)

For x = -1

y = 2x + 7

y = 2(-1) + 7

y = -2 + 7

y = 5

The other solution is (-1, 5)

The answers are (3, 13) and (-1, 5)