Question

Solve the following system of equations and show all worky = x2 + 3y = x + 5

Answer 1

The given system of equations is

`\begin{gathered} y=x^2+3\rightarrow(1) \\ y=x+5\rightarrow(2) \end{gathered}`

Substitute y in equation (1) by equation (2)

`x+5=x^2+3`

Switch the 2 sides

`x^2+3=x+5`

Subtract x from both sides

`\begin{gathered} x^2-x+3=x-x+5 \\ x^2-x+3=5 \end{gathered}`

Subtract 5 from both sides

`\begin{gathered} x^2-x+3-5=5-5 \\ x^2-x-2=0 \end{gathered}`

Factor the left side into 2 factors

`\begin{gathered} x^2=(x)(x) \\ -2=(-2)(1) \\ (x)(-2)+(x)(1)=-2x+x=-x\rightarrow(middle\text{ term\rparen} \\ x^2-x-2=(x-2)(x+1) \end{gathered}`

The factors are (x - 2) and (x + 1)

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

Equate each factor by 0 to find the values of x

`\begin{gathered} x-2=0 \\ x-2+2=0+2 \\ x=2 \end{gathered}`
`\begin{gathered} x+1=0 \\ x+1-1=0-1 \\ x=-1 \end{gathered}`

Substitute the values of x in equation (2) to find the values of y

`\begin{gathered} y=2+5=7 \\ y=-1+5=4 \end{gathered}`

The solutions of the system of equations are

(2, 7) and (-1, 4)