Question

Find the solutions of the given system of equations: x2 + y2 = 5 and y = 3x + 1.

Answer 1

Substituting the second equation in the first one we get:

`x^2+(3x+1)^2=5.`

Simplifying the above equation we get:

`\begin{gathered} x^2+(3x)^2+2*(3x)(1)+1^2=5, \\ x^2+9x^2+6x+1=5, \\ 10x^2+6x+1=5. \end{gathered}`

Subtracting 5 from the above equation we get:

`\begin{gathered} 10x^2+6x+1-5=5-5, \\ 10x^2+6x-4=0. \end{gathered}`

Now, notice that:

`10x^2+6x-4=2(5x^2+3x-2)=2(5x-2)(x+1).`

Therefore:

`10x^2+6x-4=0\text{ if and only if }x=(2)/(5)\text{ or }x=-1.`

If x=2/5:

`y=3*(2)/(5)+1=(6)/(5)+1=(11)/(5).`

Therefore

`((2)/(5),(11)/(5))`

is a solution to the given system of equations.

If x=-1:

`y=3*(-1)+1=-3+1=-2.`

Therefore (-1,-2) is a solution to the given system of equations.

Answer: Third option.