Question

how do you solve the system of linear equation y=2x-3y=x^2-3A. (0,3) and (2,0)B. (-1,-5) and (4,5)C. (3,6) and (-1,6)D. (0,-3) and (2,1)

Answer 1

We want to solve the system of equations;

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

Substituting, we have;

`\begin{gathered} 2x-3=x^2-3 \\ x^2=2x\text{ \lparen x=0 is a solution \rparen} \\ divide\text{ both sides by x} \\ x=2 \end{gathered}`

Inserting the values, we have;

`\begin{gathered} when\text{ x = 0;} \\ y=2(0)-3 \\ y=-3 \end{gathered}`
`\begin{gathered} when\text{ x = 2} \\ y=2(2)-3 \\ y=1 \end{gathered}`

Thus, the solutions are;

`(0,-3)\text{ }and\text{ }(2,1)`