Question

can someone help me using the substitution method for these two equations. first equation is y= -22² Y and the second equation is y= - 3x -4

Answer 1

We substitute the 2nd equation into first, to get:

`\begin{gathered} y=-2x^2 \\ y=-3x-4 \\ \text{Putting 2nd equation in 1st:} \\ -3x-4=-2x^2 \end{gathered}`

We re-arrange the equation into a trinomial and use quadratic formula to solve for x:

`2x^2-3x-4=0`

Quadratic Formula:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

The equation has values:

a = 2, b = -3, c = -4

So, x is:

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-(-3)\pm\sqrt[]{(-3)^2-4(2)(-4)}}{2(2)} \\ x=\frac{3\pm\sqrt[]{41}}{4} \\ x=\frac{3+\sqrt[]{41}}{4},\frac{3-\sqrt[]{41}}{4} \end{gathered}`

To find value of y, we use 1st equation and plug in value of x:

`\begin{gathered} y=-2x^2 \\ y=-2(\frac{3-\sqrt[]{41}}{4})^2=\frac{-25-3\sqrt[]{41}}{4} \\ \text{and} \\ y=-2(\frac{3+\sqrt[]{41}}{4})^2=\frac{-25+3\sqrt[]{41}}{4} \end{gathered}`

We can write the solution set as:

`\begin{gathered} (\frac{3-\sqrt[]{41}}{4},\frac{-25-3\sqrt[]{41}}{4}) \\ \text{and} \\ (\frac{3+\sqrt[]{41}}{4},\frac{-25+3\sqrt[]{41}}{4}) \end{gathered}`