Question

Y
`y = {2x}^(2) + 6 - 10 \\ y = - x + 5`find the solution to the system of equations

Answer 1

1) Solving the system of equations:

y=2x²+6-10 ⇒ y =2x² -4

y=-x+5 x (-1)

1.2 Rewriting it and Subtracting

y =2x²- 4

-y =+x-5

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0=2x²+x -9

2) Now solving that quadratic equation 2x²+x -9=0, using the Resolutive formula:

`\begin{gathered} \Delta=b^2-4ac \\ \Delta=(1)^(2)-4(2)(-9) \\ \Delta=1+72 \\ \Delta=73 \end{gathered}`
`\begin{gathered} x=\frac{-b\pm\sqrt[]{\Delta}}{2a} \\ x=\frac{-1\pm\sqrt[]{73}}{4} \\ x_1=\frac{-1-\sqrt[]{73}}{4} \\ x_2=\frac{-1+\sqrt[]{73}}{4} \end{gathered}`

3) Now plugging in back into the II equation y=-x+5

`\begin{gathered} y=-(\frac{-1-\sqrt[]{73}}{4})+5 \\ y=\frac{1+\sqrt[]{73}}{4}+5 \\ y=\frac{21+\sqrt[]{73}}{4} \end{gathered}`

3.1Plugging the x solutions into y=2x²-4

`\begin{gathered} y=2(\frac{-1-\sqrt[]{73}}{4})^2-4 \\ y=\frac{21+\sqrt[]{73}}{4} \\ y=2(\frac{-1+\sqrt[]{73}}{4})^2-4\text{ =}\frac{21-\sqrt[]{73}}{4} \end{gathered}`

After checking in both of the original equations, there is only one valid x solution.

So the solution to that system is

`x=\frac{-1-\sqrt[]{73}}{4}\text{ and }y=\frac{21+\sqrt[]{73}}{4}`