Question

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Answer 1

(3, 5)

Explanation:

Solution given by jimrgrant1 is absolutely correct. I am just offering another solution strategy without using substitution.

We have the equations

`y = 2x^2 - 7x + 8 \quad\quad [1]\\\\y = 2x^2 - 18x + 41 \quad\quad [2]`

Since the left side in both equations is y, set the right sides equal to each other;

`2x^2 - 7x + 8 = 2x^2 - 18x + 41`

Subtract
`2x^2` from both sides:

`- 7x + 8 = - 18x + 41`

`\text{Add 7x to both sides:}\\`

`- 7x + 7x + 8 = - 18x +7x + 41\\\\8 = -11x + 41\\\\\text{Subtract 41 from both sides:}:\\\\8 - 41 = -11x\\\\-33 = -11x\\\\\text{Switch sides:}\\\\-11x = -33\\\\\text{Divide by -11}:\\\\(-11x)/(-11) = (-33)/(-11)\\\\x = 3\\\\`

Looking at the answer choices, we see only one answer choice has x = 3 and this is the third option. So we need not calculate for y.

The correct answer is (3, 5)

{If curious we can plug into any of the two equations and see that y evaluates to 5

This has been done for you by user jimrgrant1 so I won't repeat

Answer 2

(3, 5 )

Explanation:

y = 2x² - 7x + 8 → (1)

y = 2x² - 18x + 41 → (2)

substitute y = 2x² - 7x + 8 into (2)

2x² - 7x + 8 = 2x² - 18x + 41 ( subtract 2x² from both sides )

- 7x + 8 = - 18x + 41 ( add 18x to both sides )

11x + 8 = 41 ( subtract 8 from both sides )

11x = 33 ( divide both sides by 11 )

x = 3

substitute x = 3 into either of the 2 equations and solve for y

substituting into (1)

y = 2(3)² - 7(3) + 8 = 2(9) - 21 + 8 = 18 - 13 = 5

solution is (3, 5 )