Answer:
1-3x+8y
Step-by-step explanation:
Multiply y and 2
Multiply y and 1
The y just gets copied along.
2*y evaluates to 2y
Multiply x and 4
Multiply x and 1
The x just gets copied along.
The answer is x
4*x evaluates to 4x
2*y-4*x evaluates to 2y-4x
The answer is 2y-4x+8
2*y-4*x+8 evaluates to 2y-4x+8
Multiply y and 6
Multiply y and 1
The y just gets copied along.
The answer is y
6*y evaluates to 6y
2y + 6y = 8y
The answer is 8y-4x+8
2*y-4*x+8+6*y evaluates to 8y-4x+8
-4x + x = -3x
The answer is -3x+8y+8
2*y-4*x+8+6*y+x evaluates to -3x+8y+8
8 - 7 = 1
The answer is 1-3x+8y
2*y-4*x+8+6*y+x-7 evaluates to 1-3x+8y
so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant
![\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx%7D%7B9%7D~~%2C~~%5Cstackrel%7By%7D%7B-3%7D%29%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bhypotenuse%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B9%5E2%2B%28-3%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B81%2B9%7D%5Cimplies%20c%3D%5Csqrt%7B90%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
70/9
7 is equivalent 63/9, and then you add 63 and 7 to get 70/9
The answer is 31. 5 goes into 15 3 times and 5 goes into 5 1 time so its 31
