The partial variation is A
The answer for this question is 36
Answer:

Step-by-step explanation:
{y = 6x − 11
{y = x² + 4x − 10
x² + 4x − 10 = 6x − 11
- [6x − 11] - [6x − 11]
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x² - 2x + 1
![{[x - 1]}^{2} = 0](https://tex.z-dn.net/?f=%7B%5Bx%20-%201%5D%7D%5E%7B2%7D%20%3D%200)
1 = x [Plug this back into both equations above to get the y-coordinate of −5]; −5 = y
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Answer:
see explanation
Step-by-step explanation:
Given the 2 equations
x + 2y = 10 → (1)
2y² - 7y + x = 0 → (2)
Rearrange (1) expressing x in terms of y by subtracting 2y from both sides
x = 10 - 2y → (3)
Substitute x = 10 - 2y into (2)
2y² - 7y + 10 - 2y = 0
2y² - 9y + 10 = 0 ← in standard form
(y - 2)(2y - 5) = 0 ← in factored form
Equate each factor to zero and solve for y
y - 2 = 0 ⇒ y = 2
2y - 5 = 0 ⇒ 2y = 5 ⇒ y = 
Substitute these values into (3) for corresponding values of x
y = 2 → x = 10 - 4 = 6 ⇒ (6, 2 ) → B
y =
→ x = 10 - 5 = 5 ⇒ (5,
) → A
1 1/3 = 4/3... 4/3 ÷2 —> 4/3 x 1/2= 2/3
4 2/3 = 14/3... 14/3 - 2/3 = 12/3= 4