X - 2y = 3
<span>4x^2 - 5xy + 6y = 3
lets solve for x the first and substitute in the second:
x = 3 + 2y
4(</span>3 + 2y)^2 - 5(3 + 2y)y + 6y = 3
4(9 + 12y + 4y^2) - 15y - 10y^2 = 3
36 + 48y +16y^2<span> - 15y - </span><span>10y^2 = 3
6y^2 + 33y + 33 = 0
we can solve using the general quadratic formula:
y = (-33 +- </span>√(33^2 - 4*6*33)<span>)/12
</span>y = (-33 +- √(297)<span>)/12
</span>so there are 2 solutions for y:
y1 = (-33 + √(297)<span>)/12
</span>y2 = (-33 - √(297)<span>)/12
</span>pick one and then substitute the y value in the first equation to find x
The answer is c because it is making it smaller
No they're not proportional, and it's four*.
4/2 = 2
20/6 = 3.333333333334 or 3 1/3
