Question

X-6 =-1/6y^2 and 2x+y=6. Solve the linear quadratic equation

Answer 1

Answer:

$(4.5,-3)\textrm{ and }(0,6)$

Step-by-step explanation:

Given:

$x-6=-\frac{1}{6}y^{2}$

$2x+y=6$

Express $x$ in terms of $y$ and then plug in first equation to solve for $y$.

$2x+y=6\\2x=6-y\\x=\frac{6-y}{2}$

Now,

$x-6=-\frac{1}{6}y^{2}\\\frac{6-y}{2}-6=-\frac{1}{6}y^{2}\\\frac{1}{6}y^{2}+\frac{6-y}{2}-6=0$.

Multiply 6 on both sides.

$6(\frac{1}{6}y^{2}+\frac{6-y}{2}-6)=6(0)\\y^{2}+3(6-y)-36=0\\y^{2}+18-3y-36=0\\y^{2}-3y-18=0\\(y-6)(y+3)=0\\y=6\textrm{ or }y=-3$

Now, for $y=-3,x=\frac{6-(-3)}{2}=\frac{6+3}{2}=\frac{9}{2}=4.5$

For, $y=6, x=\frac{6-6}{2}=0$.

Therefore, the solutions for the system of equations are $(4.5,-3)\textrm{ and }(0,6)$