Question

Solving $\frac{2}{x}+3y=-3$ and $y-4x=\frac{7}{3}$ gives (x, y)

• A. $(-\frac{1}{2},3)$
• B. $(-1,\frac{1}{3})$
• C. $(-\frac{1}{2},\frac{1}{3})$
• D. $(-2,\frac{1}{3})$

Answer 1

The correct option is C $(-\frac{1}{2},\frac{1}{3})$

Given

$\frac{2}{x}+3y=-3$ and $y-4x=\frac{7}{3}$

Multipling first equation with x, we get

$2+3xy=-3x$ -----eq(1)

From another equation, we get, $y=\frac{7}{3}+4x$

substituting y in equation 1, we get

$2+3x(\frac{7}{3}+4x)=-3x$

$2+7x+12x^2=-3x$

$2+10x+12x^2=0$

Dividing by 12 we get $x^2+\frac{5}{6}x+\frac{1}{6}=0$

By solving this quadratic equation we get $x=-\frac{1}{3}$ or $x=-\frac{1}{2}$

After looking at given options we calculate $y$ only for $x=\frac{-1}{2}$

Thus we get $y=\frac{1}{3}$