how many solutions does the system of equations have? 4x=-12y 16 and x 3y=4
We can solve for x in the first equation by dividing both sides by 4: $$x=-3y+4$$
We can substitute this expression for x into the second equation: $$(-3y+4)3y=4$$
Simplifying this equation gives us a quadratic equation in standard form: $$-9y^2+12y-4=0$$
Solving for y using the quadratic formula: $$y=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
where $a=-9$, $b=12$, and $c=-4$. Plugging in these values and simplifying gives us: $$y=\frac{2}{3},\frac{2}{3}$$
Therefore, there are two possible solutions for y. Substituting these values back into the equation we solved for x gives us: $$x=-3\left(\frac{2}{3}\right)+4=2$$
Therefore, the system of equations has two possible solutions: (2,2/3) and (2,2/3).