what is the value of y in the solution to the system of equations?\n$\frac{1}{3}x+\frac{1}{4}y = 1$\n$2x…

Explanation:

Step1: Eliminate x from the equations

First, multiply the first - equation $\frac{1}{3}x+\frac{1}{4}y = 1$ by 6 to get $2x+\frac{3}{2}y=6$. The second equation is $2x - 3y=-30$. Subtract the second equation from the new - formed first equation: $(2x+\frac{3}{2}y)-(2x - 3y)=6-(-30)$. Expand the left - hand side: $2x+\frac{3}{2}y - 2x + 3y=6 + 30$. Combine like terms: $\frac{3}{2}y+3y=36$. Rewrite $3y$ as $\frac{6}{2}y$, then $\frac{3}{2}y+\frac{6}{2}y=36$, so $\frac{3 + 6}{2}y=36$, or $\frac{9}{2}y=36$.

Step2: Solve for y

Multiply both sides of $\frac{9}{2}y=36$ by $\frac{2}{9}$: $y = 36\times\frac{2}{9}$. $y = 8$.

Answer:

8