Question

what is the value of y in the solution to the system of equations?
\\(\frac{1}{3}x + \frac{1}{4}y = 1\\)
\\(2x - 3y = -30\\)
\\(\circ -8\\)
\\(\circ -3\\)
\\(\circ 3\\)
\\(\circ 8\\)

Explanation:

Step1: Eliminate fractions in first equation

Multiply the first equation \(\frac{1}{3}x+\frac{1}{4}y = 1\) by 12 (LCM of 3 and 4) to get:
\(12\times\frac{1}{3}x + 12\times\frac{1}{4}y=12\times1\)
\(4x + 3y=12\)

Step2: Add the two equations

Now we have the system:
\(

$$\begin{cases}4x + 3y=12\\2x - 3y=-30\end{cases}$$

\)

Add the two equations to eliminate \(y\):
\((4x + 3y)+(2x - 3y)=12+(-30)\)
\(6x=-18\)

Step3: Solve for \(x\)

Divide both sides by 6:
\(x=\frac{-18}{6}=-3\)

Step4: Substitute \(x\) to find \(y\)

Substitute \(x = - 3\) into \(4x + 3y=12\):
\(4(-3)+3y=12\)
\(-12 + 3y=12\)

Add 12 to both sides:
\(3y=12 + 12=24\)

Divide by 3:
\(y=\frac{24}{3}=8\)

Answer:

8