Their is the system of linear equations \(\frac{1}{3} x+\frac{1}{4} y=1\) and \(2x-3y=-30\).
The solution of the given system of equation is 8. The correct answer is option D.
Given equations:
\(\frac{1}{3} x+\frac{1}{4} y=1\)
\(2x-3y=-30\)
Let
\(\frac{1}{3} x+\frac{1}{4} y=1\) ...........(i)
\(2x-3y=-30\) ..........(ii)
Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficients.
Multiply the equation (i) by 12 so they do have the same coefficient.
\(12(\frac{1}{3} x+\frac{1}{4} y=1)\)
\(4x+3y=12\) .............(iii)
Adding equation (ii) and equation (iii)
\(2x-3y=-30\\4x+3y=12\)
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\(6x=-18\\x=\frac{-18}{6}\\x=-3\)
Substitute x = -3 into one of the original equations to find y.
\(2(-3)-3y=-30\\-6-3y=-30\\-3y=-30+6\\y=\frac{-24}{-3} \\y=8\)
Therefore, the answer is the option D
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