Equations: \(x+y=7\), \(2x+y=12\)
Both equations are simultaneous linear equations in two variables because both linear equations have two similar variables.
Solve: \(x + y = 6\) and \(2x + y = 8\)
From the graph above, the point of intersection is \((2,4) \).
Thus, the solution is \(x=2\) and \(y=4\).
Solve: \(x – 3y = 7\) and \(5x + 2y = 1\)
\(\begin{aligned} x-3y&=7....\boxed{1} \\\\5x+2y&=1....\boxed{2} \end{aligned}\)
From \(\boxed{1}\), \(x=7+3y....\boxed{3}\).
Substitute \(\boxed{3}\) into \(\boxed{2}\),
\(\begin{aligned} 5(7+3y)+2y&=1 \\\\35+15y+2y&=1 \\\\35+17y&=1 \\\\17y&=1-35 \\\\17y&=-34 \\\\y&=-2. \end{aligned}\)
As \(y=-2\),
\(\begin{aligned} x&=7+3(-2) \\\\&=1. \end{aligned}\)
Solve: \(x + 2y = 9\) and \(3x – 2y = 15\)
\(\begin{aligned} x+2y&=9....\boxed{1} \\\\3x-2y&=15....\boxed{2} \end{aligned}\)
Eliminate the variable \(y\) by adding \(\boxed{1}\) and \(\boxed{2}\).
\(\begin{aligned} x+2y&=9 \\\\+\quad3x-2y&=15 \\\underline{\hspace{2cm}}&\underline{\hspace{2cm}} \\4x+0&=24 \\\\4x&=24 \\\\x&=6. \end{aligned}\)
Substitute \(x=6\) into \(\boxed{1}\),
\(\begin{aligned} 6+2y&=9 \\\\2y&=9-6 \\\\2y&=3 \\\\y&=\dfrac{3}{2}. \end{aligned}\)
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