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Consider the following system of equations:

\n

\\begin{align} 
\\simplify{{a}x+{b}y}&=\\var{c}\\text{,}\\\\
\\simplify{{a1}x+{b1}y}&=\\var{c1}\\text{.}
\\end{align}

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Solve to find the values of $x$ and $y$. 

\n

$x=$ [[0]]

\n

$y=$ [[1]]

\n

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Multiply the first equation by $\\var{b1}$ and the second equation by $\\var{b}$ so they both have the same $y$ coefficient:

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\\begin{align}
\\simplify{{a*b1}x+{b*b1}y} &= \\var{c*b1} \\\\
\\simplify{{a1*b}x+{b1*b}y} &= \\var{c1*b}
\\end{align}

\n

Next, subtract the second equation from the first to get

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\\[ \\simplify[std]{{a*b1-a1*b}x} = \\var{c*b1-c1*b} \\]

\n

So $x = \\displaystyle \\simplify[std]{{(c*b1-c1*b)/(a*b1-a1*b)}}$.

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Substitute this value of $x$ into the first equation and rearrange to obtain $y$:

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\\begin{align}
\\simplify[std]{{a}*{(c*b1-c1*b)/(a*b1-a1*b)} + {b}y} &= \\var{c} \\\\
\\simplify[std]{{b}y} &= \\simplify[std]{{c}-{a*(c*b1-c1*b)/(a*b1-a1*b)}} \\\\
y &= \\simplify[std]{{(c-a*(c*b1-c1*b)/(a*b1-a1*b))/b}}
\\end{align}

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