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Solve for $x$ and $y$:  \\[ \\begin{eqnarray} a_1x+b_1y&=&c_1\\\\   a_2x+b_2y&=&c_2 \\end{eqnarray} \\]

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The included video describes a more direct method of solving when, for example, one of the equations gives a variable directly in terms of the other variable.

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Solve the following simultaneous equations for $x$ and $y$. Input your answers as fractions or integers, not as decimals.

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Input as a fraction or an integer not as a decimal

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Input as a fraction or an integer not as a decimal

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\\[ \\begin{eqnarray} \\simplify[std]{{a}x+{b}y}&=&\\var{c}\\\\ \\simplify[std]{{a1}x+{b1}y}&=&\\var{c1} \\end{eqnarray} \\]

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$x=\\phantom{{}}$[[0]], $y=\\phantom{{}}$[[1]]

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Input your answers as fractions or integers, not as decimals.

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\\[ \\begin{eqnarray} \\simplify[std]{{a}x+{b}y}&=&\\var{c}&\\mbox{ ........(1)}\\\\ \\simplify[std]{{a1}x+{b1}y}&=&\\var{c1}&\\mbox{ ........(2)} \\end{eqnarray} \\]
To get a solution for $x$ multiply equation (1) by {this} and equation (2) by {that}

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This gives:
\\[ \\begin{eqnarray} \\simplify[std]{{a*this}x+{b*this}y}&=&\\var{this*c}&\\mbox{ ........(3)}\\\\ \\simplify[std]{{a1*that}x+{b1*that}y}&=&\\var{that*c1}&\\mbox{ ........(4)} \\end{eqnarray} \\]
Now {aort} (4) {fromorto} equation (3) to get
\\[\\simplify[std]{({a*this}+{s6*a1*that})x={this*c}+{s6*that*c1}}\\]
And so we get the solution for $x$:
\\[x = \\simplify{{c*b1-b*c1}/{b1*a-a1*b}}\\]
Substituting this value into any of the equations (1) and (2) gives:
\\[y = \\simplify{{c*a1-a*c1}/{b*a1-a*b1}}\\]
You can check that these solutions are correct by seeing if they satisfy both equations (1) and (2) by substituting these values into the equations.

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