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There are several ways to solve linear simultaneous equations, here we will focus on equalling the coefficients of $x$ and then subtracting the equations.

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To get the same coefficient of $x$ in both equations we can multiply the first by $\\simplify{{b1}}$ and the second by $\\simplify{{a1}}$:

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\\[\\simplify{{a1}{b1}x+{a2}{b1}y={a3}{b1}} \\quad \\text{and} \\quad \\simplify{{b1}{a1}x+{b2}{a1}y={b3}{a1}}\\]

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Subtracting the second from the first gives:

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\\[ \\simplify{({a2}{b1}-{b2}{a1})y=({a3}{b1}-{b3}{a1})}\\]

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Dividing through by $\\simplify{{a2}{b1}-{b2}{a1}}$ gives:

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\\[y=\\simplify{{y}}\\]

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Subbing this value of $y$ into either of the original equations gives:

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\\[x= \\simplify{{x}}\\]

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Solve the following simultaneous equations:

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\\[ \\simplify{{a1}x+{a2}y={a3}}\\] \\[\\simplify{{b1}x+{b2}y={b3}}\\]

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$x =$ [[0]]

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$y =$ [[1]]

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