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The forces $F_1$ and $F_2$ in a system are described by the simultaneous equations given below.

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Solve for $F_1$ and $F_2$, giving your answers to $1$ decimal place where necessary.

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$\\simplify{{a}F_1 + {b}F_2 = {c}}$

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$\\simplify{{a1}F_1 + {b1}F_2 = {c1}}$

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$F_1$=  [[0]]

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$F_2$=  [[1]]

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Have a look at this youtube video to understand simultaneous equations better: Simultaneous Equations

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Multiply the first equation by $\\var{b1}$ and the second equation by $\\var{b}$ so they both have the same $F_2$ coefficient:

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\\begin{align}
\\simplify{{a*b1}F_1+{b*b1}F_2} &= \\var{c*b1} \\\\
\\simplify{{a1*b}F_1+{b1*b}F_2} &= \\var{c1*b}
\\end{align}

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(It is also fine to multiply the first equation by $~\\var{a1}$ and the second equation by $~\\var{a}$ in order to make the $F_1$ coefficients the same.)

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Then, subtract the second equation from the first to get

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\\[ \\simplify[std]{{a*b1-a1*b}F_1} = \\var{c*b1-c1*b} \\]

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So $F_1 = \\simplify[std]{{(c*b1-c1*b)/(a*b1-a1*b)}}$.

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Substitute this value of $F_1$ into the first (or second, but here we use the first) equation and rearrange to obtain $F_2$:

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\\begin{align}
\\simplify[std]{{a}*{(c*b1-c1*b)/(a*b1-a1*b)} + {b}F_2} &= \\var{c} \\\\
\\simplify[std]{{b}F_2} &= \\simplify[std]{{c}-{a*(c*b1-c1*b)/(a*b1-a1*b)}} \\\\
F_2 &= \\simplify[std]{{(c-a*(c*b1-c1*b)/(a*b1-a1*b))/b}}
\\end{align}

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