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Solve a system of three simultaneous linear equations

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Input the value of \\(x\\) that satisfies the three equations.

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\\(x = \\) [[0]]

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Input the value of \\(y\\) that satisfies the three equations.

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\\(y = \\) [[1]]

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Input the value of \\(z\\) that satisfies the three equations.

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\\(z = \\) [[2]]

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(i)    \\(\\var{a1}x+2y+4z=\\var{r1}\\)

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(ii)   \\(2x+\\var{b1}y+3z=\\var{r2}\\)

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(iii)  \\(5x+6y+\\var{c1}z=\\var{r3}\\)

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First reduce the three equations in three unknowns to a two equations in two unknowns problem by eliminating one of the variables.

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We can eliminate \\(x\\) using equations (i) and (ii)

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2*(i)     \\(\\simplify{2*{a1}}x+4y+8z=\\simplify{2*{r1}}\\)

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\\(\\var{a1}\\)*(ii)    \\(\\simplify{2*{a1}}x+\\simplify{{a1}*{b1}}y+\\simplify{3*{a1}}z=\\simplify{{a1}*{r2}}\\)

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Subtracting gives us a new equation

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(iv)    \\(\\simplify{(4-{a1}{b1})y+(8-3*{a1})z}=\\simplify{2*{r1}-{a1}*{r2}}\\)

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We can also eliminate \\(x\\) using equations (ii) and (iii)

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5*(ii)    \\(10x +\\simplify{5*{b1}}y+15z=\\simplify{5*{r2}}\\)

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2*(iii)   \\(10x+12y+\\simplify{2*{c1}}z=\\simplify{2*{r3}}\\)

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Subtracting gives us another new equation

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(v)     \\(\\simplify{(5*{b1}-12)y+(15-2*{c1})z}=\\simplify{5*{r2}-2*{r3}}\\)

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We could then eliminate the \\(y\\) from these two new equations

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\\(\\simplify{5*{b1}-12}\\)*(iv)    \\(\\simplify{(5*{b1}-12)*(4-{a1}{b1})y+(5*{b1}-12)*(8-3*{a1})z}=\\simplify{(5*{b1}-12)*(2*{r1}-{a1}*{r2})}\\)

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\\(\\simplify{4-{a1}{b1}}\\)*(v)    \\(\\simplify{(4-{a1}{b1})*(5*{b1}-12)y+(4-{a1}{b1})*(15-2*{c1})z}=\\simplify{(4-{a1}{b1})*(5*{r2}-2*{r3})}\\)

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Subtracting gives us

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\\(\\simplify{(5*{b1}-12)*(8-3*{a1})-(4-{a1}{b1})*(15-2*{c1})}z=\\simplify{(5*{b1}-12)*(2*{r1}-{a1}*{r2})-(4-{a1}{b1})*(5*{r2}-2*{r3})}\\)

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Thus

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\\(z=\\frac{\\simplify{(5*{b1}-12)*(2*{r1}-{a1}*{r2})-(4-{a1}{b1})*(5*{r2}-2*{r3})}}{\\simplify{(5*{b1}-12)*(8-3*{a1})-(4-{a1}{b1})*(15-2*{c1})}}=\\simplify{decimal{((5*{b1}-12)*(2*{r1}-{a1}*{r2})-(4-{a1}*{b1})*(5*{r2}-2*{r3}))/(  (5*{b1}-12)*(8-3*{a1})-(4-{a1}*{b1})*(15-2*{c1}))}}\\)

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We can now back substitute this value for \\(z\\) into equation (iv) to find the correct value for \\(y\\) and then back substitute both these values into equation (i) to calculate \\(x\\). 

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", "statement": "

Solve the following system of three simultaneous linear equations:

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\\(\\var{a1}x+2y+4z=\\var{r1}\\)

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and

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\\(2x+\\var{b1}y+3z=\\var{r2}\\)

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and

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\\(5x+6y+\\var{c1}z=\\var{r3}\\)

", "type": "question", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}