// Numbas version: finer_feedback_settings {"name": "Solve a system of three simultaneous linear equations", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"advice": "
(i) \\(\\var{a1}x+2y+4z=\\var{r1}\\)
\n(ii) \\(2x+\\var{b1}y+3z=\\var{r2}\\)
\n(iii) \\(5x+6y+\\var{c1}z=\\var{r3}\\)
\nFirst reduce the three equations in three unknowns to a two equations in two unknowns problem by eliminating one of the variables.
\nWe can eliminate \\(x\\) using equations (i) and (ii)
\n2*(i) \\(\\simplify{2*{a1}}x+4y+8z=\\simplify{2*{r1}}\\)
\n\\(\\var{a1}\\)*(ii) \\(\\simplify{2*{a1}}x+\\simplify{{a1}*{b1}}y+\\simplify{3*{a1}}z=\\simplify{{a1}*{r2}}\\)
\nSubtracting gives us a new equation
\n(iv) \\(\\simplify{(4-{a1}{b1})y+(8-3*{a1})z}=\\simplify{2*{r1}-{a1}*{r2}}\\)
\nWe can also eliminate \\(x\\) using equations (ii) and (iii)
\n5*(ii) \\(10x +\\simplify{5*{b1}}y+15z=\\simplify{5*{r2}}\\)
\n2*(iii) \\(10x+12y+\\simplify{2*{c1}}z=\\simplify{2*{r3}}\\)
\nSubtracting gives us another new equation
\n(v) \\(\\simplify{(5*{b1}-12)y+(15-2*{c1})z}=\\simplify{5*{r2}-2*{r3}}\\)
\nWe could then eliminate the \\(y\\) from these two new equations
\n\\(\\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})}\\)
\n\\(\\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})}\\)
\nSubtracting gives us
\n\\(\\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})}\\)
\nThus
\n\\(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}))}}\\)
\nWe 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\\).
\n", "variable_groups": [], "ungrouped_variables": ["a1", "b1", "c1", "r1", "r2", "r3"], "functions": {}, "extensions": [], "variablesTest": {"condition": "", "maxRuns": 100}, "type": "question", "statement": "Solve the following system of three simultaneous linear equations:
\n\\(\\var{a1}x+2y+4z=\\var{r1}\\)
\nand
\n\\(2x+\\var{b1}y+3z=\\var{r2}\\)
\nand
\n\\(5x+6y+\\var{c1}z=\\var{r3}\\)
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\n\\(x = \\) [[0]]
\nInput the value of \\(y\\) that satisfies the three equations.
\n\\(y = \\) [[1]]
\nInput the value of \\(z\\) that satisfies the three equations.
\n\\(z = \\) [[2]]
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", "licence": "Creative Commons Attribution 4.0 International"}, "name": "Solve a system of three simultaneous linear equations", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}