// Numbas version: exam_results_page_options {"name": "Jo\u00ebl's copy of Solving linear simultaneous equations by elimination", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [{"name": "Part b", "variables": ["a", "b", "c", "d", "f", "g", "x2", "y2"]}, {"name": "part a", "variables": ["h", "j", "k", "l", "m", "n", "y1", "x1", "numerator", "divisor"]}], "extensions": [], "variablesTest": {"condition": "lcm(a,d)Solve this set of simultaneous equations and give your answers for $x$ and $y$ below.

\\begin{align}
\\simplify{{h}x+{k}y} &= \\var{m} \\text{,} \\\\
\\simplify{{j}x+{l}y} &= \\var{n} \\text{.}
\\end{align}

$x =$ [[0]]

$y =$ [[1]]

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Solve this set of simultaneous equations and give your answers for $x$ and $y$ below.

\\begin{align}
\\simplify{{a}x + {b}y} &= \\var{c} \\text{,} \\\\
\\simplify{{d}x + {f}y} &= \\var{g} \\text{.}
\\end{align}

$x =$ [[0]]

$y =$ [[1]]

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a)

\\begin{align}
\\var{h}x+\\var{k}y&=\\var{m}\\text{,}\\\\
\\var{j}x-\\var{l}y&=\\var{n}\\text{.}\\\\
\\end{align}

To find the solution to these equations, we need to cancel one of the unknowns.

Notice that $\\var{h}x$ in the first equation can be multiplied by $\\var{j/h}$ to match $\\var{j}x$ in the second equation. This means that we will only have to manipulate the first equation and can leave the second equation as it is.

We have to multiply the entire first equation by $\\var{j/h}$, not just the $x$ term to ensure the equation still holds.

$\\var{h}x+\\var{k}y=\\var{m}$ multiplied by $\\var{j/h}$ gives $\\var{j}x+\\var{k*(j/h)}y=\\var{m*(j/h)}.$

We now have a common $x$ term and we can cancel this by subtracting one equation from the other to find the $y$ term.

\\begin{align}
&&\\var{j}x+\\var{k*{j/h}}y&=\\var{m*(j/h)}\\\\
-&&\\var{j}x-\\var{l}y&=\\var{n}\\\\
&&\\overline{\\qquad} & \\overline{\\qquad}\\\\
&&0x+\\var{k*(j/h)+l}y&=\\var{m*(j/h)-n}\\\\[1em]
&&y&=\\frac{\\var{m*j/h-n}}{\\var{k*j/h+l}}\\\\
&&y&=\\var{y1}
\\end{align}

We can find the corresponding value of $x$ by substituting this value for $y$ back into either of the original equations.

\\begin{align}
\\var{h}x+(\\var{k}\\times\\var{y1})&=\\var{m}\\text{,}\\\\
\\var{h}x+\\var{k*y1}&=\\var{m}\\text{,}\\\\
\\var{h}x&=\\var{m-(k*y1)}\\text{,}\\\\
x&=\\var{x1}\\text{.}\\\\
\\end{align}

Therefore, $x=\\var{x1}$ and $y=\\var{y1}$.

b)

\\begin{align}
\\var{a}x+\\var{b}y&=\\var{c}\\text{,}\\\\
\\var{d}x+\\var{f}y&=\\var{g}\\text{.}\\\\
\\end{align}

To be able to solve the equations, we need to cancel one of the unknowns by manipulating the two equations so that the variable we wish to cancel is of the same value in each equation.

Although we can choose to cancel either variable, $x$ or $y$, a good rule of thumb is to look at the lowest common multiples of the coefficients for each variable and cancel the variable with the lowest LCM.

The LCM of the coefficients of the $x$ terms is $\\var{lcm(a,d)}$.

The LCM of the coefficients of the $y$ terms is $\\var{lcm(b,f)}$.

Therefore, we will choose to cancel the $x$ terms.

We need to multiply the equations individually to achieve the lowest common multiple identified.

\\begin{align}
\\simplify{ {a}x + {b}y } &= \\var{c} &\\text{multiply by } \\var{lcm(a,d)/a} \\text { to obtain } && \\simplify{ {lcm(a,d)}x + {b*lcm(a,d)/a}y} &= \\var{c*lcm(a,d)/a} \\\\
\\simplify{ {d}x + {f}y } &= \\var{g} &\\text{multiply by } \\var{lcm(a,d)/d} \\text { to obtain } && \\simplify{ {lcm(a,d)}x + {b*lcm(a,d)/d}y} &= \\var{c*lcm(a,d)/d}
\\end{align}

We now have a common $x$ term, and can cancel this by subtracting one equation from the other.

\\begin{align}
&& \\simplify{ {lcm(a,d)}x+{b*lcm(a,d)/a}y } = \\var{c*lcm(a,d)/a} \\\\
- && \\simplify{ {lcm(a,d)}x + {f*lcm(a,d)/d}y } = \\var{g*lcm(a,d)/d} \\\\
&& \\overline{\\simplify[]{ 0x+{b*lcm(a,d)/a-f*lcm(a,d)/d}y} = \\var{c*lcm(a,d)/a-g*lcm(a,d)/d}}
\\end{align}

\\begin{align}
\\var{(b*lcm(a,d)/a)-(f*lcm(a,d)/d)}y &= \\var{(c*lcm(a,d)/a)-(g*lcm(a,d)/d)}\\text{,}\\\\
y &= \\var{y2}\\text{.}
\\end{align}

We can find the corresponding value of $x$ by substituting thsi value of $y$ value back into either of the original equations.

\\begin{align}
\\simplify[]{ {a}x + {b}{y2}} &= \\var{c} \\\\
\\simplify[]{ {a}x + {b*y2}} &= \\var{c} \\\\
\\var{a}x&=\\var{c-b*y2} \\\\
x &= \\var{x2} \\text{.}
\\end{align}

Therefore, $x=\\var{x2}$ and $y=\\var{y2}$.

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This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve.

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