Maths Support: Simultaneous equations

Question 1

Solve:

\[\begin{eqnarray*} \simplify{{a}x+{b}y}&=&\var{c}\\\\\simplify{{a1}x+{b1}y}&=&\var{c1}\end{eqnarray*}\]

$x=$ $\displaystyle{}$ question.score feedback.none Expected answer:

$y=$ $\displaystyle{}$ question.score feedback.none Expected answer:

input your answers as fractions and not as decimals.

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Advice

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

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

Next, subtract the second equation from the first to get

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

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

Substitute this value of $x$ into the first equation and rearrange to obtain $y$:

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

Question 2

Solve the following simultaneous equations for $x$ and $y$. Input your answers as fractions or integers, not as decimals.

\[ \begin{eqnarray} \simplify[std]{{a}x+{b}y}&=&\var{c}\\ \simplify[std]{{a1}x+{b1}y}&=&\var{c1} \end{eqnarray} \]

$x=\phantom{{}}$ $\displaystyle{}$ question.score feedback.none Expected answer: , $y=\phantom{{}}$ $\displaystyle{}$ question.score feedback.none Expected answer:

Input your answers as fractions or integers, not as decimals.

Click Show steps for a video that describes a more direct method of solving when, for example, one of the equations gives a variable directly in terms of the other variable.

(Your score will not be affected.)

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Advice

\[ \begin{eqnarray} \simplify[std]{{a}x+{b}y}&=&\var{c}&\mbox{ ........(1)}\\ \simplify[std]{{a1}x+{b1}y}&=&\var{c1}&\mbox{ ........(2)} \end{eqnarray} \]
To get a solution for $x$ multiply equation (1) by 1 and equation (2) by 2

This gives:
\[ \begin{eqnarray} \simplify[std]{{a*this}x+{b*this}y}&=&\var{this*c}&\mbox{ ........(3)}\\ \simplify[std]{{a1*that}x+{b1*that}y}&=&\var{that*c1}&\mbox{ ........(4)} \end{eqnarray} \]
Now take away the equation (4) from equation (3) to get
\[\simplify[std]{({a*this}+{s6*a1*that})x={this*c}+{s6*that*c1}}\]
And so we get the solution for $x$:
\[x = \simplify{{c*b1-b*c1}/{b1*a-a1*b}}\]
Substituting this value into any of the equations (1) and (2) gives:
\[y = \simplify{{c*a1-a*c1}/{b*a1-a*b1}}\]
You can check that these solutions are correct by seeing if they satisfy both equations (1) and (2) by substituting these values into the equations.

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Maths Support: Simultaneous equations

Question 1

Question 1

Advice

Question 2

Question 2

Advice