Solving linear simultaneous equations by elimination

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From users who are members of Transition to university :

Christian Lawson-Perfect	said	Ready to use	7 years, 4 months ago
Lauren Richards	said	Needs to be tested	7 years, 8 months ago

From users who are not members of Transition to university :

Deirdre Casey

said

Has some problems

7 years, 4 months ago

History

○

Christian Lawson-Perfect 7 years, 4 months ago

Gave some feedback: Ready to use

○

Christian Lawson-Perfect 7 years, 4 months ago

Saved a checkpoint:

You're right! I've fixed it, and improved the TeX. Worried that I let that slip by!

Restore this checkpoint

○

Deirdre Casey 7 years, 4 months ago

Gave some feedback: Has some problems

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Deirdre Casey commented 7 years, 4 months ago

I think there is a sign error in the second equation in part b.

○

Christian Lawson-Perfect 7 years, 8 months ago

Gave some feedback: Ready to use

○

Christian Lawson-Perfect 7 years, 8 months ago

Saved a checkpoint:

Set up the randomisation again so that all numbers are integers.

In part a, the $x$ coefficient of the second equation is a multiple of the coefficient in the first equation.

In part b, the LCMs of the $x$ and $y$ coefficients are strictly larger than any of the coefficients.

Restore this checkpoint

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Elliott Fletcher 7 years, 8 months ago

Published this.

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Lauren Richards 7 years, 8 months ago

Gave some feedback: Needs to be tested

○

Lauren Richards commented 7 years, 8 months ago

I've made sure that the solutions are integers but that has meant that the question is not randomised.

I'm also having a little trouble in centering a section of part a) which includes an overline and a subtraction.

Otherwise, I've made all the changes you requested.

○

Christian Lawson-Perfect 7 years, 8 months ago

Gave some feedback: Has some problems

○

Christian Lawson-Perfect 7 years, 8 months ago

Saved a checkpoint:

I'd prefer it if you made sure that the solutions are always integers. Rounding errors are a distraction here. Generate the solutions first, then come up with a pair of independent equations.

I'm not sure about the way you've set out subtracting one equation from another - it's not obvious that the minus signs cancel out when there's a negative coefficient. This is one of the big problems that trips people up when doing this.

Be careful with, for example, "We have to multiply the entire first equation by 3, not just the x term to ensure we do not change the value of the equation." - an equation doesn't have a 'value'. You could instead say that you want to keep the same solution.

Right at the end of the advice, I don't like $8(5)$ for multiplication - $8 \times 5$ is much more conventional.

Restore this checkpoint

○

Lauren Richards 7 years, 8 months ago

Gave some feedback: Needs to be tested

○

Lauren Richards commented 7 years, 8 months ago

It won't let me put visibility conditions on it. :(

○

Lauren Richards 7 years, 8 months ago

Created this.

Name	Status	Author	Last Modified
Solving linear simultaneous equations by elimination	Ready to use	Lauren Richards	20/11/2019 14:42
Christopher's copy of Solving linear simultaneous equations by elimination	draft	Christopher boyd	17/01/2018 10:54
Marte's copy of Solving linear simultaneous equations by elimination	draft	Marte Bråtalien	12/04/2018 14:12
Solving linear simultaneous equations by elimination	draft	Paul Hancock	24/07/2018 02:39
Solving linear simultaneous equations by elimination	draft	Paul Hancock	24/07/2018 02:39
Solving linear simultaneous equations by elimination [L4 Randomised]	Needs to be tested	Matthew James Sykes	25/07/2018 10:50
Tom's copy of Solving linear simultaneous equations by elimination [L4 Randomised]	draft	Tom Carroll	27/07/2018 10:22
Ann's copy of NUMBAS - Simultaneous equations	draft	Ann Smith	07/08/2018 16:30
Thomas's copy of Ann's copy of NUMBAS - Simultaneous equations	draft	Thomas Huettemann	13/08/2018 11:21
Joël's copy of Solving linear simultaneous equations by elimination	draft	Jo Cohen	02/12/2020 15:42
Simultaneous equations: elimination	Ready to use	heike hoffmann	14/11/2018 22:36
Basic Algebra : Simultaneous Equation	draft	Johnny Yi	02/12/2020 13:05
Maria's copy of Solving linear simultaneous equations by elimination	Ready to use	Maria Aneiros	11/09/2019 10:45
Solving linear simultaneous equations by elimination	draft	Xiaodan Leng	11/07/2019 01:04
Solving linear simultaneous equations by elimination	draft	sean hunte	11/09/2019 10:46
Musa's copy of Solving linear simultaneous equations by elimination 3	draft	Musa Mammadov	27/06/2023 01:44
Musa's SIT316 Week XXX Q1	draft	Musa Mammadov	20/07/2023 01:46
Musa's SIT316 Week XXX Q2	draft	Musa Mammadov	20/07/2023 01:46
Musa's SIT316 Week XXX Q3	draft	Musa Mammadov	20/07/2023 01:47
Musa's SIT316 Week XXX Q4	draft	Musa Mammadov	20/07/2023 01:47
Musa's SIT316 Week XXX Q5	draft	Musa Mammadov	20/07/2023 01:47
SIT316 MO-Geometric Method	draft	Musa Mammadov	20/07/2023 01:42
SIT316 MO-Geometric Method-Q3	draft	Musa Mammadov	20/07/2023 01:43
SIT316 MO-Geometric Method-Q1	draft	Musa Mammadov	20/07/2023 01:42
Musa's SIT316 MO-Geometric Method-Q2_1	draft	Musa Mammadov	20/07/2023 01:42
Musa's SIT316 MO-Simplex Method	draft	Musa Mammadov	20/07/2023 01:56
Musa's copy SIT316 Week 7 - Branch and Bound for integer programming V2	draft	Musa Mammadov	20/07/2023 02:06
SIT316-Constraint Programming - Q3	draft	Musa Mammadov	20/07/2023 02:12
Musa's SIT316 MO-Simplex Method-2	draft	Musa Mammadov	20/07/2023 01:56
Musa's copy SIT316 Week 7 - Branch and Bound for integer programming V2-Q2	draft	Musa Mammadov	20/07/2023 02:06
SIT316-Constraint Programming - Q4	draft	Musa Mammadov	20/07/2023 02:12
Musa's SIT316 MH-Single state Week 8 - V2021	draft	Musa Mammadov	20/07/2023 02:09
Musa's SIT316 MH-Population based - Week 9 - V2021	draft	Musa Mammadov	20/07/2023 02:10
Simple simultaneous equations	Ready to use	Shaheen Charlwood	27/10/2022 18:51
Ugur's copy of Solving linear simultaneous equations by elimination	Ready to use	Ugur Efem	14/01/2022 15:06

There are 88 other versions that do you not have access to.

Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
b	integer	5
c	number	-2
d	integer	8
f	integer	8
g	number	-16
x2	integer	-4
y2	number	2

			Name	Type	Generated Value

h	integer	3
j	integer	6
k	integer	3
l	integer	3
m	number	-12
n	number	-27
y1	number	1
x1	integer	-5
numerator	number	3
divisor	rational	9

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..5)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

Advice
Variable testing condition
"Part b)" - prompt

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add
Part b) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

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

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

$x =$ Gap 0.

$y =$ Gap 1.

Advice

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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Solving linear simultaneous equations by elimination

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Taxonomy: mathcentre

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Taxonomy: Context

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Christian Lawson-Perfect 7 years, 4 months ago

Christian Lawson-Perfect 7 years, 4 months ago

Deirdre Casey 7 years, 4 months ago

Deirdre Casey commented 7 years, 4 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Lauren Richards 7 years, 8 months ago

Lauren Richards commented 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Lauren Richards 7 years, 8 months ago

Lauren Richards commented 7 years, 8 months ago

Lauren Richards 7 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

a)

b)

Loading...

Error

Solving linear simultaneous equations by elimination

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 7 years, 4 months ago

Christian Lawson-Perfect 7 years, 4 months ago

Deirdre Casey 7 years, 4 months ago

Deirdre Casey commented 7 years, 4 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Lauren Richards 7 years, 8 months ago

Lauren Richards commented 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Lauren Richards 7 years, 8 months ago

Lauren Richards commented 7 years, 8 months ago

Lauren Richards 7 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

Generated value: `integer`