Solving linear simultaneous equations by elimination

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sean hunte 5 years, 6 months ago

Created this as a copy of Maria's copy of Solving linear simultaneous equations by elimination.

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

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Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	2
b	integer	4
c	number	14
d	integer	3
f	integer	11
g	number	46
x2	integer	-3
y2	number	5

			Name	Type	Generated Value

h	integer	2
j	integer	6
k	integer	4
l	integer	4
m	number	0
n	number	8
y1	number	-1
x1	integer	2
numerator	number	-8
divisor	rational	16

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..5)

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Description

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Generated value: `integer`

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Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part b) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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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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This question is used in the following exam:

Numeracy practice test MN1105 RHUL by Jose-Rodrigo Cordoba-Pachon in Jose-Rodrigo's workspace.

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Solving linear simultaneous equations by elimination

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sean hunte 5 years, 6 months ago

Error in variable testing condition

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Generated value: `integer`

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b)

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Solving linear simultaneous equations by elimination

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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History

Comment

Checkpoint description

sean hunte 5 years, 6 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

Generated value: `integer`