Algebraic simultaneous equations

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Question statement

Use the {method} method to find the point of intersection of the following two lines.

$y = \var{m1}x \var{b1sign} \var{b1}$

$y= \var{m2}x \var{b2sign} \var{b2}$

Give any introductory information the student needs.

			Name	Type	Generated Value

m1	integer	-5
m2temp	integer	2
m2	integer	2
b1	integer	6
b2	integer	-43
px	integer	7
py	integer	-29

			Name	Type	Generated Value

b1sign	string	$+$
b2sign	string
b1signneg	string

			Name	Type	Generated Value

method

string

elimination

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(-1,1)*random(1..6)

JME function reference

Description

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

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Generated value: `integer`

-5

→ Used by:

This variable doesn't seem to be used anywhere.

Parts

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1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Gap-fill

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The point of intersection occurs at

$x=$ Gap 0.

$y=$ Gap 1.

Advice

To use the substitution method, we set the two expressions that are equal to $y$ equal to each other.

That is,

$\var{m1}x \var{b1sign} \var{b1} = \var{m2}x \var{b2sign} \var{b2}$

Then we subtract $\var{m2}x$ from both sides:

$\var{m1}x - \var{m2}x \var{b1sign} \var{b1} = \var{b2}$

In your working it is fine to replace ' $--$ ' with ' $+$ ' and ' $+-$ ' with ' $-$ ' if they ever appear.

The we subtract $\var{b1}$ from both sides.

$\var{m1}x -\var{m2}x = \var{b2} - \var{b1}$

Now all the terms with $x$ are on one side, and all the terms without $x$ are on the other side.

Factorise out the $x$ on the left hand side:

$(\var{m1} -\var{m2})x = \var{b2} - \var{b1}$

Calculate $(\var{m1} -\var{m2})$ and $(\var{b2} - \var{b1})$

$(\var{m1-m2})x = \var{b2-b1}$

Finally, divide both sides by $\var{m1-m2}$ to get

$x=\var{(b2-b1)/(m1-m2)}$

To use the elimination method, let's start by giving the equations names so we can refer to them:

$y = \var{m1}x + \var{b1} \qquad \qquad$ (1)

$y = \var{m2}x + \var{b2} \qquad \qquad$ (2)

Let's calculate equation (1) - equation (2). This will eliminate the $y$ variable.

First just write out the subtraction. Make sure to put brackets around the two parts on the right hand side.

In your working it is fine to replace ' $--$ ' with ' $+$ ' and ' $+-$ ' with ' $-$ ' if they ever appear.

$y - y = (\var{m1}x + \var{b1}) - (\var{m2}x + \var{b2})$

Then rearrange the right hand side:

$0 = \var{m1}x + \var{b1} - \var{m2}x - \var{b2}$

$0 = \var{m1}x - \var{m2}x + \var{b1} - \var{b2}$

Now calculate $\var{b1} - \var{b2}$

$0 = \var{m1}x - \var{m2}x + \var{b1-b2}$

Add $\var{b2-b1}$ to both sides of the equation:

$\var{b2-b1} = \var{m1}x - \var{m2}x$

Factorise the $x$ out on the right hand side.

$\var{b2-b1} = (\var{m1} - \var{m2})x$

Calculate $(\var{m1} - \var{m2})$

$\var{b2-b1} = (\var{m1-m2})x$

Divide both sides by $(\var{m1-m2})$

$\var{(b2-b1)/(m1-m2)} = x$

So $x=\var{px}$

Now substitute this value for $x$ back into one of the original equations:

$y = \var{m1} \times \var{px} \var{b1sign} \var{b1}$

$y = \var{m1*px+b1}$

The point of intersection is $(\var{px},\var{py})$ .

We can check that this point is correct by substituting it back into both of the original equations and seeing that they are both correct.

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Taxonomy: mathcentre

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Merryn Horrocks 4 years, 7 months ago

Merryn Horrocks 4 years, 7 months ago

Merryn Horrocks 4 years, 7 months ago

Error in variable testing condition

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Generated value: `integer`

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Algebraic simultaneous equations

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Merryn Horrocks 4 years, 7 months ago

Merryn Horrocks 4 years, 7 months ago

Merryn Horrocks 4 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`