Simultaneous Equasions

How to solve simultaneous equasions using the substitution and elimination methods, plus a few questions to try for yourself.

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  • Created by: Conor
  • Created on: 11-04-10 22:18

The substitution method

When using the substitution method, you rearrange one of the equasions to make either x or y the subject. Then you substitute this into the other equation, which allows you to work it out.

Example:

(1) 3x - 2y = 0

(2) 2x + y = 7

First, you rearrange one of the equations to make either x or y the subject. I am going to make y the subject of equation (2): y = 7 - 2x

Next, you substitute this into equasion (1): 3x - 2 (7 - 2x) = 0

Now there are no y values in the equation, so you can work it out:

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Substitution cont.

3x - 2 (7 - 2x) = 0 (expand first of all)

3x - 14 + 4x = 0 (simplify)

7x - 14 = 0 (move the 14 over)

7x = 14 (divide by 7)

x = 2

Once you have a value for either x or y, substitute this value into the other equation:

4 + y = 7

y = 3

And that's all there is to the substitution method...

There are some practice questions on the last card to try out.

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The elimination method

The elimination method is normally quicker, but can be more confusing on harder questions. It involves adding or subtracting one equation from the other, in order to eliminate either the x or y values.

Example:

(1) 2x + 3y = 5

(2) 5x - 2y = -16

First, you have to make either the x or y values of both equations the same. On this question, I am going to multiply equation (1) by two, and equation (2) by three, so the y values are both 6y:

(1) 4x + 6y = 10

(2) 15x - 6y = -48

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Elimination cont.

Because one 6y is plus, and the other is minus, we need to add them together to eliminate them. If they were both plus, the would have to be taken from each other. Once you add them, you are left with:

19x = -38 (divide by 19)

x = -2

Now you have the x value you can put it into one of the equations to get y:

-4 + 3y = 5 (move the -4 over)

3y = 9 (divide by 3)

y = 3

And thats it...

The next card has some practice questions.

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Practice Questions

Use the substitution method for these questions:

1) 2x + y = 5 and x + 3y = 5

2) x + 2y = 8 and 2x + 3y = 14

3) 2x + y = 13 and 5x - 4y = 13

4) 3m - n = 5 and 2m + 5n = 7 *challenge question*

Use the elimination method for these questions:

1) 5x + 2y = 13 and 2x + 6y = 26

2) 2x + 5y = 24 and 4x + 3y = 20

3) x + 3y - 7 = 0 and 2y - x - 3 = 0

4) 2x - y = 5 and x/4 + y/3 = 2 *challenge question*

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Comments

Emma

This is ace! helped loads, thanks! :-)

emily collins

they are great notes! thanks but, you could do with the answers on them as well

Shyan Ahmed

agree with emily, answers would be great

Matthew

<3

Kiitz

the answers are on the next page all you have to do is click next :p

daviesg

Really useful demonstration of both elimination and substitutin methods

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