Revision:Simultaneous Equations

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Revision:Simultaneous Equations

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Simultaneous Equations

Simultaneous equations are two or more equations which are true for two or more unknowns. For example, x + y = 4 and x - 2y = 1 are simultaneous equations which are true for x = 3 and y = 1 . When there are two unknowns, as there are here (x and y), then two equations are needed to find the unknowns. When there are 3 unknowns, 3 equations are needed, etc.

1 Example
2 Method 1: elimination
3 Method 2: Substitution
4 Harder simultaneous equations
- 4.1 Example
5 Using Graphs
- 5.1 Example
6 Comments

Example

A man buys 3 fish and 2 chips for £2.80 A woman buys 1 fish and 4 chips for £2.60 How much are the fish and how much are the chips?

There are two methods of solving simultaneous equations. Use the method which you prefer.

Method 1: elimination

First form 2 equations. Let the cost of fish be and the cost of chips be .

We know that:

$\displaystyle 3f + 2c = 280 \quad \quad \quad \quad (1)$

$\displaystyle f + 4c = 260 \quad \quad \quad \quad (2)$

Doubling (1) gives:

$\displaystyle 6f + 4c = 560 \quad \quad \quad \quad (3)$

Then, $\displaystyle (3)-(2) \Rightarrow 5f = 300$

$\displaystyle \Rightarrow f = 60$ .

Therefore the price of fish is 60p

Substitute this value into (1):

$\displaystyle 3(60) + 2c = 280$

$\displaystyle 2c = 100$

$\displaystyle c = 50$

Therefore the price of chips is 50p

Method 2: Substitution

Rearrange one of the original equations to isolate a variable.

Rearranging (2): $\displaystyle f = 260 - 4c$

Substitute this into the other equation:

$\displaystyle 3(260 - 4c) + 2c = 280$

$\displaystyle 780 - 12c + 2c = 280$

$\displaystyle 10c = 500$

$\displaystyle c = 50$

Substitute this into one of the original equations to get f = 60 .

Harder simultaneous equations

To solve a pair of equations, one of which contains x^2 , y^2 or , we need to use the method of substitution.

Example

$\displaystyle2xy + y = 10 \quad \quad \quad \quad (1)$

$\displaystyle x + y = 4 \quad \quad \quad \quad (2)$

Take the simpler equation and get y = .... or x = ....

from (2), $\displaystyle y = 4 - x \quad \quad \quad \quad (3)$

this can be substituted in the first equation. Since y = 4 - x , where there is a in the first equation, it can be replaced by 4 - x .

sub (3) in (1):

$\displaystyle 2x(4 - x) + (4 - x) = 10$

$\displaystyle 8x - 2x^2 + 4 - x - 10 = 0$

$\displaystyle 2x^2 - 7x + 6 = 0$

$\displaystyle (2x - 3)(x - 2) = 0$

either $\displaystyle 2x - 3 = 0$ or $\displaystyle x - 2 = 0$

therefore x = 1.5 or .

Substitute these x values into one of the original equations.

When x = 1.5, y = 2.5 ;

when x = 2, y = 2 .

Using Graphs

You can solve simultaneous equations by drawing graphs of the two equations you wish to solve. The x and y values of where the graphs intersect are the solutions to the equations.