# 6 - Systems of linear equations

## Problems with two variables

Lucy has bought two doughnuts and a chocolate cake. She has paid 2.10 dollars. In the same bakery, Jeremy has bought a doughnut and three chocolate cakes, he has paid 3.05 dollars. How much is a doughnut and a chocolate cake in this bakery?

To find out, we will need a System of linear equations.

1. Set: x= "what we are looking for". Here we are looking for two things, the price of a doughnut and the price of a chocolate cake. Set x="the price of a doughnut" and y="the price of a chocolate cake". We want to find x and y.

2. You write equations corresponding to the problem. It is the most difficult step but you will be able to make it, sure! You write the two equations between brackets

It is a system of two equations with two variables. We are going to learn how to solve it.

3. We solve the system. There are two methods to do that, you can pick up the one you understand best!

## Solving a system of linear equations

1st method: You isolate one variable in an equation then you replace its value in the other equation.

=> => => => => => => => => (phew!).

2nd method : You multiply an entire equation by the same number, you must manage to have the same number in front of x in both equations. Then you subtract both equations.

=>

The subtraction is , so Finally we are looking for x and take one of the two equations we had first :
You can really find the same solutions!
Conclusion: A doughnut costs 0.65 dollar and a chocolate cake costs 0,80 dollar.
You can also write :.

Another system :
Find the value of x and y knowing that

With the 1st method :

=> => => => => => =>

With the second one :

=>
(you make an addition this time) so , and

Remark : With the second method, if the system is like that :

You have to multiply both the first line by 5 and the second line by 3 to get 15 in front of x in both equations and then you can subtract them.