A level revision

Simultaneous equations

To solve simultaneous equations you need to find values for x and y which satisfy both of the equations. You need to know how to solve both linear (highest power is 1) and nonlinear (e.g. quadratic) simultaneous equations.

Linear equations

Linear equations typically take the form ax + by = c, the highest power is 1. Two linear equations can be solved simultaneously by using two methods.

By elimination

Solution by elimination means trying to eliminate one of the variables, be it x or y. To do this you must subtract one equation from the other. However for this to eliminate one variable you typically will have to multiply one equation.

Example question: Solve the following simultaneous equations: 2x + 8y = 30 and x + 2y = 4

Worked answer: Start by looking at which term, the x or the y, will be easiest to eliminate. In this example multiplying the second equation by 2, so the x term becomes 2x, will result in the x terms being eliminated when you subtract the first equation from the other. So to start multiply the second equation by 2 to get 2x + 4y = 8. You can subtract this from the first equation:

2x + 8y = 30 - 2x + 4y = 8 = 4y = 22

To find the value for y simply divide 22 by 4 to get 5.5. To find x you should go back to the original second equation and substitute y = 5.5 back in:

x + 2(5.5) = 4 so x + 11 = 4, hence x = 4 - 11 = -7

Now you have your x and y values you can check that you're correct by substituting them both into the remaining original equation:

2x + 8y = 30, 2(-7) + 8(5.5) = 30 so -14 + 44 = 30

And that is solution by elimination!

By substitution

To solve by substitution you need to rearrange on of the equations so either x or y is the subject, then substitute that equation for x or y into the other equation.

Example question: Solve the following simultaneous equations: 2x + 8y = 30 and x + 2y = 4

Worked answer: This is the same example used earlier but this time we will use the substitution method. Firstly look for the easiest equation to rearrange into the form x = .. or y = .. In this example x + 2y = 4 can quickly be arrange into x = 4 - 2y. Once we have our new equation we substitute into the other equation:

2x + 8y = 30 and x = 4 - 2y
therefore: 2(4 - 2y) + 8y = 30
multiplying out: 8 - 4y + 8y = 30
collecting like terms: 8 + 4y = 30
rearrange for y: 4y = 22 therefore y = 5.5

Horray! We got the same answer as before. Now it's just a matter of substituting y = 5.5 into the original second equation to find out x. As this was done in the previous example we won't bother doing it again.

Linear and quadratic equations

You may have to solve two simultaneous equations where on is quadratic, i.e. has an x² term. To do this you follow the same steps as solving by substitution but you have check your work carefully as expanding the brackets can get messy.

Example question: Solve the following simultaneous equations: x² + 3x - y = -4 and y - x = 1

Worked answer: Firstly we need to rearrange the linear equation to get x as the subject so that when it is substituted into the other equation all of the terms will be y. y - x = 1 so x = y - 1. Now we can substitute this into the quadratic equation:

x² - 3x - y = -4 and x = y - 1
therefore: (y - 1)² - 3(y - 1) - y + 4 = 0
multiplying out the brackets:y² - 2y + 1 - 3y + 3 - y + 4 = 0
collecting like terms: y² - 6y + 8 = 0

We now have a quadratic equation which needs to be factorised (see: solving quadratic equations to revise). The quadratic factorises to give (y - 4)(y - 2) = 0. Therefore y is either +4 or +2 (as either y - 4 is 0 or y - 2 is 0 for (y - 4)(y - 2) to be 0). To find the x terms we do as with the other methods and substitute our y values back into our equations:

y = +4 or +2 and y - 1 = x
therefore: 4 - 1 = 3 and 2 - 1 = 1
so either y = 4 and x = 3 or y = 2 and x = 1

As you can see simultaneous equations involving quadratics typically result in two pairs of answers. Remember that the order does matter: y = 4 and x = 1 is not a correct answer, you must specifically say which y value goes with which x value!

Graphical solutions

You can also solve simultaneous equations by drawing the graphs of both equations on the same axes. The points at which they intersect give you the x and y values that satisfy both equations. Therefore if you are asked to find the points at which two lines intersect you should simply solve the equations of the two lines simultaneously using the methods on this page.