Solving simultaneous equations algebraically

Watch this video to learn about how to solve simultaneous equations using the elimination method.

This video can not be played

To play this video you need to enable JavaScript in your browser.

Simultaneous equations can be solved by plotting them on a graph but this can be time consuming.

When solving simultaneous equations algebraically, the first step is to try to eliminate one of the unknowns.

Example

Solve these simultaneous equations and find the values of \(x\) and \(y\).

\(2x + y = 7(Equation\,1)\)

\(3x - y = 8(Equation\,2)\)

Answer

In order to eliminate one of the unknowns, we need to have either the same coefficient (number) of \(x\)s or the same number of \(y\)s in the equations. Here, we have the same coefficient of \(y\)s, and also one is positive and one is negative. To eliminate the \(y\)s, we need to add Equation 1 and Equation 2 together to get \(y + ( - y) = 0\).

\(2x+y=7\)

\(3x-y=8\)

Adding these equations we get:

\(5x = 15\)

\(x = 15 \div 3\)

\(x = 3\)

At the point of intersection the x-coordinates are the same and the y-coordinates are the same. This means that we can substitute this value for \(x\) into one of the original equations to find the value of \(y\).

It is usually easier to substitute \(x\) into the equation for which \(y\) is positive.

Substitute into Equation 1: \(2x + y = 7\)

When \(x = 3\)

\(2 \times 3 + y = 7\)

\(6 + y = 7\)

\(y = 7 - 6\)

\(y = 1\)

Therefore \(x = 3\) and \(y = 1\)

Hence the point of intersection is (3,1)

91�ȱ�

Working with simultaneous equationsSolving simultaneous equations algebraically