### Video Transcript

How many solutions are there to the simultaneous equations π₯ plus seven π¦ equals 20 and two π₯ plus 14π¦ equals 40?

Our usual method to solve a pair of linear simultaneous equations is to make either the π₯ or the π¦ coefficients the same. In this case, we could multiply the first equation by two or, alternatively, we could divide the second equation by two. Multiplying the first equation by two gives us two π₯ plus 14π¦ equals 40.

At this stage you will notice that the two equations are actually identical. So, initially we thought we had a pair of simultaneous equations. But in reality in this case, we just need to solve one equation. If we consider the simplified version of the equation π₯ plus seven π¦ equals 20, we can immediately see that there are many integer value solutions.

For example, when π₯ equals 13 and π¦ equals one, π₯ plus seven π¦ equals 20, as 13 plus seven multiplied by one is equal to 20. The values π₯ equals six and π¦ equals two also solve the equation as six plus seven multiplied by two is equal to 20. We can extend this to negative solutions. When π₯ equals negative eight and π¦ equals four, negative eight plus seven multiplied by four equals 20, as negative eight plus 28 is equal to 20.

The three solutions we have seen so far are all integer solutions for π₯ and π¦. But when we extend this to fractions or decimals, we can see that there an infinite number of solutions to the equation π₯ plus seven π¦ is equal to 20. As π₯ plus seven π¦ equals 20 has an infinite number of solutions, the pair of simultaneous equations in this question must also have an infinite number of solutions.

We can extend this one step further to say that any linear equation of the form ππ₯ plus ππ¦ equals π where π, π, and π are constants has an infinite number of solutions. In terms of simultaneous equations, there will be an infinite number of solutions if and only if the two equations are identical. We could also have solved this equation graphically.

Rewriting the two equations in the form π¦ equals ππ₯ plus π gives us π¦ equals minus a seventh π₯ plus 20 sevenths and the second equation π¦ equals negative two 14th π₯ plus 40 14ths. As the two gradients are equal, the two lines would definitely be parallel. Furthermore, as the interceptor are also equal, the two lines will set exactly on top of one another.

Both lines have a gradient of negative one- seventh and a π¦ intercept of 20 sevenths. This means that they will have an infinite number of solutions.