Simulation equations with a quadraticWatch
So have I rearranged the second equation correctly? I have done: 3y=1-x
Whenever we solve two equations simultaneously, we are effectively looking for a solution(s) which satisfy both equations at once. Graphically, these are the coordinates where the two equations intersect each other. At the point of intersection, the y-coordinates are the same, and the x-coordinates are the same.
Since we know that the y-coordinates are the same, this means that in is exactly the same as the in . This means if you can rearrange the second equation for on its own, then you can substitute that into the first equation and obtain a relation that solely must satisfy.
So far, you have found what is, but you need to know what on its own is. So what's one more step??
Also, as a side note, you may be used to 'rearranging for y and subbing it into the other equation' type of approach, but I also said that the x-coordinates are the same. So there is nothing stopping you from rearranging the second equation for instead and subbing that into the first equation. In fact, this is a much simpler alternative than that which you are doing because we avoid fractions this way and the algebra is less messy hence making less room for error.
So x = 1 - 3y.
You can then substitute this expression for x in your first equation and solve for your possible values for x.
From there you can find y.
Me - Yeah, yeah, sure you are.