Anyone good at maths here?
- Thread starter savanna
- Start date
for part a) first you need to understand what is going on.
Each line is defined by an equation. That equation defines where each point of that line lies.
When you solve a simultaneous equation, you're finding a solution that is valid for both equations. In other words, you are finding a point that lies on both lines. Do you see what you need to do now? Note that the question tells you to use the diagram, and is only worth one mark. This means the solution should be very easy to spot and shouldn't require any working out.
Each line is defined by an equation. That equation defines where each point of that line lies.
When you solve a simultaneous equation, you're finding a solution that is valid for both equations. In other words, you are finding a point that lies on both lines. Do you see what you need to do now? Note that the question tells you to use the diagram, and is only worth one mark. This means the solution should be very easy to spot and shouldn't require any working out.
Mega_Cabbage
Meh
You can use the point-slope formula (Y-Y1)=M(X-X1), where M is the slope. So Y-4=.5(X-0) for question B. I'm sure you can do the basic algebra from there.
