Mymaths Quadratic Simultaneous Equations Help!

[Silvermist887]

https://www.dropbox.com/s/r25h7k6hd3u1v6x/oma.png
The link to the questions is above, I desperately need help with the answers I already completed the first part but Im stuck on all of these and I've been at it for hours, since I have no one else to ask for help Im hoping that someone can help me Please

[elliemay1]

I wish I could help. In the first set x=1 and y=6 but not sure how to find co-ordinates of where the 2 graphs meet.

[bibblebub]

in each case, rearrange the 2nd equation to be of the form y = ax+b

then substitute that value of y in the 1st equation, expand the terms and rearrange so that you have a quadratic in x of the form ax² + bx + c = 0 (not the same a and b from the first step)

then use the standard solution for x in such a quadratic

[bibblebub]

...and for each value of x that you get for each set use the 2nd equation to get the corresponding y value

[elliemay1]

Bibblebub, in the 1st set of eqns x=1, y=6, the graphs cross at (1,6). Why are they looking for 2 answers? Do they cross again? Please explain.

[jim360]

Quadratic equations have zero, one, or two real solutions - in this case that will mean zero, one or two crossing points. So if x=1 and y=6 is one solution, there may be another. Hence the second set of brackets.

Bibblebub has described the method well -- if you still need help silver then I'll try to make a more detailed post later.

[elliemay1]

Jim, is there another solution here?

[bibblebub]

(1,6) and (-6,-1) for the first pair

[elliemay1]

Thanks, bb, hadn't thought of the minus ones!

[Prudie]

All the x² + y² equations are circles and the y=x ones straight lines so the line may cut the circle in 2 places. As said you need to substitute the y value (for example) into the quadratic and the solve for x getting 2 values for x and then substituting these values back to find the y giving you 2 sets of co-ordinates.