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Use algerbra to find the point on the graph

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1. Use algerbra to find the point on the graph
Use algerbra to find the points where the line y=-2x+7 intersects the curve y=x^2-3x+5

I've never done this before, could you direct me at a suitable exam solutions tutorial or offer some help please? Thanks
2. Re: Use algerbra to find the point on the graph
You put the two equations equal to each other (since they both equal y), rearrange, you end up with a quadratic which you solve for x. You then put the x value(s) back into one of the equations, that gives you the y value, and you have a point (x,y).
3. Re: Use algerbra to find the point on the graph
Hey Okay so you want the point where the Y and X coordinates are equal to each other. So the first thing I would do in this case is get rid of the Y's by writing them both together (i.e. if the Y's are the same then both write sides are equal) so its -2x+7=x^2-3x+5. Then gather the terms on to one side (best to do the side with x^2 so you dont end up with negative x squared) so you get
x^2-3x+5+2x-7. Then simplify to give x^2-x-2 then see if you can solve for x at that point. When you have the two x values substitue both in to either equation and solve for the two corresponding y values
4. Re: Use algerbra to find the point on the graph
(Original post by multiplexing-gamer)
Use algerbra to find the points where the line y=-2x+7 intersects the curve y=x^2-3x+5

I've never done this before, could you direct me at a suitable exam solutions tutorial or offer some help please? Thanks
OP, where the curves cross are where the equations both equal each other. y = y. So -2x +7 = x^2 -3x +5

by the way, in your sig, why have you put up all that stuff about universities, Oxford, History and UNIQ, when you've only just finished your GCSEs? I mean, think about that.
5. Re: Use algerbra to find the point on the graph
(Original post by flown_muse)
You put the two equations equal to each other (since they both equal y), rearrange, you end up with a quadratic which you solve for x. You then put the x value(s) back into one of the equations, that gives you the y value, and you have a point (x,y).
Could you please show me the working in terms of Maths?
(Original post by Pride)
OP, where the curves cross are where the equations both equal each other. y = y. So -2x +7 = x^2 -3x +5

by the way, in your sig, why have you put up all that stuff about universities, Oxford, History and UNIQ, when you've only just finished your GCSEs? I mean, think about that.
They tell me at College that we apply in about a year so I'm preparing, UNIQ applications aren't too far away..
6. Re: Use algerbra to find the point on the graph
(Original post by multiplexing-gamer)
Use algerbra to find the points where the line y=-2x+7 intersects the curve y=x^2-3x+5

I've never done this before, could you direct me at a suitable exam solutions tutorial or offer some help please? Thanks

When you say that you have never done this before ... this is on the GCSE syllabus

Solving simultaneous equations algebraically, including a linear and quadratic

Then someone gave you the equation

then you asked for this to be explained using maths????

Can you solve the above quadratic
7. Re: Use algerbra to find the point on the graph
(Original post by TenOfThem)
When you say that you have never done this before ... this is on the GCSE syllabus

Solving simultaneous equations algebraically, including a linear and quadratic

Then someone gave you the equation

then you asked for this to be explained using maths????

Can you solve the above quadratic
Well I did GCSE Maths in Year 10 and am starting Year 12 next month, so I probably did do it, two years ago. I'm playing catch up now over the GCSE concepts.
i can solve quadratics yes, you'd rearrange to get it into one big equation then just solve it like a normal quadratic?
I asked for it "in maths" as he/she was writing a lot and I was finding it a little difficult to understand, so was hoping for a stage by stage explanation using numbers... But thank you anyway.
8. Re: Use algerbra to find the point on the graph
(Original post by multiplexing-gamer)
Well I did GCSE Maths in Year 10 and am starting Year 12 next month, so I probably did do it, two years ago. I'm playing catch up now over the GCSE concepts.
This is the problem with schools that do early entry and then do not think about the maths that needs to be done in Y11

Yes, re-arrange and solve as normal

9. (Original post by TenOfThem)
This is the problem with schools that do early entry and then do not think about the maths that needs to be done in Y11

Yes, re-arrange and solve as normal

I've sat it early, but I'll be doing AS maths next year, so it's not always bad
10. (Original post by multiplexing-gamer)
Use algerbra to find the points where the line y=-2x+7 intersects the curve y=x^2-3x+5

I've never done this before, could you direct me at a suitable exam solutions tutorial or offer some help please? Thanks
Full working:
Spoiler:
Show

-2x + 7 = x^2 - 3x + 5
x^2 - x - 2 = 0
(x+1)(x-2) = 0
x = -1, 2

When x = -1, y = 9
When x = 2, y = 3

Points of intersection:
(-1,9) and (2,3)

11. Re: Use algerbra to find the point on the graph
I've sat it early, but I'll be doing AS maths next year, so it's not always bad
I did not say it was always bad

I know many schools that do early entry and have a sensible programme of Y11 maths available ... clearly this is not the case in the OPs school
12. (Original post by TenOfThem)
I did not say it was always bad

I know many schools that do early entry and have a sensible programme of Y11 maths available ... clearly this is not the case in the OPs school
I see, my school is to raise the league points... (they won't admit that) they expect everyone to get Cs but I made them enter me.do higher and everyone else will resit in November, quite cunning...

There was a boy in my school who did his AS in his January of year 10, he got C1 98 C2 100 M1 89
Sorry, random.story
13. Re: Use algerbra to find the point on the graph
Full working:
Spoiler:
Show

-2x + 7 = x^2 - 3x + 5
x^2 - x - 2 = 0
(x+1)(x-2) = 0
x = -1, 2

When x = -1, y = 9
When x = 2, y = 3

Points of intersection:
(-1,9) and (2,3)

Thank you very much!
14. (Original post by multiplexing-gamer)
Thank you very much!
No problem! I can always help, or try!

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Last updated: August 1, 2012
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