coordinates

**coolubi** · February 13th 2010, 04:29 AM

http://www.emaths.co.uk/sats/2009/2009_68_Paper1.pdf

please go to question 24 and help me answer

**mathemagister** · February 13th 2010, 04:41 AM

Originally Posted by coolubi

http://www.emaths.co.uk/sats/2009/2009_68_Paper1.pdf

please go to question 24 and help me answer

This looks a lot like a test, so here is just a hint:
Notice they have told you the two equations you need to deal with. You have 2 variables and 2 equations. Do you know how to solve simultaneous equations? When you solve for x and y, make sure you pair up the right numbers with each other for the 2 points.

Feel free to ask if you need any more help.

**coolubi** · February 13th 2010, 04:52 AM

Originally Posted by mathemagister

This looks a lot like a test, so here is just a hint:
Notice they have told you the two equations you need to deal with. You have 2 variables and 2 equations. Do you know how to solve simultaneous equations? When you solve for x and y, make sure you pair up the right numbers with each other for the 2 points.

Feel free to ask if you need any more help.

pls could you help me a little more as i am new to it(help with some working out) and only have one more week to my exam thanx

**mathemagister** · February 13th 2010, 05:44 AM

Originally Posted by coolubi

pls could you help me a little more as i am new to it(help with some working out) and only have one more week to my exam thanx

The information tells you that at the points A and B, [b]both[/both] the equations $y=x-2$ and $y=x^2-2$ are correct for the same x and y values.

So, to solve for the intercepts, you need to simultaneously solve the equations
$y=x-2$ and $y=x^2-2$ .

You know that $y=x-2$ , so anywhere you see a $y$ , you are allowed to substitute it for $x-2$ because they are both the same thing. (Think of it like if you know $3=5-2$ , you can put $5-2$ wherever you see a $3$ .

Well, you see a $y$ in the second equation; you can substitute that $y$ for something else that equals it: $x-2$ .

This gives you:
$x-2=x^2-2$

$x=x^2$

$x-x^2=0$

$x(1-x)=0$

$x=0$ and $1-x=0$
$x=0,1$

to find the y value for these points simply plug them into one of the original equations:
$y=x-2$
$y(0) = 0 - 2 = -2$
$y(1) = 1 - 2 = -1$

These give you the points (0,-2) and (1,-1).

Since you can see on the graph that point A lies on the y-axis (where x is always 0), you know that point A is (0,-2) and point B has to therefore be (1,-1).

Hope that helps

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