solving equation

• October 8th 2012, 03:03 AM
Tweety
solving equation
The graphs with equations http://mapleta7.eps.surrey.ac.uk:808...kipaopgngm.gif and http://mapleta7.eps.surrey.ac.uk:808...plafpdjmch.gif meet at two points. At one of these points, http://mapleta7.eps.surrey.ac.uk:808...jijlbkpdha.gif. What is the value of http://mapleta7.eps.surrey.ac.uk:808...gmiemniapm.gif at the other point?

I dont know how to do this question, I think i should set them equal to each other,

any help appreciated.
• October 8th 2012, 03:11 AM
MarkFL
Re: solving equation
I recommend substituting for y as given by the first equation into the second equation. This will given you a quadratic in x. Solve this, then plug the root not equal to 2 into the first equation for x to determine the value of y at the other point.
• October 8th 2012, 03:58 AM
Tweety
Re: solving equation
Quote:

Originally Posted by MarkFL2
I recommend substituting for y as given by the first equation into the second equation. This will given you a quadratic in x. Solve this, then plug the root not equal to 2 into the first equation for x to determine the value of y at the other point.

Am a bit confused, i did what you said

$9x^{2} - 4 = 1$

9x^{2} -20 = 0 [/tex]

and got $x = \frac{2\sqrt{5}}{3}$

so i plug this answer into the first equation to get an answer for y ?
• October 8th 2012, 10:04 AM
MarkFL
Re: solving equation
If you substitute for y using the first equation, into the second equation, you get:

$x\left(\frac{9}{4}x-4 \right)=1$

Distribute the x on the left side:

$\frac{9}{4}x^2-4x=1$

Multiply through by 4:

$9x^2-16x=4$

$9x^2-16x-4=0$