i was wondering what the method is to figuring this out:

$\displaystyle (\frac {36-4}{8})^2$ am i right in thinking the answer is 16?

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- Sep 21st 2009, 09:50 AMmarksquaring brackets
i was wondering what the method is to figuring this out:

$\displaystyle (\frac {36-4}{8})^2$ am i right in thinking the answer is 16? - Sep 21st 2009, 10:00 AMmasters
- Sep 21st 2009, 10:05 AMmark
ah thanks masters, so i've got that part of it right then. do you reckon you could look at the last question i posted on here please? if you could could you explain everything i need to do to complete it in detail becase there must be something i'm missing thats making it hard to understand. thanks

ps its the one called "equation" - Sep 21st 2009, 11:47 AMmasters
Sure, mark,

(1) $\displaystyle 5x + 3y = 4$

(2) $\displaystyle 5x^2 - 3y^2 = 8$

Solve the first equation for x.

(1) $\displaystyle x=\frac{4-3y}{5}$

Substitute this in (2).

$\displaystyle 5\left(\frac{4-3y}{5}\right)^2-3y^2=8$

Expand and simplify:

$\displaystyle 5\left(\frac{16-24y+9y^2}{25}\right)-3y^2=8$

$\displaystyle \left(\frac{16-24y+9y^2}{5}\right)-3y^2=8$

Multiply all terms by 5.

$\displaystyle 16-24y+9y^2-15y^2=40$

Combine terms and simplify.

$\displaystyle -6y^2-24y-24=0$

$\displaystyle y^2+4y+4=0$

Factor.

$\displaystyle (y+2)(y+2)=0$

$\displaystyle y=-2$

Now, substitute this back into (1) to find the corresponding x value.

$\displaystyle 5x+3(-2)=4$

$\displaystyle 5x-6=4$

$\displaystyle 5x=10$

$\displaystyle x=2$

The intersection is (2, -2). The line is tangent to the hyperbola at this point. - Sep 21st 2009, 11:54 AMmark
ahhh thankyou very much, that was a really good explanation