1. ## equation

hi i've been trying to solve a simultaneous equation and i'm sure i've gone wrong somewhere and would be grateful if someone could point out where:

$\displaystyle 5x + 3y = 4$
$\displaystyle 5x^2 - 3y^2 = 8$

i started by making x the determinant in the first equation then subbing it into equation 2 like so

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

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

then multiplied everthing by 5 to remove the fraction
$\displaystyle 25(4 - 3y)^2 - 15y^2 = 40$ then expanded the squared bracketed term:
$\displaystyle 25(16 - 24y + 9y^2) - 15y^2 = 40$ then carried on with
$\displaystyle 400 - 600y + 225y^2 - 15y^2 = 40$ then
$\displaystyle 360 - 600y + 210y^2 = 0$

i really don't think this is right, could someone explain this to me please? thanks

2. Originally Posted by mark
hi i've been trying to solve a simultaneous equation and i'm sure i've gone wrong somewhere and would be grateful if someone could point out where:

$\displaystyle 5x + 3y = 4$
$\displaystyle 5x^2 - 3y^2 = 8$

i started by making x the determinant in the first equation then subbing it into equation 2 like so

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

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

then multiplied everthing by 5 to remove the fraction
$\displaystyle 25(4 - 3y)^2 - 15y^2 = 40$ then expanded the squared bracketed term:
$\displaystyle 25(16 - 24y + 9y^2) - 15y^2 = 40$ then carried on with
$\displaystyle 400 - 600y + 225y^2 - 15y^2 = 40$ then
$\displaystyle 360 - 600y + 210y^2 = 0$
$\displaystyle \not5(\frac{4-3y}{\not5}) - 3y^2 = 8$
Next step. $\displaystyle ({4-3y}) - 3y^2 = 8$

3. , you forgot to SQUARE the parenthetical expression

(4 - 3y)^2 - 15y^2 = 40, expand

16 - 24y + 9y^2 - 15y^2 = 40, sum up

16 - 24y - 6y^2 = 40, re-arrange

6y^2 + 24y - 34 = 0, DBS by 2

3y^2 + 12y - 17 = 0,

the quadratic formula can bash this . . . .

4. what does the symbol replacing the 5's mean? and what happened to the square? i don't understand...
by the way i missed out the square symbol in the original post but then realised i had and added it in later, i still carried on the workings out though as if the square symbol was there

the answer the book gives is x = 2 and y = -2

5. see the graph

6. do you remember this?

the 5 in the denominator is squared, but you only have 5 in the numerator,

in order to eliminate that 25 in the denominator, you multiply 5 by 5

5()

25[(4 - 3y)^2]/25 - (5)(3y^2) = 5(8),

(4 - 3y)^2 - 15y^2 = 40, i know you can finish this now

7. i've only been doing maths for about 6 weeks or so now and i don't really know what that graph was supposed to mean. do you think you could explain it in a different way?

8. i still don't really follow, you said "the 5 in the denominator is squared, but you only have 5 in the numerator", where is the 5 in the numerator? i thought the numerator was the number(s) above the line in a fraction?

then you multiplied everything by 5, which i understand but divided part of it by 25 like this:

$\displaystyle 25[(4 - 3y)^2]/25 - (5)(3y^2) = 5(8)$

the only thing i have trouble understand is dividing the first 25 by 25 again...

9. ,

(4 - 3y)/5 is a squared quantity, it means (4 - 3y) is squared and 5 in the D is squared

now look , in the perenthesis term it is precede by 5 right?

then you have 5^2 in the D, what will you multiply with the N to eliminate 25 in the D?

5 right?

so, 5[5/5^2][4-3y]^2 - 15y^2 = 40, that red thing cancels