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# C1 Q watch

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1. solve these equations simultaneously
x+2y=6
4y^2-5x^2=36

and i have no idea how to do this (can any 1 describe the method and soluton for me please )
Find the values of A B and C
3x^3-16x-1 = Ax(x-1)(x+2) + B(x-1)^2 + C(4x^2+3)
2. x+2y=6
4y^2-5x^2=36

x=6-2y

substitute in to 4y^2-5x^2=36

4y^2-5(6-2y)(6-2y)=36

multiply out the brackets
collect like terms on the left hand side
do a quadratic, you will have two values for Y
substitute back into
x=6-2y
for each value of Y you will get a X value

you will have the answer to be
x= ?, y = ? or x=??,y=??

sry i didn't expand brackets for you, i have to just finish my media cwk off before rush hour 2 is on
3. Yeah, as above. Heres the working.

4y^2 - 5(6-2y)(6-2y) - 36 = 0

4y^2 - 5(36-24y+4y^2) - 36 = 0

4y^2 - 180 + 120y - 20y^2 - 36 = 0

4y^2 - 20y^2 + 120y - 180 - 36 = 0

-16y^2 + 120y - 216 = 0

Factor of 8

-2y^2 + 15y - 27 = 0

Using the forumla;

Discriminant = b^2 - 4ac

15^2 - 4 x -2 x - 27 = 9

Y = -15 + 3 / -4 or -15 - 3 / -4

Y = 3 or 4.5

Sub into x = 6 -2y

When Y = 3

x = 6 - 6
x = 0

When Y = 4.5

x = 6 - 9
x = -3

Answers x = -3, y = 4.5 or x = 0, y = 3
4. how do you do the second Q mentioned please?
5. 3x^3-16x-1=Ax(x-1)(x+2) + B(x-1)^2 + C(4x^2+3)

Expanding the brackets on the R. H. S. gives:

Ax((x^2)+x-2)+B((x^2)-2x+1)+C(4(x^2)+3)

=>A(x^3)+A(x^2)-2Ax+B(x^2)-2Bx+B+4C(x^2)+3C

The coefficients, on the R. H. S. of:

(x^3):A

(x^2):A+B+4C

(x):-2A-2B

Constants on the R. H. S.:

(c):B+3C

Now equating these coefficients to what is on the L. H. S.:

A=3 (i)

A+B+4C=0 => B+4C=-3 (ii)

-2A-2B=16 => -6-2B=-16 (iii)

B+3C=-1 (iv)

Soving (iii):

2B=10 => B=5

Substituting into (ii):

5+4C=-3

Therefore:

4C=-8

=> C=-2

A=3, B=5, C=-2

Newton.

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Updated: January 2, 2005
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