Ext1 Maths HELP! (1 Viewer)

[bored.of.u]

Hey guys i need help with these two questions (they from cambridge 3u)

Q1.
Consider the equations 12x^2 - 4xy + 11y^2 = 64 and
16x^2 - 9xy + 11y^2=78

(a) by letting y = mx show that 7m^2 + 12m - 4 = 0
(b) hence, or otherwise, solve the two equations simultaneously

Q2.
(a) Find x in terms of c, given that (2/3x-2c) + (3/2x-3c) = 7/2c
(b) Find x in terms of a and b if, a<SUP>2</SUP>b/x<SUP>2</SUP> + (1 + b/x)a = 2b + a<SUP>2</SUP>/x

thanx in advance =D

 

[Trebla]

Hey guys i need help with these two questions (they from cambridge 3u)

Q1.
Consider the equations 12x^2 - 4xy + 11y^2 = 64 and
16x^2 - 9xy + 11y^2=78

(a) by letting y = mx show that 7m^2 + 12m - 4 = 0
(b) hence, or otherwise, solve the two equations simultaneously

(a)
12x² - 4xy + 11y² = 64
(1) => 12x² - 4mx² + 11m²x² = 64
16x² - 9xy + 11y² = 78
(2) => 16x² - 9mx² + 11m²x² = 78
Subtracting the two equations:
4x² - 5mx² = 14
x² = 14/(4 - 5m)
Sub into (1)
168 / (4 - 5m) - 56m / (4 - 5m) + 154m² / (4 - 5m) = 64
154m² - 56m + 168 = 256 - 320m
=> 154m² + 264m - 88 = 0
.: 7m² + 12m - 4 = 0

(b)
Solving:
7m² + 12m - 4 = 0
(m + 2)(7m - 2) = 0
.: m = -2 or m = 2/7

Now using x² = 14/(4 - 5m):

When m = - 2
x² = 1
.: x = ± 1
Since y = -2x
y = 2 when x = -1
y = - 2 when x = 1

When m = 2/7
x² = 14/(4 - 10/7)
= 49/9
.: x = ± 7/3
Since y = 2x/7
y = - 2/3 when x = - 7/3
y = 2/3 when x = 7/3