maths f4

(a) Combine them into one quadratic equation in "y":

In (1), rewrite it as x = 2y + 20 ----- (3)

Put (3) into (2): ( 2y + 20 )^2 + 3y = 15

4 y^2 + 83y + 385 = 0
( 4y + 55 )( y + 7 ) = 0
y = -55/4 or y = -7

By (3): When y = -55/4, x = -15/2; When y = -7, x = 6

( x, y ) = ( -15/2, -55/4 ) or ( 6, -7 )
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(b) xy - 2x - 2y = 20 ===> ( xy ) - 2( x - y ) = 20 --- (1) x^2 y^2 + 3x +3y = 15 ===> ( xy )^2 + 3( x + y ) = 15 ---- (2)In (a), put x of (a) = ( xy ) here, put y of (a) = ( x + y ) here

Case 1:
xy = -15/2, x + y = -55/4
( x, y ) ~ ( -14.28, 0.53 ) or ( 0.53, -14.28 )
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Case 2:

xy = 6, x + y = -7
( x, y ) = ( -1, -6 )
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(a) x=2y+20
So, x^2 + 3y=15
(2y+20)^2 + 3y =15
(4y^2+80y+400)+3y=15
4y^2+83y+400=15
4y^2+83y+385=0
(y+7)(4y+55)=0
y=-7 or -55/4
So, x=6 or -7.5 respectively

The solutions are (1) x=6, y=-7; (2) x=-7.5, y=-55/4

(b) xy-2x-2y=20
xy-2(x+y)=20...(1)

x^2y^2+3x+3y=15
(xy)^2+3(x+y)=15...(2)

Now, we set x=xy and y=x+y.xy=6x+y=-7

y=-x-7
So, xy=6
x(-x-7)=6
-x^2-7x=6
x^2+7x+6=0
(x+1)(x+6)=0
x=-1 or -6
So, y=-6 or -1 respectively

Solution: (1) x=-1, y=-6; (2) x=-6, y=-1

xy=-7.5x+y=-55/4
y=-x-55/4

So, xy=-7.5
x(-x-55/4)=-7.5
-x^2-55/4 x=-7.5
x^2+55/4 x-7.5=0
x= { -55/4 +/- V[(55/4)^2 - 4(1)(-7.5)] } / 2(1)
x=0.525380057 or -14.27538006
So, y=-14.27538006 or 0.52538006

Solution: (1) x=0.525, y=-14.3; (2) x=-14.3, y=0.525




The solutions are
x=-1, y=-6x=-6, y=-1x=0.525, y=-14.3x=-14.3, y=0.525