聯立方程 f.4 maths

1.
2x-3y=2 => x=(2+3y)/2...(1)
x^2+y^2+3x-2=0...(2)

Sub (1) into (2):
[(2+3y)/2]^2+y^2+3[(2+3y)/2]-2=0
(4+12y+9y^2)/4+y^2+(6+9y)/2-2=0
(4+12y+9y^2)+4y^2+2(6+9y)-8=0
(4+12y+9y^2)+4y^2+12+18y-8=0
13y^2+30y+8=0
(13y+4)(y+2)=0
y=-4/13 or y=-2

When y=-4/13, x=(2+3(-4/13))/2=7/13
When y=-2, x=(2+3(-2))/2=-2

2.
-3x+4y=25 => y=(25+3x)/4...(1)
x^2+y^2=25...(2)

Sub (1) into (2):
x^2+[(25+3x)/4]^2=25
x^2+(625+150x+9x^2)/16=25
16x^2+(625+150x+9x^2)=400
25x^2+150x+225=0
x^2+6x+9=0
(x+3)^2=0
x=-3

When x=-3, y=(25+3(-3))/4=4

3.
Let the length of DE and DC be x and (x+4)

x^2+(x+4)^2=400
x^2+x^2+8x+16=400
2x^2+8x-384=0
x=12 or x=-16 (rejected)

So, the side of the smaller square is 12cm and the side of the larger square is 16cm.

4.
Let the length and width of each rectangle be x and y respectively.

(4x)(2y)=96 => xy=12 => x=12/y...(1)
12x+10y=78 => 6x+5y=39...(2)

Sub (1) into (2):
6(12/y)+5y=39
72+5y^2=39y
5y^2-39y+72=0
(5y-24)(y-3)=0
y=24/5 or y=3

When y=24/5, x=5/2 and when y=3, x=4
Hence, the dimensions are 5/2cm*24/5cm and 4cm*3cm