有人知道怎計嗎?

1. 3y-3=4x .... i
y=7x-16 ......ii
sub ii into i
3(7x-16)-3-4x=0
17x=51
x=3

3y-3=4x
3y-3=12
y=5

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2. y-2=(1/4)(x-2)
y-(3/2)-(1/4)x=0
Eqn required : 4y-x-6=0
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3. cos(170+θ) = cos 60 or cos300
170+θ = 60 or 300
θ = -110 (rej.) or 130


希望幫到你啦

3y - 3 = 4x(1)
y - 7x = -16(2)

由(2):y=7x-16(3)
把(3)代入(1)
3(7x-16)-3=4x
21x-48-3=4x
-51=-17x
x=3
把x=3代入(3)
y=7(3)-16
y=21-16
y=5

1)
3y-3=4x
3y=4x 3
y=(4x 3)/3 --- (1)
Sub (1) into y-7x=-16
(4x 3)/3-7x=-16
4x 3=-48 21x
17x=51
x=3
Sub x=3 into 3y-3=4x
3y-3=4x3
3y-3=12
3y=15
y=5
So x=3, y=5.

sorry,,,我只係可以幫到你第一題,,,=(

我淨係識第一咋

(1) 3y-3=4x------(1)
y-7x=-16--------(2)

From (1):x=3y-3over4--------(3)

sub (3) into (2)

y-(3y-3over4)=-16

4(3y-3over4)=(-16-y)4

3y-3=-64-4y

3y+4y=-64+3

7y=-61

y=-9

Sub y=-9 into (1)

3(-9)-3=4x

-27-3=4x

-30=4x

-30 over4=x

x=-30 over4

so The solution is y=-9, x=-30 over4.