數學問題[[方程]]

1.兩直線 L1:2x+5y-4=0
及 L2:3x+py+17=0
相交於點(q,2)，求p和q的值。

answer:(q,2)代入L1及L2得 2q+6=3q+p+17
p=-q-11-------(1)
由L1得: 2q+6=0
q=-3
p=-8

2.若(x,y)=(2,-3)是聯立方程
4x+ay+b=0-----1
{
x-ay+b=0-----2
的一個解，則b=

answer:(2,-3)代入1和2 得:
8-3a+b=0-----3
{
2+3a+b=0-----4

4-3得:-6+6a=0
a=1
b=-5

1.
相交於點(q,2)，代入y=2,
L1: 2x+5(2)-4=0
X=-3
代入y=2,X=-3,
L2:3(-3)+p(2)+17=0
P=???
相交於點(q,2)，L1:2x+5y-4=0
2(q)+5(2)-4=0
q=???

2.
(x,y)=(2,-3) 代入 方程,你就搵到答案la..
要自己試試做先識咖

1. sub (q,2) into L1, we have
2q+5(2)-4 = 0
2q = -6
q = -3
sub (-3,2) into L2, we have
3(-3)+p(2)+17 = 0
-9+2p+17 = 0
2p = -8
p = -4

2. sub (2,-3) into 4x+ay+b = 0, we have
4(2)+a(-3)+b = 0
8-3a+b = 0
b = 3a-8 ---------(1)
sub (2,-3) into x-ay+b = 0, we have
2-a(-3)+b = 0
2+3a+b = 0 ------------- (2)

sub (1) into (2), we have
2+3a+3a-8 = 0
6a-6 = 0
a = 1
sub a =1 into (1), we have
b = 3(1)-8
b = -5