數學(急!)

(y-x)²=16........1
{
y=2x-1.........2

sub 2 into 1
(2x-1-x)²=16
(x-1)²=16
x-1=±4
x=5 or x= -3.........3

sub 3 into 2
y=2(5)-1 or y=2(-3)-1
y=9 or y= -7

so the solution is (5,9) or (-3,-7)
唔知對不對

y-x^2=16 ------------------(1)
y=2x-1 ------------------(2)

Put (2) into (1) ,we have

(2x-1) - x^2 = 0
2x - 1 - x^2 = 0
x^2 - 2x +1 = 0
(x-1)(x+1) = 0

Therefore, x = 1 or x = -1

When x =1, or when x = -1
y = 2(1)-1 y = 2(-1) - 1
y=1 y = -3


Hence ,the solution is (1,1) or (-1,-3)

y-x^2=16 ___(10
y=2x-1 ___(2)
(2)式代入(1)式
(2x-1)-x^2 = 16
x^2 – 2x + 1 + 16 = 0
x^2 – 2x + 17 = 0
判別式大於零所以沒有實根。
但用複數計算

x = {2±√[2^2 – 4(1)(17)]}/2
x = {2±√[4 – 68]}/2
x = {2±√[– 64]}/2
x = {2±8i}/2
x = 1±4i
y = 2x – 1
y = 2(1±4i) – 1
y = 2±8i – 1
y = 1±8i

y-x^2=16...........(1)
{
y=2x-1..............(2)

from (1), y=x^2 +16...............(3)

Sub. (3) into (2)

x^2+16 = 2x-1
x^2 - 2x + 17 =0
delta=(-2)^2 -4(1)(17)
= -64<0
∴no real roots

(y-x)²=16........1
{
y=2x-1.........2

sub 2 into 1
(2x-1-x)²=16
(x-1)²=16
x-1=±4
x=5 or x= -3.........3

sub 3 into 2
y=2(5)-1 or y=2(-3)-1
y=9 or y= -7

so the solution is (5,9) or (-3,-7)

Put (2) into (1)
(2x-1)-x^2=16
-x^2+2x-1=16
x^2-2x+1=-16
x^2-2x+17=0
by qua. equation
x = no solution

don't know if I count wrongly......