A bug landed at a point P0 on the number axis. For the first time, it jumped 1 unit to the left from P0 to P1, and for the second time, it jumped 2 units to the right from P1 to P2, Jump 3 units from P2 to P3 for the third time, and jump 4 units from P3 to P4 for the fourth time. After jumping 100 times according to the above rule, it falls at P100 on the number axis, and the number represented by P100 is exactly 2003. What is the number represented by P0?

Let the initial position be A0, A1 = a0-1, A2 = a1 + 2, a2k = a (2k-1) + 2K, a (2k + 1) = a2k - (2k + 1), then a (2k + 1) = a (2k-1) - 1, a (2k + 2) = (a2k) + 1, odd term is the arithmetic sequence with tolerance of - 1, even term is the arithmetic sequence with tolerance of 1, A100 = 2003, A100 = A0 + 50, then A0 = 2003-50 =

Solving the equation X-2 / 3x = 3

x-2/3x=3
1/3x=3
x=3÷1/3
x=9

3 / 2-x / 3-3x = - 2 to solve the equation

The solution is 3 / 2-x / 3-3x = - 2
Multiply both sides by six
9-2x-18x = - 12
That is - 20x = - 21
That is, x = 21 / 20

The solution equation is y-0.8y = 14.2

y-0.8y=14.2
0.2y=14.2
y=14.2÷0.2
y=71

0.95+6x=1.1 5.8y-y=14.4

0.95+6x=1
0.95+6x-0.95=1-0.95
6x=0.05
x=1/120
15.8y-y=14.4
14.8y=14.4
y=34/37

3 (y + 3) - 8y = 14 what is the value of Y

3y+9-8y=14
-5y=5
So y = - 1

Solution equation: 1 / 4x (X -- 0.75) = 13 / 16

4.75

Solution equation: (600 + 6x) △ 30 × 13 = 148 + 4x = = please tell me more about it!
It is: (600 + 6x) △ 30 × 13 = 148 + 4x

（20+6X）*13=148+4x
260+13/5X=148+4x
X=90

4X + 6x + x = 12 - (x + 4),

4X + 6x + x = 12 - (x + 4)
4x+6x+x＝12-x-4
4x+6x+x+x＝12-4
12x=8
x=8/12
x=2/3

①1/4x-4x=1/2 ②50%x-5/6=5/12 ③6x+4(20-x)
①1/4x-4x=1/2 ②50%x-5/6=5/12 ③6x+4(20-x)=90

The sixth grade mathematics problem solving equation:
①1/4x-4x=1/2
-15/4x=1/2
x=-2/15
②50%x-5/6=5/12
0.5x=5/12+5/6
0.5x=5/4
x=5/2
③6x+4(20-x)=90
6x+80-4x=90
2x=90-80
x=5