A car runs 255 kilometers in 3 hours. According to this calculation, how many kilometers can the car run in 5 hours?

A car runs 255 kilometers in 3 hours. According to this calculation, how many kilometers can the car run in 5 hours?

The two cars of a and B leave from the two stations at the same time. Car a travels 51.5 kilometers per hour. The speed of car B is 1.2 times that of car a. the two cars pass through 4.2 hours

Is the title still incomplete?
According to the known conditions, if the two stations meet after 4.2 hours, the distance between them can be calculated
（51.5+51.5×1.2）×4.2
=113.3×4.2
=475.86 (km)

Simple operation of 43 and 1 / 3x1 / 4 + 51 and 1 / 4x4 / 5

It should be 51 and 1 / 4
41 + 1 / 3x3 / 4 + 51 + 1 / 4x4 / 5 + 61 + 1 / 5x5 / 6
=（40+4/3）x3/4+（50+5/4）x4/5+（60+6/5）x5/6
=40x3/4+4/3x3/4+50x4/5+5/4x4/5+60x5/6+6/5x5/6
=30+1+40+1+50+1
=123
If it's 51 and 3 / 4
41 + 1 / 3x3 / 4 + 51 + 3 / 4x4 / 5 + 61 + 1 / 5x5 / 6
=（40+4/3）x3/4+（50+7/4）x4/5+（60+6/5）x5/6
=40x3/4+4/3x3/4+50x4/5+7/4x4/5+60x5/6+6/5x5/6
=30+1+40+7/5+50+1
=122+7/5
=123 and 2 / 5

X-2 / 5 = 1 / 2-3 / 10

x-2/5=1/2-3/10
x-2/5=1/5
x=1/5+2/5
x=3/5

Is to solve the equation, (x / 3) * 10 + (115-x / 3) * 5 = what?
115 students from two classes of grade seven in a school actively participated. It is known that one third of the students in class a donated 10 yuan each, while the rest of the students in the two classes donated 5 yuan each. If there are X students in class A, what is the total amount of donations of the two classes?

No need to solve the equation. He has told you that there are X students in class A
The answer is to use X to represent the total amount of donations of the two classes and simplify the equation

Solving equation 8 × (x + 5 / 1) = 3.25 / 3 × (x-50%) = 10 / 1

8 × (x + 5 / 1) = 3.2, 1 / 5, writing 1 / 5,
8×（x+1/5）=3.2
8×（x+1/5）÷8=3.2÷8
(x+1/5)=0.4
(x+1/5)×5=0.4×5
5x+1=2
5x+1-1=2-1
5x=1
5x/5=1/5
X = 1 / 5, [1 / 5]
5 / 3 × (x-50%) = 10 / 1, 3 / 5 writing 3 / 5, 1 / 10 writing 1 / 10,
3/5×（x-50％）=1/10
3/5×（x-1/2）=1/10
3x/5-3/10=1/10
3x/5=1/10+3/10
3x/5=4/10
3x=2
X = 2 / 3, [2 / 3]

Solving equation 3 (X-2) + 1 = x - (2-1)

3x-6-1=x-2+1
2x=-2+1+6+1
2x=6
x=3

False score: 4 out of 7, 23 out of 9, 13 out of 3, 11 out of 6, 17 out of 8
Is a false score into a score with ah, don't make a mistake... OK to give points····

23 out of 9 = 2 and 5 out of 9
13 out of 3 = 4 and 1 out of 3
6 out of 5 = 1 and 1 out of 5
11 out of 6 = 1 and 5 out of 6
17 out of 8 = 2 and 1 out of 8
Pseudo fractional with fraction: divide the numerator by the denominator, the quotient is the integral part of the fraction, the remainder is the numerator of the fraction, and the denominator remains unchanged

Solution equation: 2x △ (5 / 6 + 7 / 8) = 7 / 10

2x ÷ (5 / 6 + 7 / 8) = 7 / 10
2x ÷ (20 / 24 + 21 / 24) = 7 / 10
2x △ 41 / 24 = 7 / 10, both sides multiply by 41 / 24 at the same time
2X = 7 / 10 × 41 / 24
2X = 287 out of 240 divided by 2 at the same time
X = 287 out of 480

A problem of solving equation, do me a favor!
54=1.8x

54=1.8x
1.8x=54
x=54/1.8
x=30