A batch of coal is transported to a factory. It is planned to burn 5 tons per day for 90 days. How many days can it burn 4.5 tons per day

A batch of coal is transported to a factory. It is planned to burn 5 tons per day for 90 days. How many days can it burn 4.5 tons per day

5:4.5=90:X

There is a batch of coal that is planned to burn 15 tons a day and can be burned out in 100 days. In fact, it burns one fifth less than the plan every day. How many days did it actually burn?
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100X5/4=125

The school plans to use 45 tons of coal in October, which is 18% less than the plan?

45 × (1-18) = 45 × 78 = 710 tons; answer: 710 tons of coal are actually used

A pile of coal, the factory originally planned to burn 60 days, 15 tons a day, the actual daily savings of 20% than the original plan, this batch of coal actually burned how many days?

60 × 15 = 900 (tons); 15 × (1-20%) = 12 (tons); 900 ／ 12 = 75 (days). Answer: this batch of coal actually burned for 75 days

Equations: half x-one-third y = 1, negative one-third X-Y = two-thirds, solve it (have process)

Original formula: 1 / 2x-1 / 3Y = 1 ①
-1/3x-y=2/3 ②
① Formula multiplied by 6 3x-2y = 6 ③
② The formula is multiplied by 3 - x-3y = 2
④ Formula multiplied by 3 - 3x - 9y = 6 5
③+⑤ y=-12/11,x=14/11

Equations x + y = 1, X-Y = half
The score is not very many, but requests the experts to tell me the process and the answer
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x+y=1 ①
x-y=1/2②
① + 2
2x=3/2
x=3/4
Substituting x = 3 / 4 into (1) yields
3/4 +y=1
y=1/4
To sum up, x = 3 / 4, y = 1 / 4

Solve the equation system half x-4y = 30 y-2x = 15

In this paper, a new system of equations is constructed by sorting out x-8y = 60 of half x-4y = 30 and y-2x = 15,
y-2x=15
We get y = 15 + 2x
Take y = 15 + 2x into x-8y = 60,
X=-12
Take x = - 12 into y = 15 + 2x,
Y=-9
Therefore, the solution of the original equations is x = - 12 and y = - 9

Equation group 1: 2 of X + y = 6-3 of X-Y. equation group 2: 4 (x + y) - 5 (X-Y) = 2 to solve the equation group, I don't know how to use addition, subtraction and elimination
This equation

The equation 1 3 (x + y) = 36-2 (X-Y)
3x＋3y＝36－2x＋2y
3x＋3y＋2x－2y＝36
5x＋y＝36③
The equation (2) 4x + 4y-5x + 5Y = 2
－x＋9y＝2
x＝9y－2④
④ Substituting 3, we get 5 (9y-2) + y = 36
45y－10＋y＝36
46y＝46
∴y＝1
Substituting y = 1 into 4 gives x = 7
∴x＝7,y＝1

The system of equations x-3 / 2 + Y-2 / 3 = 4, x-3 / 3 / 2 = 2 (addition and subtraction elimination method)?

(x-3)/2 + (y-2)/2 = 4 (1)
(x-3) - (y-2)/2 =2 (2)
(1)+(2),
(x-3)=6,x=9
(1)-(2)
(y-2)=2,y=4
So the final result is x = 9, y = 2

The following equations X / 2 + Y / 3 = 7 3 (X-2) = 2 (Y-9) are solved by the method of addition, subtraction and elimination

695221770 ：
x/2+y/3=7
3(x-2)=2(y-9)
It's time to tidy up
3x+2y=42①
3x-2y=-12②
① + 2, get it
6x=30
x=5
Substituting x = 5 into 1, we get
3×5+2y=42
15+2y=42
2y=27
y=27/2
The solution of this system of linear equations of two variables is: x = 5, y = 27 / 2