Math problems, equation solving, columnar formulas, no solutions

Unary equations: Let the math book x, then the language book x + yuan, then 28x+30 (x + binary system of equations: let the math book x yuan, the language book y yuan, then: 28x+30y=

Solution: Let the math book be x-yuan then the language book (x+ yuan.)

28x+30（x+

Solution: Let the language book have x yuan, and the math book is (yuan.

then 30x+28 (

Unary quadratic equations.

Set the language of the text x yuan. Math book y yuan.

30x+28y=

x=y + unary linear equation.

Set the language of the text x yuan. Mathematics Book (Meta.

30x+28(

If each book of language is $ x, then each book of mathematics ** is available.

30*x+28*(

Calculate x math ** then is.

If the language book is x, then the math book is the yuan.

30x+28(

When an ornamental stone sinks into the water, it can be seen that the volume of the ornamental stone is the same as the volume of the rising water.

Volume of ornamental stones: 80 60 2 9600 cm

The volume of the ornamental stone is the volume of the falling water, and the formula is: 80 60 2 = 9600 cm.

Assuming that the number of vats is the same as the kegs, then the number of vats and kegs is 15000 (1000-750) = 60 kegs, which adds up to 120 kegs, which is 70 more than 50 kegs, and since the large kegs can hold 1000 and the kegs can hold 750, the number of 3 kegs can hold = the number of kegs that can be filled with 4 kegs.

So to reduce 3 large barrels, you need to reduce 4 small barrels, and since addition, you have to subtract 7 barrels (3 large and 4 small) at the same time to ensure that the large barrels contain 15,000 grams more than the small barrels 70 7 = 10 that is to say, you have to subtract 10 times to add large barrels to small barrels = 50 barrels, so the number of large barrels is 60 - (3 * 10) = 30 barrels The number of small barrels is 60 - (4 * 10) = 20 barrels.

Set x small barrels and 50-x large barrels.

1000（50-x）-750x=1500050000-1000x-750x=150001750x=35000

x = 20.

The large barrel is 50-20 = 30 barrels.

The kegs are 20 pcs.

Each vat contains 250 grams more than the keg.

If a total of 1,500 grams of oil is added, then there are 6 more vats than the kegs, and then the 6 are deducted from the total 50, which is 44, and the number of vats and kegs is the same.

On average, 22 each.

Then 6 more large barrels than small barrels is 28 and 22 small barrels.

The extra weight of the large barrel than the small barrel plus the weight of the assumption that all the small barrels are 15000 + 750 * 50 = 52500 grams.

52,500 (1,000+750) = 30 barrels.

This is the number of vats.

So, the number of kegs is 50-30 = 20 kegs.

Actually, I'm taking the equations apart. The column equation is much simpler.

Let's assume that a cow eats 1 unit of grass in a day.

Then according to the title, the grass that grows every day is (17*30-19*24) (30-24)=9

Let's assume that the four cows are not sold, but they are given an extra 2*4=8 units of grass for the last two days to eat.

Then a total of (6+2)*9+2*4=80 units of grass were eaten.

So there are a total of 80 (6+2)=10 cows.

ps: For primary school math problems, you don't always need to solve the problem with one formula, sometimes you can divide it into multiple formulas, so that the problem solving process is clearer. Hope it helps.

If x tickets are sold for $40 and $50 each, $30 tickets will be sold.

40x+50x+30(200-2x）=7800

40+50=90;

The number of tickets for both $40 and $50 is 60

The number of votes for $30 is 80

Set 30 yuan to buy x sheets, 40 yuan to sell y sheets. 50 yuan for selling Z sheets.

x+y+z=200

30x+40y+50z=78000, y=z30 yuan for 80 sheets, 40 yuan for 60 sheets.

400 30 - 8880) (Hui Zheng Lao 100 + 30).

24 (buried in front of the box).

24 boxes of cargo were lost.

If there is no knowledge of the loss of goods, the foundation should get 400 * 30 = 12000 freight

The actual loss of freight for a box is 30 + 100 = 130

So the loss of ambush cargo (12000-8880) 130 = 27 boxes.

Solution: (1) 45 (3 7) (5 6) = 126 (person) Answer: There are 126 students in the sixth grade. (2) Small ball 63 9 * (9 + 14) = 161 (pieces) Football 63 9 * 14 = 98 (pieces) Answer: 161 balls and 98 balls and soccer balls respectively.

3) 32 * (1 8) = 4 (tons) A This year plans to increase production by 4 tons (4) 2 [(2 7)-(1 4)] = 56 (km) A: The total length of the highway is 56 km. (5) The length of the rectangle is [840 2] * [4 (4+3)] = 240 (m) The width is 240 * (3 4) = 180 (m) So the area of the rectangle is 240 * 180 = 43200 (m2) A: The area of the rectangle is 43200 square meters. (6) Because 5+7=12 The multiple of 12 between 40 and 50 is 48, so there are 48 students in this class, of which 48 are boys * [5 (5+7)] = 20 (people) Girls account for 48-20 = 28 (people) (7) This sugar water has sugar 363 * [1 (1 + 100)] = 363 101 (kg) Water 363 * 100 101 = 36300 101 (kg) A: 363 101 kg and 36300 101 kg of water are needed.

Fill in the blanks (15):(8).

1,45/(3/7)/(5/6)=126

2,63 (9 14) = 98 small balls.

98 + 63 = 161 small footballs.

5，840/2/（3+4）*3*4=7206，

Tens of digits of 2 3, 1-2 3 = 1 32 (1 3) = 6 digits.

The number of tens is 6-2 = 4

So it's 46

There are only 2 3 4 6 6 9 in 2:3.

Because ten digits plus 2 equals single digits, that equals 46