Junior high school math problems, urgent, tomorrow

Set to buy a fruit x kilograms, the unit price is y yuan.

to get a system of equations. xy＝100

x+10）（y+

Solution: x1 40 ; y1＝

x2＝50 ；y2＝2

According to the first case x1 40; y1, he sold a total of yuan.

Lost the yuan. According to the second case x1 50;y1 2, he sold a total of: yuan.

Earned 294-250 44 yuan.

This is the result of one of my classmates who is studying at South China Normal University, and there is nothing wrong with the question he has written, not to mention that this is your junior high school question, which is a bit deep for you now, but it is still difficult for him, I believe him, and it depends on you whether you believe it or not

First of all, let the B fruit be xkg, then the A fruit is x-10kg, and the result of the solution is two x=60 or x=50, x=60.

The formula is 60*

x=50.

The formula is 50*

Assuming that the ** of the fruit is X yuan and the weight is y kg, then:

x*y＝100

x+ finds x, y, and then calculates how much he sells.

Suppose this person buys x fruits A and y fruits B.

Available Equations:

x=y-10

100/x=150/

After calculating x y, the following is a matter of course.

It's difficult, which professor will answer it?

When point C is in the second quadrant, the right triangle BMC and the right triangle OAB are congruent when the cm x axis and M are done through point C.

bc=ba=13, so cm=bo=2, bm=oa=3;So the coordinates of point c at this time are (-5,2);

Point C is in a similar situation in the third quadrant, please practice it yourself, and you will be familiar with it later.

Point c is at any point in the line where y= is located.

First find the slope of ab, =2 3, and then find the straight line y=-3 2x-9 2 that is perpendicular to ab and passes b;

Just a few... It's all right.

It's best to find a two-quadrant and a three-quadrant.

There are an infinite number of such dots.

What else does the question require for C?

There are countless points, what else is there to ask for?

If there are x trucks in A, then 20-x in Truck B, 4x+12(20-x)>=100 x<=

--15<=x<=

11x+7(20-x)>=200

Because x is the number of trucks, it is an integer. Therefore, there are two schemes: x=15 or x=16, which is 115 trucks A and 5 trucks B; 2.There are 16 trucks A and 4 trucks B.

Set up x car A, y car B.

then 4x+12y 100

11x+7y≥200 ②

x≤20 ..3y is less than or equal to 20...4

You can draw a graph to find the logarithm of x,y, or you can do it directly algebraically by x25-3y 0, so y 8

Take y=8 and substitute 20 x 14 to get 14,15,,18,19,20 There are 7 kinds.

Take y=7, substitute 2, get 20 x 14, take 14, 15, 16 17 18 19 20 has 7 kinds.

Take y=6 and substitute it into 2 to get 20 x 15, there are 6 kinds.

Take y=5 and substitute it into 2 to get 20 x 15, there are 6 kinds.

Take y=4 and substitute it into 2 to get 20 x 16, there are 5 types.

Take y=3 and substitute it into 2 to get 20 x 17, there are 4 kinds.

Take y=2 and substitute it into 2 to get 20 x 17, there are 4 types.

Take y=1 and substitute it into 2 to get 20 x 18, there are 3 types.

There are 50 types of the above, and the scheme is as above.

Drawing a shape is the fastest, just count the whole number of points.

If x car A is used, then B will use (20 x) car.

From the title: 11x+7(20 x) 200 4x+12(20 x) 100

Solution: 15 x

Because x is an integer, x

Therefore, there are three schemes: 1. 15 vehicles for A and 5 vehicles for B; 2. 16 vehicles of A and 4 vehicles of B; 3. 17 cars A and 3 cars B.

This is a question for junior high school students, and is it clearly a linear programming problem for high school students?

As long as the following three conditions are met (x, y is an integer, x is A, y is B), 4x+12y 100

11x+7y≥200 ②

x=20 ..3

The method is similar to that of the upstairs, so I'll summarize the results. A 15, B 5, A 16, B 4, A 17, B 3. In other words, there are three kinds of results.

Solution: Let the water injection speed of the three water pipes A, B and C be as follows: a, b, c; Then there are: 6a = 8b = 12c, that is, 6a is the volume of this pool; Replace b and c with the formula containing a to obtain: b = 3a 4;c=a/2

If the pool can be filled with water after opening the C pipe x hours, then the question can be obtained: (2+x)(a+b)-cx=6a

x=2Answer: The pool can be filled with water after opening the C pipe for 2 hours.

[1-2(1 6+1 8)] 1 6+1 8-1 12) = 2 hours solution: The pool can be filled after opening the C pipe x hours.

1 6 + 1 2-1 12) x+2 (1 6 + 1 8) = 1x = 2 Answer: The pool can be filled up after opening the C tube for 2 hours.

Analysis: A tube can be filled in 6 hours, then it is injected into the pool in 1 6 2 hours, 2 6 hours, and a 6 in an hour

Pipe B can be filled in 8 hours, so it will be filled into the pool in 1 hour, 8 pipes C can be emptied in 12 hours, then it will be discharged from the pool in 1 hour, and A 12 will be discharged in 1 12 hours

Set the C pipe to fill the pool with water after a few hours.

2 6 + 2 8 + a 6 + a 8-a 12 = 1 solution Multiply both sides by 48 16 + 12 + 8a + 6a - 4a = 4810a = 20

a=2 (hours).

A.

[1-2(1 6+1 8)] 1 6+1 8-1 12)=2h solution: The pool can be filled after opening the C pipe x hours.

1 6 + 1 2-1 12) x + 2 (1 6 + 1 8) = 1 Solution: x = 2

Answer: The pool can be filled up after 2 hours after opening the C tube.

A's speed is one-sixth and B's speed is one-eighth of the speed of driving together for 2 hours is the sum of one-sixth plus one-eighth Multiply 2 equals 7/12 Because C has a speed of 1/12 so the difference between 1 minus 7/12 divided by one-sixth plus one-eighth minus 1/12 = 2 hours.

Solution: The pool can be filled after opening the C pipe for x hours.

2（1/6+1/8）+x(1/6+1/8-1/12)=1

x=2A: The pool can be filled up after opening the C tube for 2 hours.

Set x hours to fill up, and turn the full pool of water into 1;

2 6 + 2 8 + x 6 + x 8-x 12 = 1 A bet 1 6 per hour, B bet 1 8 per hour, C 1 12 per hour, so, the equation is.

2 6+2 8+x 6+x 8-x 12=1 solves the equation to get x=2

Solution: If you open the C pipe for x hours, you can fill the pool with water.

2/6+2/8+（1/6+1/8-1/12）x=1

x=2A: The water is full for two hours.

acres, 2 acres.

2. A: y1=5x

B: y2=kx+b

Points (1,7) and points (5,23) are substituted.

7=k+b23=5k+b

The solution is k=4 and b=3

i.e. y2 = 4x+3

When 5x=4x+3

i.e. when x=3.

y1=y2Answer: After 3 hours, the number of acres irrigated by team A is equal to the number of acres irrigated by team B.

(1)5 2

2) B's function: y=4x+3, condition: x>=1A's function: y=5x

Let's say x hours have passed, both are the same.

That is, 4x+3=5x

then x=3 so it is 3 hours.

I hope to help the landlord, it is not easy to answer the question.

1）7，2

2) The intersection of the two lines of A and B is a perpendicular line to the x-axis, and the x-coordinate of the perpendicular foot is about 3) The visual inspection is about 3) You can also set a function to solve: (mark the inflection point as a and the end point as b) A: Y1=5X

B: y2=kx+b (17=k+b.)

23=5k+b

The solution is k=4 and b=3

i.e. y2 = 4x+3

When the number of irrigated acres is equal, it is y1=y2

That is, 5x = 4x+3

i.e. when x=3.

Answer: After 3 hours, the number of acres irrigated by team A and the number of acres irrigated by team B is equal In fact, strictly speaking, zero point is also counted, but the number of irrigated acres is 0

(1) In the first 1 hour, Team B irrigated 7 acres per hour, and after 5 hours, Team A irrigated 2 acres more than Team B.

2) Team A 25 5=5, Team B (23-7) (5-1)=4, 7-5=2, 2 (5-4)=2, 2+1=3After 3 hours, the number of acres irrigated by team A is equal to the number of acres irrigated by team B.

Just look at the picture. (1) 7 (look at the x-axis, find the y-value corresponding to 1) 2 (look at the x-axis to find the y-value corresponding to 5, subtract the two y-values) (2) 3 (equal represents the intersection of two straight lines).

7 acres; The difference between the two areas).

y1 = 5xy2 = 7x (04 x + 3 (1 solution to the system of equations.