Who will help me solve 3 unary quadratic equation application problems? Thank you

1: Let's use the equation to solve it.

The original solution can make x sets of computer desks, which are: each set of computer desks needs to use 6 x m of wood. After the improvement, you only need to use 6 x-1 55m wood without a computer desk, and the following equation can be listed according to the question:

6 x-1 55)(x+3)=6, from which x can be solved. I'll leave the calculation to yourself, hehe.

2: Let A and B meet, A's velocity is x, B's velocity is y, and when they meet money, A's velocity is x, and B's velocity is y-10. The time from the beginning to the meeting can be calculated:

t=600/(x+y).So when they meet, A has traveled a distance of 600x (x+y) and B has traveled a distance of 600y (x+y). In the period of time between the encounter and the arrival at the destination, A has traveled 600x(x+y), so A has traveled 600 (x+y) time.

In the same way, B walked 600y [(x+y)(y-10)].Second, according to the meaning of the topic, we can know that the time used at home is one hour more than the time used by B, that is, there is 600 (x+y)=600y [(x+y)(y-10)]+1. You can solve the answer from this equation.

3: If the price of each piece is reduced by x yuan, the profit of each piece will be (40-x) yuan. Now it's a matter of figuring out how many pieces we can sell each day.

If the price of each piece is reduced by 4 yuan, you can sell 8 more pieces, so if we reduce the price by x yuan, we can sell 4x more keys, that is, we can sell 20 + 4x pieces. So, according to the meaning of the title there are:

20+4x)(40-x)=1200, and then it's time to solve the equation, I believe you have no problem. If you still have questions about solving equations, you can continue to ask me, and I will do my best to answer them for you.

1. The original plan can make x sets of computer desks.

6/x-1/55)(x+3)=6

x=-33 rounded.

x=30

1. Solution: The fabric of the tailoring jacket and pants is x m, y m x + y = 15 respectively

x=2y (1 shirt or 2 pants can be cut per 1 meter of fabric, in order to make the jacket and pants match, the fabric of the shirt is 2 times that of the pants).

The solution is x=10

y=5, so the fabric of the shirt and pants is 10 meters, 5 meters2, and the solution: let the speed of A and B be x kilometers and Y kilometers respectively, 5 (x+y) = 25

5x+3(x+y)=25 (the sum of distances equals the total distance) gives x=2

y=3 so the speed of A and B is 2 kilometers per hour and 3 kilometers per hour respectively.

2x+2（x+y）=350

2y+3（x+y）=350

The solution is x=75

y=25, so two people make 75 and 25 parts each day.

1.Design: Cut the fabric x meters. Pants y m, then:

x+y=15

x=2y simultaneous solution: x=10, y=5

2.Let A's velocity be x and B's velocity y. Then:

5x+5y=25

5+3)x+3y=25

Simultaneous solving: x+y=5 8x+3y=25 5x=15 x=3 y=2

3.Let's set: A does x per day, and B makes y per day. Then:

2+2）x+2y=350

3x+(2+3)y=350

Simultaneous solving: 4x+2y=350

3x+5y=350

Add the two formulas to yield: x+y=100

Substituting the above equation, we get: 2x=150 x=752y=50 y=25

A garment factory processes a batch of suits, every 1 meter of fabric can cut a jacket or cut 2 pants, the existing fabric is 15 meters, in order to make the jacket and pants match, how many meters of fabric should be cut for the jacket and pants?

x tops, y pants.

x=y15=x+(1/2)y

Therefore: 15=x+(1 2)x=(3 2)x

Therefore: x=10

A and B start from two places 25 kilometers apart at the same time, walk like each other, meet at 5 o'clock, if A starts at 5 o'clock first, and then B sets off again, so that after 3 o'clock, the two meet, and find the speed of A and B.

A speed a km hour, B speed b km hour 25 = 5 * a + 5 * b --1).

25=(5+3)*a+3*b --2)(1)-(2)：3a=2b

Therefore: b=3a 2

Therefore: 25 = 5 * a + 5 * (3a 2) = (25 2) * a therefore: a = 2, b = 3

A batch of 350 machine parts is produced, A does it for 2 days, B joins the cooperation, and completes the task after 2 days, if B and B do it for two days first, and A joins the cooperation and needs to complete the task after 3 days, how many parts do the two people make each day?

A makes a part every day, B b.

2*a+2*(a+b)=350 --3)2*b+3*(a+b)=350 --4)(3): 2a+b=175 --3')(4): 3a+5b=350 --4')5*(3):

10a+5b=875 --5)(5)-(4):7a=525

a=525/7=75

b=175-2a=25

1、x+y=15

x=2y to get x=10 (top) y=52, set the armor to x 5x+5y=25

8x+3y=25 to get x=2 y=33, let the armor be x 4x+2y=350

3x+5y=350 gives x=75 y=25

Solution: Set the walking speed of Class A to be x kilometers per hour.

10 divided by x is equal to 10 divided by (x plus six) minus.

Set the width to x meters.

then a+2x=35 1).

ax=150 2)

From equation (2): a=150 x

x2=10，a=15

Because a is greater than or equal to 20m, so x2=10 is not in line with the topic, so the length of the chicken farm is 20 meters, and the width is meters.

If the width of the chicken farm is xm, the length is (35-2x)m and the equation is x(35-2x)=150

Finishing, get 2x2-35x+150=0

Solve the equation to get x1=10, x2=

So when x=10, 35-2x=15;

When x=, 35-2x=20

Therefore, when the width of the chicken farm is 10m, the length is 15m; When the chicken farm is wide and 20m long, I'm doing a task, if this answer is okay, you can help set it up to a satisfactory answer, thank you, help.

Solution: If the chicken farm is x meters long, it will be (35-x) 2 meters wide.

x*(35-x)/2=150

x^2-35x+300=0x=

Let the width be x, the lawn area is 760, then the total area of the two roads = 40*22-760=120

The area of the two roads is 40x+22x-x2=120, which gives x=60 or x=2, and because the rectangular ground is 22m, x<22 gets x=2

Set the width to x....Then we can see that 40x+22x-x*x=4o*22 40x represents the area occupied by one road and 22x represents another!! Since the width of the two strips is the same and vertical, subtract the square of the overlapping part x!! It is equal to the total area of 40*22, I won't solve it, I will solve it myself....

Price increase in this case:

Let the average percentage of the price increase be x, then: the current ** = original ** number type (1 + x) number of price increases.

Price increase this Heng Brother Situation:

Let the average percentage of price increase be x, then: now**=original** (1-x) number of price reductions.

Set a price reduction of x percent each time

300 (1-hundredth x) 2 = 243 to unravel.

Price increase in this case:

Let the average price increase percentage be x, then: current price cong faction = original ** (1+x) number of price increases.

Price increase in this case:

Set the average percentage of price increase to x, and dismantle Zheng Hui.

Then: now**=original** (1-x) is reduced to x300 (1-hundredth x) 2=243 for each price reduction.

The price of each New Year's card should be reduced by X yuan.

The amount of money reduced by the price of x yuan is yuan, then the profit of each piece after the reduction is (yuan, because when the sales price is reduced by each yuan, its sales volume will be sold by 100 more, so the average daily profit after the price reduction x yuan can be sold (500 + 100x), so the average daily profit after the price reduction is (yuan.

So the equation (

Solution: The price of each New Year's card should be reduced by X yuan.

The answer is or should be a positive value, so the price should be reduced.

If you want to know how to solve it, you can add.

If there is a problem with the conditions of the question, the profit is 150 yuan when there are 500 tickets, and 120 yuan when there are 600 tickets.