Please help solve two math problems for the first grade of junior high school

Set x people to produce screws, 100-x people to produce nuts.

16x*2=18*(100-x)

32x=1800-18x

x=36100-36=64 people.

36 people for screws, 64 people for nuts.

Let the velocity of A be x and the velocity of B be y

4x-4y=16

x+y=16

x=10y=6

A has a speed of 10 kmh and B has a speed of 6 kmh.

1) Equipped with x individual production screws, (100-x) individual production nuts.

List the equation: 2*16x=18(100-x)50x=1800

x=36, so 100-x=64

Answer: 36 people produce screws, 64 people produce nuts to make the screws and nuts just matched.

2) Let the speed of A be xkm h, and the speed of B is ykm h.

List the system of equations: 4*(x-y)=16

1*（x+y）=16

Solution: x=10 y=6

A: The speed of A is 10km h, and the speed of B is 6km h.

1。There are x people to produce screws and Y people to produce nuts.

x+y=100

16x*2=18y

x=36，y=64

36 people produce screws and 64 people produce nuts.

2。The speed of each of them is x,y kilometers per hour.

x-y=16/4

x+y=16/1

x=10，y=6

The speed of A and B is 10.6 km/h each.

A set of assigned to make screws have x people, nuts y people:

x+y=100

2×16x=18y

x=36y=64

2. A velocity is x, B is y:

1×(x+y)=16

4x-4y=16

x=10 y=6

1.The solution is set up by person A to make screws and person B to produce nuts.

a+b=100

16a*2=18b

Solution. a=36

b=642."It takes 4 hours for A to catch up with B, and if he travels in the same direction"What does that mean.

I did what I had in mind.

The speed of dissolving A is a, and the speed of the dissolution is b

a-b)4=16

a+b)1=16

Solution. a=10b=6

Answer 1: After all of them, combine the same terms to get =-6x +18+18 and bring in x=1 3 to get the formula equal to 70 3

Answer 2: Merge the formula after (x +3x+2) (ax+b) to get ax +(3a+b) x +(2a+3b)x+2b and because the quadratic coefficient is 4, 3a+b=4; If the coefficient of the primary term is -1, then 2a+3b=-1; After synthesis, we get a = 13 7 and b = -11 7

Joke! First Question! It's not so difficult in the second year of junior high school! I'm in the top 50 in math!

The two steamers were driven out of the two places at the same time, their respective speeds are fixed, the first encounter is 700m away from the first bank, continue to move forward after the encounter, and return to the opposite bank immediately after the time is not counted), the second encounter is 400 meters away from the second bank, how many meters is the river wide?

The first encounter, 700 meters away from the first bank, then the boat from the first place traveled 700 meters, the second encounter, a total of 3 whole journeys, then the boat from the first place traveled 700 3 = 2100 meters.

At this time, the distance from B is 400 meters, then the whole process is 2100-400 = 1700 meters, if the distance from the first bank is 400 meters, then the whole process (2100 + 400) 2 = 1250 meters, the minute hand goes one minute and walks 6 degrees, that is, the angular velocity of the minute hand is: 6 degrees, and the hour hand goes one minute, that is, the angular velocity is: degrees.

At first, the minute hand was 110 degrees behind the hour hand, and later the minute hand was 110 degrees in front of the hour hand, which was a catch-up problem.

Set shared x points.

x[x=40

That is, a total of 40 minutes out.

Question 1:

The width of the river is set to be x meters.

Then the distance of the two ships in the first meeting is: x-700 700, and the distance of the second encounter is: (2x-400) (x+400) so the columnable equation is:

x-700) 700=(2x-400) (x+400), can solve x=1700m.

Question 2: 40 minutes.

X minutes out, the minute hand to the 6 o'clock position angle 180-6x, the hour hand to the 6 o'clock position angle x 60 * 30

180-6x+

x=140/11

Y minutes come back, the minute hand to the 6 o'clock position angle 6Y-180, the hour hand to the 6 o'clock position angle Y 60*30

y=580/11

y-x=440/11=40

Both questions test the mastery of the perfect squared formula.

As long as you can understand (a+b) 2=a 2+b 2+2ab; (a-b) 2=a 2+b 2-2ab, the problem is very good.

1.Solution: Because a+1 a=8, a 2+(1 a) 2=(a+1 a) -2=62

2.Solution: Because a+b=3, ab=2So.

a^2+b^2=(a+b)^2-2ab=5(a-b)^2=a^2+b^2-2ab=(a+b)^-4ab=1.

1. A+1 A=8 (A+1 A) 2=64 A 2+2+(1 A) 2=64 So A2+(1 A) 2=62

2、a+b=3 (a+b)^2=3^2 a^2+b^2+2ab=9 a^2+b^2=9-4=5

a-b) 2=a 2+b 2-2ab=5-4=1 Note A 2 is the square of a.

(A + 1/a) squared = a square + 2 + 1/2 + 1/2 a square = 64, so a square + a square = 62

The square of a + the square of b = the square of (a + b) - 2ab = 9-4 = 5 (a-b) square = the square of a - 2ab + the square of b = 5-2 * 2 = 1

1 in 10 (8-a).

a-b) = 2/3 (3-b).

It's so complicated, I'm tired

1、∵（a+b)^2=m，（a-b)^2=n∴a^2+2ab+b^2=m ①

a^2-2ab+b^2=n ②

2(a 2+b 2)=m+n a 2+b 2=(m+n) 2

4ab=m-n ab) 3= (m-n) 4 3

Isn't question 2 a bit of a problem, too easy, right? If so, it is equal to 10 squared, 100.

1.(a+b)*2=m, that is, 2a+2b=m, in the same way, 2a-2b=n, the two formulas are added, 4a=m+n

Subtract the two formulas, 4b=m-n, needless to say in the future, right?

2.Isn't it 20 ......

a 2 ten b 2 = (m+n) 2

ab）^3=〔（m-n）/4〕^3

a, ten 1, a=10

The squares of the two sides get: a 2 ten 2 ten (1 a) 2 = 100, a 2 ten (1 a) 2 = 98

（1）a^2+b^2= [（a+b)^2+（a-b)^2]/2=(m+n)/2

2) Because (a+b) 2-(a-b) 2=4(ab) 2=m-n, ab=root(m-n).

So (ab) 3=(m-n)*root number (m-n)(3) from a+a/1=10, get (a+a/a) 2=100, are you wrong in the last one? Should I ask for a 2 + a 2 out of a 2?

In this case, the answer that can be reached by squarering both sides at the same time should be 98