Solve 4 junior high school problems and 4 junior high school math problems

The first mentions: set up x number of students.

kx+14=9x-3

9-k）x=17

So k = 6 then x = 17

So there are 17 students, and there are 17*9-3=150

The second is the same as the first.

Fourth. After adding the three formulas, divide by 2 to calculate x+y+z=45 and subtract the three formulas respectively to find xyz

The third question comes out after adding and subtracting.

Solve equation (1): 3x+4y=11 from +. 3x-y=9 from +

From - we get 5x=2 then x=

Substituting x=, we get y=, we get x=, y=, we get z=so x=, y=, z=

1.There are x number of students.

kx+14=9x-3

9-k）x=17

Because 17 is a prime number.

So (9-k) = 1 x = 17

Number of trees = 17 * 9-6 = 150 trees.

2.There are x names for men and y names for women.

15y/(x+y)=6

y=2/3x

The efficiency of girls is 2 3 of boys

So the man should plant 10 trees.

3.：x+2y-z=3

2x+2y+z=8

x-y-z=1

From the third one: x=1+y+z

Add it to the first one: y=2 3

Multiply the third left and right by 2: 2x-2y-2z=2

After subtracting from the second: 4y+3z=6, z=10 9, bring the values of x and y into the third, and get x=25 9

4.Similarly. x=12

y=15z=18

Solving a system of equations Solving a system of equations.

x+2y-z=3 (1) x+y=27 (1)2x+2y+z=8 (2) y+z=33 (2)x-y-z=1 (3) x+z=30 (3)2)-(1)get (2)-(1).

x+2z=5 (4) z-x=6 (4)

2*(3)+(2)(3)+(4).

4x-z=10 (5) z=18

4) +2*(5) to substitut z=18 into (4).

x=25/9 x=12

2) + (3) to substitute x=12 into (1).

3x+y=9 (6) y=15

2) +(1) So this system of equations is solved.

3x+4y=11 (7) x=12

7)-(6) gives y=15

y=2/3 z=18

Substitute x=25 9 y=3 2 into (3).

z=10/9

So this system of equations is solved as.

x=25/9

y=2/3z=10/9

1. Set the purchase price to x

What is improved.

The remainder is 10, which means that 1998 is divisible by these numbers, and there are 11 divisors greater than 10 in the divisor of 1998).

3. Because the sum of 2 numbers is 1576

So there is a maximum of 4 digits and no more than 1576: a=1 because the sum of 2 numbers is 4375 upside down

All have at least one 4-digit number.

Let one be ABCD

The other is xyz

abcd+xyz=1576

dcba+zyx=4375

a+z=5>x=4x+b=5

b=1b+y=7

y=6c+y=7

c=1c+z=3

z=2d+z=6

d=44、s=2-(1/2-3/2)-(1/3-4/3)-(1/4-..2008/2007-1/2008)=2-(-1)*2007=2+2007=2009

(1) Suppose the income of the previous year is $x. From the title: x*(1+(2) set x months later.

From the title: 800x+12000=20800 (3) Let the radius of the inner circle be x cm. From the meaning of the title:

10-x）^2π=200(4) x-2y=-2x+y

x+2x=2y+y

3x=3yx=y

1.Let it be made into x tabletops and y table legs to get the system of equations:

x/y=1/4

1/50)x+(1/300)y=5

i.e. y=4x.........1）

6x+y=1500………2）

Substitute (1) into (2) to obtain:

4x+6x=1500

i.e. 10x=1500

So x=150

Substituting x=150 into (1) yields:

y=4*150=600

So x=150

y=600, so the square table is 150.

2.Let the whole length s be long, and the velocities of Li Ming and Wang Yun are x and y, then there is a system of equations:

s/(x+y)=80………1）

s-60x)/(x+y)=40………2) Change (1) to x+y=s 80 and substitute (2) to get:

s-60x=(1/2)s

i.e. s=120x

So s x = 120 .........3）

That is, it takes 120 minutes for Li Ming to walk alone.

Change (3) to s=120x and substitute (1) to obtain.

120x=80(x+y)

That is, 80y=40x, i.e., y=(1 2)x

So s y = 2 (s x) = 240

That is, it takes 240 minutes for Wang Yun to walk alone.

3.Suppose the yield of the two fields last year was x, y(kg), then there is a system of equations:

x+y=470………1）

x(1+16%)+y(1+10%)=523………2) Change (1) to y=470-x and substitute (2) to simplify and find value:

x=100y=370

So x(1+16%)=116

y(1+10%)=407

Therefore, the yields of each field were 100 and 370 kg respectively before the improvement of the two fields.

Modified 116 and 407 kg.

4.From A to Party B, the flat road, uphill and downhill are x, y, and z (km) respectively. Then there is a system of equations:

x+y+z=70………1）

x/30+y/20+z/40=

x/30+z/20+y/40=

2) + (3) get:

x/15+(y+z)*(1/20+1/40)=x/15+(y+z)*(3/40)=

x 15 + (70-x) * (3 40) = change (1) to y + z = 70-x substitution].

Solving this unary equation of x yields x=18

2) + (3) get:

y-z)(1/20)+(z-y)(1/40)=(z-y)(1/40-1/20)=

z-y=4………4）

Substituting x=18 into (1) yields: y+z=52......(5) Simultaneous (4) and (5) to obtain a system of binary linear equations:

z-y=4y+z=52

Solution: y=24

z=28 so in total:

x=18y=24

z=28

It's too complicated, don't talk about math for many years. Think for yourself. Beneficial development

Teacher Wu in the sixth grade talked about La

1. You can add two equations to get: x squared + 2xy y squared = 2 + 5 and 2 on both sides of the equation at the same time gives 1 2x squared + 1 2y squared + xy=

2. Suppose the sales volume increases by x, and the original selling price is y, then the column equation is 1 y=(1+x), and both sides y eliminate the unknowns y and become 1= to get x=

3. The pursuit problem of this question The speed difference of the pursuit distance = the pursuit time is 36 (70-52) = 2

4. A student walked for 1 minute, and in that minute, the team walked 120 meters, and with the length of the line, the student walked (120+a) meters.

Landlord: Choose me...

x2+xy=2,y2+xy=5,2 are added together, then (x+y) is divided equally=7 answers.

1+x)* get x=

Taking the team leader as a reference, this student walked a meter more than the team leader, and the team leader walked 120*1=120m. Then this student walks 120+A meters.

Set the time to t

When T 4, M is on OA and N is on AP, at this time: PN=8-2T, AM=6-T

s△pmn=pn*am/2=(4-t)(6-t)=t^2-10t+24=

There is no integer solution to the equation.

When 4 t 6, point M is on OA and point N is on OP, at this point:

om=tpn=2t-8, on=10-pn=18-2t pmn, the height on the om edge = on*cosp=(18-2t)*4 5 s pmn=t*(18-2t)*4 5 2=after finishing: t 2-9t+18=0

Solution: x=6 (x=3 rounded).

There is an integer time, i.e., after 6 seconds, the PMN area is .

Are there integers? What is it?

Where's the picture??? There is no picture, how do I see it.

1.Set the original brine x grams, and evaporate y grams.

Solution x=500

2.Let the hundreds, tens, and units be x, y, and z respectively

x+z=y (1)

7x-(y+z)=2,7x-y-z=2 (2)x+y+z=14 (3)

8x=16x=22+z=y

2+y+z=14

2+2+z+z=14

z=5,y=7

So the number is 275

3.It takes X weeks to complete A alone, and it costs 10,000 yuan per week;

It takes Y weeks for B to complete alone, and it costs B million yuan per week.

From the inscription column:

1/x+1/y=1/6

4 x+9 y=1 (the specific steps to solve the equation are omitted) x=10 y=15

6a+6b=

4a+9b=

a= b=4/15

A alone requires 10*10,000.

B needs 15 * 4 15 = 40,000 to complete alone.

Company B. 4.(1)x+y=-7-a...1)

x-y=1+3a...2)

2x=2a-6

x=a-3<=0 a<=3

2y=-8-4a

y=-4-2a<0 a>-2

So -2-5

5-5-2(3) is obviously x=-3+a, y=-4-2a

So -3+a 0, -4-2a<0

So -2 so a-3 + a+2 3-a+a+2 5 and 2a+1<0

So a<-1 2

So a is -1

1.(a 2*b 3) 4=m 4 so m a 2*b 3

2.（a^n•b^3)^3•a^2•b^2n-7-〔a^n+1b〕^2•(ab^2〕^n

a^3n*b^9*a^2*b^2n-7-(a^2n+2a^n*b+b^2)*a^2n*b^2n

a^(3n+2)*b(2n+9)-7-a^4n*b^2n-2a^3n*b(2n+1)-a^2n*b^(2n+2)

There should be a symbol or data copied incorrectly in this question, otherwise you can only do this step.

That's 2002 zeros, which is 2003 digits.

3*7=21, the last digit is 1, no matter how many power-to-power last digits or 1, the last digit of several powers of 101 is also 1, so that there is 7 left after multiplication, so the last digit is 7

, both sides open to the power of 4 at the same time, and get m=a 2b 32I'm sorry I can't read the question.

There are 2002 zeros, which is 2003 digits.

To the nth power is the cycle;

7 to the nth power is the cycle;

The end of 11 to the nth power is 1.

The end of 3 to the power of 99 is 7, the end of 7 to the power of 100 is 1, the end of 11 to the power of 101 is 1,7 1 1=7, so it is 7.

I've worked hard to type out the drops! Always give some points!

It looks so complicated, and the above two should really be rewarded.

Correct the above 2 digits:

1。Knowing that m 4 = a 8 times b 12, find m: m a 2 * b 3