Solve a few math problems in the first year of junior high school, solve a math problem in the first

1) d2) a = 1 or 0

3)x+(1+k)y=0 ……

1-k)x+ky=1+k ……

1+k)x+(12-k)y=-(1+k) …

Obtained by +.

2x+12y=0

x+6y=0 ……

Lianli is available. 1+k)y=6y

So, y=0 or k=-1

Substituting y=0 into the system of equations yields x=0 and k=-1

then k=-1 has the same result.

So there is only one set of solutions, x=y=0, k=-1.

4)x+y=a ①

5x+3y=31 ②

2x=31-3a

x=(31-3a)/2

x>031-3a)/2>0

31-3a>0

3a<31

a<31/3 ③

2y=5a-31

y=(5a-31)/2

y>05a-31)/2>0

5a-31>0

5a>31

a>31/5 ④

Simultaneous 31 5 satisfies the integer a of .

x is a positive integer and x=(31-3a) 2;y is a positive integer and y=(5a-31) 2

A=7, x=(31-3*7) 2=(31-21) 2=10 2=5, x is a positive integer.

y=(5*7-31) 2=(35-31) 2=4 2=2, y is a positive integer.

a=7 meets the requirements of the question.

a=8, x=(31-3*8) 2=(31-24) 2=7 2, x is not a positive integer, discard a.

A=9, x=(31-3*9) 2=(31-27) 2=4 2, x is a positive integer.

y=(5*9-31) 2=(45-31) 2=14 2=7, y positive integer.

a=9 satisfies the requirements of the question.

When a=10, x=(31-3*10) 2=(31-30) 2=1 2, x is not a positive integer, discard a.

In summary, when a = 7 or a = 9, the solution of the system of equations {x+y=a 5x+3y=31 is a positive integer.

5) Solveable x=2

Substituting y=(3-a) 2

Because y is a positive integer and a is a positive integer.

So a=1

1.6x+3y=4 can be reduced to 2x+y=4 3 so choose d2 ∵x+ay=1 ， 3x-y=1

The solution gives x= 1+a, 3a+1, y=2, 3a+1, because the solution is an integer, and from y=2 3a+1 gives 3a+1= 1 or 2, and solves a, 3, -1

a is an integer and a takes 0 or -1

Take a or -1 and substitute it into x= 1+a 3a+1, x=1 or 0 is an integer.

So a=0 or -1

3. x+（1+k）y=0 ①

1-k）x+ky=1+k②

I'll make it up later. ，5x+3y=31

The solution gives x=31-3a 2 y=5a-31 2 and the solution is positive 31-3a>0 5a-31>0

31 50 and 3-a is a multiple of 2.

The solution a<3 is known from the question a>0

In summary, a=1

1 should be d because both equations are the same

1) There is only one solution to a system of binary linear equations

Solution: If point A is X kilometers away from Beishan Station, then point A is 18-X kilometers away from the departure point, and according to the equal time spent by the two groups, it can be obtained that the two teams have the same distance on foot and by car. Team A travels by car (18-x) km, walking trip x km Team B:

If team B reaches point B when team A starts walking, and team B reaches point C when they meet the car, then the equation pc=x is obtained. x=4 [(18-x)+(18-x-x)] 60 solution. x=2

Answer. Point A is 2 km away from Kitayama Station.

1) Root number.

2)x=143500

4) Let h=t=196 49

t=25) solution: Let the width be x meters.

x²=24000

x=155 and set the radius to y meters.

y = y = i.e. the radius is meters.

Question 1, question 2, 143500

Question 3, Question 4, Question 2s

Question 5 160000/ ；Root number 2400=

Open the root number again.

One, two, 143500

Three, four, 2 seconds.

5. (1) m (2) m.

1.Henry accompanied his girlfriend Nancy to shop in a store, Nancy picked out four items, one of which was only one yuan, Henry calculated it by heart, a total of yuan, so he hurriedly took out the money. Suddenly, he found that when the shopkeeper used the computer to calculate the total price, he pressed the multiplication key, and he was about to negotiate, strangely, the shopkeeper calculated the total price was also yuan, do you know what the unit price of these four trinkets is?

The unit price of the four trinkets is 2 1

dissolution; One item is X;

3x+1=3x=x= 23/12.

a^2xy-2abxy+b^2xy

a^2-2ab+b^2）xy

a-b)^2 xy

For such questions, pay attention to extracting common factors.

xy can be eliminated by substituting the value of xy

Then substituting a=6-b into the equation will give you a polynomial sum form about b, and then you can complete the question.

Solution: Let's go up to x kilometers.

The solution is x=6 and travels up to 6 km.

Let the function relation: y=kx+b

y=6+ (x>=2)

x=10 km.

12-6=6 yuan traveled 2 kilometers.

6 4*1=4 Traveled 4 km.

It traveled up to 6 km.

（12-6）/

Then he took a taxi for a maximum of 6 kilometers.

1.Let it be: its volume is v, when the gold content is 90%, there is a volume a*, the gold content is 100%, and there is a volume a, then the volume of this jewelry is between a* and a.

2.Solution: Set up from A to B uphill x meters, flat road y meters, downhill z meters. then, x+y+z= (1).

20x+15y+12z=51 (2)

12x+15y+20z= (3)

The solution yields y=z=x

1.2.Set the uphill distance x km, the downhill y km, and the flat road z km (A to B).

Get x+y+z=

x÷3+y÷5+z÷4=

x÷5+y÷3+z÷4=

The solution is x=y=z=

1.The larger the proportion, the smaller the volume, so the content is not less than 90% If it is only understood as the proportion of the mass to the whole, then it can be inferred that the maximum volume is when the content is 90%, and the minimum volume is when the content is 100%.

So the volume range can be expressed as follows.

a Volume v<90%*a

2.Solution: Set up from A to B uphill x meters, flat road y meters, downhill z meters. then, x+y+z= (1).

20x+15y+12z=51 (2)

12x+15y+20z= (3)

The solution yields y=z=x

Will do code words code code does not come out on the second floor is very strong.