Ask for help for a math problem in the first year of junior high school, ask for help in the math pr

If you've ever studied binary equations, take a look at the following:

Solution: Let one addition be x and the other add y.

According to the set, the system of equations can be listed:

10x+y=242……（1）

x+10y=341……（2）

x = 21 brings x into (1):

y=32 In summary, the two additions are 21 and 32, respectively.

The sum is 21 + 32 = 53

341-242] [10-1] = another number *

242-*】 10 = an additive.

The two additions are a, b

10a+b=242

a+10b=341

Equation summing: 11(a+b)=583

a+b=53

In the first year of junior high school, do you have to learn x,y?

Let one number be x and the other be y, then:

10x+y=242……Type 1.

x+10y=341……Type 2.

Formula 1 + 2: 10x+y+x+10y=242+341 => 11(x+y)=583 => x+y=53

The sum of the two additions is 53

Let the first addition be x and the other add y

10x+y=242①

10y+x=342②

100y 10x 10 x y 3410 242 solution: y 32

Put substitution in, got.

10x＋32＝242

Solution: x 21

Author: 2010-7-12 12:23

Correct solution, depressed ......I forgot to press login, forget it, it became anonymous.

Solution: I know a 2 ten a = one 1, find a 2005 ten a 3006 ten a 4007 from the known a 2 ten a ten 1 20 know a ≠ 0 equation both sides of the equation multiply a 1 to get (a 1 1) (a 2 ten a + 1) 20 a 3 a 1 = 0 a 3 21 a 2005 two (a 3) 668 x a 2 (a 3) 1002 two 1 and a 4007 two (a 3) 1335 a 2 a 2 a 2 a 2005 a 3006 a 4007 a ten 1 ten a 2 20

It can be seen from the question.

11 sets of matches with a maximum of 11 points.

The total score of each set is 1, that is, the total score of 11 points for two people in 11 sets = points of Russia + 3 times more points of Russian players = 1 point more than 4 times of Russian players = 11 points.

Then the Russian player's points = (11-1) 4 = points.

Then Xie Jun's points = points.

1.Write it out with scientific notation.

2.a=-1 b=1 a+b=0 b a=-1, then the 3rd power of 2b is 8

3b squared 9

The final result.

1.Add all the reps and divide them by the reps.

49 + 50 + 64 + 58 + 53 + 56 + 55 + 57) 8 = vehicles rounded up to 55 vehicles 3 minutes.

2.60 minutes per hour 3 minutes = 20

So 1 hour average through the vehicle vehicle.

Let's talk about the idea, when encountering this kind of problem, first read the question, don't be afraid of the long question, look at what the question type is, and refine the main data, for example, the first question of this question is about finding the "average", then I think of what problems to pay attention to in finding the average, clarify the quantitative relationship, what is the total amount, how many times, how to calculate, etc., the second question is based on the first question, in fact, the first question is also a guide question, which should be used in the second question.

1) Every three minutes on average (49 + 50 + 64 + 58 + 53 + 56 + 55 + 57) 8=

2) In the afternoon, the entrance passed through an average of cars per hour, of which 20 means an hour and 20 3 minutes, and so much is said to be almost.

∵|a-1/2|+|b-3|=0

Again|a-1/2|Greater than or equal to 0

And|b-3|Greater than or equal to 0

a-1/2|=0

And|b-3|=0

a-1/2=0

b-3=0∴a=1/2,b=3

a+b=

|a-1/2|+|b-3|=0

According to the meaning of the absolute value, then |a-1/2|=0，|b-3|=0, so a=1 2, b=3

a+b=7/2

Simple, because the absolute value of a-1 2 + the absolute value of b-3 = 0 so the absolute value of a-1 2 is 0, and the absolute value of b-3 is equal to 0, so a=1 2, b=3then a+b=3+1 2=3 7

First of all. The absolute value is greater than or equal to 0 drops.

The absolute value of a-1 2 is greater than or equal to 0

The absolute value of b-3 is greater than or equal to 0

a-1/2=0

b-3=0∴a=1/2 b=3

then a+b=1 2+3=7 2