A few math problems in the first year of junior high school are in a hurry! Give it to me at 22:50!

The first question can be set up as A to make x and B to make Y.

Get. x+y=15

25x+15y=275

The solution is x=5 y=10

In the second question, you can set the speed of A x km and the speed of B ykm

Get. The solution gives x=5 y=3

The third question can be set to be X years old A and B y years old.

Get. 2y-x=4

2x-y=61

The solution is x=42 y=23

Give it to me!!

1；Set X for type A and Y for type B, OK;

25x+15y=275

x+y=15

The solution is : x=5 y=10, i.e. a 5 b 10.

1. If the manufacturing of A parts x and B parts Y, then:

25x+15y=375 (4 hours and 35 minutes).

x+y=15

Solve the equation to obtain: x=5, y=10

1.Solving the two solutions to the equation a-b=4 a+b=2 gives a=3, b=-1

Substituting x=4, y=6 into 3*4+(-1)*6=6, then x=4 y=6 is not the solution of the equation ax+by=4.

2 equation丨2x-3y+4丨+(x+2y-5) =0 , from the nonnegativity of the absolute value and the square, we can know that 2x-3y+4=0, x+2y-5=0

The solution is x=1, y=2

The system of equations 3bx-ay=4 bx+ay=12 is 3b-2a=4, b+2a=12 and the solution is a=4, b=4

a²-2ab+b²=(a-b)(a-b)=0

1.The binary linear equations ax+by=4 of x=1 y=-1 and x=2 y=2 are used to obtain the binary linear equations about a and b, and the values of a and b are obtained by solving the equation (a=3, b=-1), and then x=4 y=6 is brought into as+by=4 to test (the answer is not).

2.This is a typical problem by saying that the sum of several non-negative numbers is 0, then they are zero, so 2x-3y+4=0

x+2y-5=0

Solve x,y().

Bringing x=1,y=2 into the system of equations 3bx-ay=4 bx+ay=12 gives the value of a,b (a=b=4).

So a -2ab + b = (a-b) 2 = 0

1, x=1, y=1 and x=2, y=2 are all solutions of the equation ax by=4, then there is: a b=4 and 2a 2b=4, the solution is a=3, b= 1, then this equation is 3x y=4, and when x=4 and y=6, this equation is not satisfied, that is, this is not the solution of the equation;

2、|2x－3y＋4|(x 2y 5) =0, we get: 2x 3y 4=0 and x 2y 5=0, the solution gives x=1, y=2, and this solution also satisfies the system of equations, then there is: 3b 2a=4 and b 2a=12, and the solution gives a=4, b=4, then a 2ab b b =(a b) =0.

1: From the question, we can see that a=1 b=-3

ax+by=4 is x-3b=4

So x=4 y=6 is not a solution to the equation ax+by=4, and 2 can be seen from the problem.

2x-3y+4=0 x+2y-5=0 gives x= 1 y=2

Substitute 3bx-ay=4 bx+ay=12

a=4 b=4

a²-2ab+b²=0

2a+2b=4

a=3,b=-1

3x-y=4

3*4-6=6 is not equal to 4

So x=4,y=6 are not solutions to the original equations.

a-b=4

2a+2b=4

a=3,b=-1

So it's not a solution to this equation.

2x-3y+4=0

x+2y-5=0

x=1y=2

Substituting yields 3b-2a=4

b+2a=12

The solution is a=b=4

a²-2ab+b²=(a-b)²=0

I'm sorry, I didn't understand it.

2x-3y+4=0

x+2y=5

2x-3y=-4

x=5-2y

2(5-2y)-3y=-4

10-4y-3y=-4

y=14x=5-2×14=-23

3bx-ay=4

bx+ay=12

3b×(-23)-14a=4

23b+14a=12

69b-14a=4

14a=4-69b

23b+4-69b=12

4-92b=12

b=a=10

a*-2ab+b*=(a-b)*

The front is what I did seriously, and the back is uncertain.

1：-2009x^2009 2010x^2010 nx^n

2：m=n=

3:2 Probably not.

4：2n+1 6n-3

5: Because the denominator numerator can be reduced to x at the same time, there is no need to use the value of x.

1)s/(a+3)+s/(a-3)=2as/a^2-92)(-2)^(n-1)a^n

3) Store A: y=300+

B shop: z=200+ let y=z, solution: x=600 so, x>600, choose A, x<600, choose B.

4）x^2y,5x^2y

5)4(y-x)=-4(x-y)=-36）a

When the former form is greater than the latter style, home A is discounted; When the latter style is greater than the previous style, B is preferential; When the two formulas are equal, A and B are the same.

Dumbfounded to read three to the young idfsfd

A:1. ∠bof=∠aoe=∠eod-90°=38°19′∠aof=180=141°41′

That is, 903*(3+2)=150°

3。(1) If 1= 2, con= aom=90°, then nod=180°- con=90°

2) 1=, then aoc=90°-30°=60°;

mod=90°+60°=150°

Do you want to go to QQ space to see pictures? Shrimp? Make no mistake? What kind of plane are you doing, depressed, fluttering, drifting.

Too lazy! Like the second question, you can see that it is 90 plus 60 for a total of 150. You're so unserious. The first question is simpler only to look at the picture. Take a look at it again.

1、∵∠eod=128°19′

ab⊥cd∠aod=90°

aoe= 128°19′—90°=28°19′∴∠bof=∠aoe=28°19′

That's it, you can see the rest for yourself, I don't have time. Excuse me, landlord.