There is a math problem to help me solve the next math problem, please help me solve it

The image of the primary function y1=3x-2k intersects the image of the inverse proportional function y2=k-3 x (k-3 of x), where the ordinate of one of the intersection points is 6

Suppose this intersection is (a,6).

So. 6=3a-2k (1)

6=(k-3)/a (2)

6a=k-3

k=6a+3

Substitute in (1).

6=3a-2(6a+3)

9a=12a=-4/3

k=-5 The two functions are parsed as.

y1=3x+10

y2=-8/x

2) Find the coordinates of the intersection of the two images and combine the images to find y10, at x>0, y2<0 and y1 and y21 intersect at (-4, 3, -6).

x<0 time.

3x+10<-8/x

3x^2+10x+8>0

3(x+5/3)^2-1/3>0

So x<-2 or x>-4 3

x>0.

3x+10<-8/x

3x^2+10x+8<0

3(x+5/3)^2-1/3<0

2x>-4/3

6=3x-2k (1)

6=(k-3) x,6x=k-3,3x=(k-3) 2 is substituted into (1).

k/2-3/2-2k=6

k-3-4k=12

k=-5, 3x=-4,x=-4/3

y1=3x+10,y2=-8/x

y=-4/3*3+10=6

The coordinates of the intersection are: (-4 3,6).

y2-y1>0

8 x-3x+10>0,3x-10+8 x<0When x>0,3x2-10x+8<0,3(x 2-10x 3+100 36)-100 12+8<0

3(x-10/6)^2<1/3

x-5/3)^2<1/9

x-5/3)^2>1/9

x-5 3>1 3 or, x-5 3<-1 3x>2 (unreasonable because x<0), x<4 3

So take x<0

So the value range of x is: 4 3< x<2 and x<0< p>

Bring y=6 in.

6=3x-2k

6=（k-3）/x

k=-5y1=3x+10

y2=-8/x

2) the coordinates of the intersection of the two images (-4, 3, 6), (2, 4).

6=3x-2k

6=(k-3) x system of force equations.

k = -5 coordinates of two points (-4 3,6) (-2,4) 3x=10<-8 x.

infinity, -2) and (-4 3,0).

40-x)(20+2x) =1200

Divide both sides by 2 at the same time

40-x)(10+x) =600

400 + 40x - 10x - x = 600 move items. x² -30x + 200 = 0

Cross multiplication.

x -10x -20

So x1 = 10 ; x2 = 20

1:1/8

2: Set up x hectares of sown wheat.

2x+4/5=54

3: Set up this aqueduct with a total length of x meters.

3/8x=39+3

4: Let B make x parts per hour.

3/2(60+x)=150

5: Set rape to weigh x kg.

The barrel weighs y kilograms.

1/3x+y=8

2/3x+y=20

(Answer with elementary school knowledge) (Note: 8 out of 13 writing.)

2) (54-4 5) 2=ha).

3) (39+3) 3 8 = 112 (m).

4) 150 3 2-60 = 40 (pcs).

5) (20-8) (1-1 3) = 18 (kg).

Set up treasury bonds of x yuan and deposits of y yuan.

then x+y=30000

y=30000-x

So the interest due is 3*

So y=30000-x

So x=18000

y=12000

Answer: The treasury bond is 18,000 yuan, and the deposit is 12,000 yuan.

If you buy x yuan treasury bonds, you will deposit (30,000-x) yuan in the bank.

x+x* solve it yourself, it's a little annoying.

It's so hard to talk about the topic.

Powerless t t

Summary: 1 This question mainly examines the practice of auxiliary lines when angular bisector lines appear. Looking at the graph, we can see that we can easily make auxiliary lines:

Considering that the point F is on the angular bisector of DBC and BCE, i.e., F is used as FK AD, FI AE, FJ BC, so as to obtain the new condition FK=Fi. Using these new conditions, we can transform the conclusion that the required point f is a point on the angular bisector of DAE, thus linking the known with the to be proven.

2 There are generally two ways to do auxiliary lines with angular bisectors. Perpendicular lines from one point on the bisector of the angle to both sides; Use angular bisector lines to construct symmetrical shapes. Under normal circumstances, when there are conditions such as right angles or verticals, it is generally considered to be a perpendicular line; In other cases, consider constructing symmetrical figures.

As for which method to choose, we should combine the problem graph and the known conditions, in this problem we make perpendicular lines from the point F on the bisector of the angles of DBC and BCE to both sides of DBC and BCE, please think about which method is used.

1. I won't say that the real estate market is correct.

From the problem, the opening of this quadratic function must be upward. (Imagine an image, if it goes down, it won't be 0 forever).

So there is m+1>0 and the number of intersections with the x-axis is less than or equal to 1

0 （-2（m-1))^2 - 4·(m+1)·3(m-1)≥0