Spiders have 8 legs, dragonflies have 6 legs and 2 pairs of wings, and cicadas have 6

There are x cicadas.

According to the number of wings, the number of dragonflies is (14-x) 2.

Then the spider has 16-(14-x) 2-x individuals.

Spider*8 + Dragonfly*6 + Cicada*6=110

Solution. x=4

So 4 cicadas.

5 dragonflies. 7 spiders.

If there are x cicadas, there will be dragonflies (14-x) 2 and only spiders (110-6x-6(14-x) 2) 8.

Then add them up to 16 and you're done.

This question you can have dragonflies x only. This way we can conclude that the cicadas have (14-x) only. At this point, we can't count their legs.

Dragonflies are 6*x legs. The cicada has 6*(14-x) legs, here we assume that x is a known number, and we can calculate the number of spiders (13+3x 4) from the sum of the legs of the three insectsThe sum of the three insects is 16

That is, x+(14-2x)+(13+3x)=16 The unary equation brother will always calculate x=5 14-2x=4 13+3x 4=7 That is, there are 7 spiders, 5 dragonflies, and 4 cicadas.

16*8=128 128-110=18 18 (8-6)=9 16-9=7 spiders.

9*2=18 18-14=4 4 (2-1)=4 cicadas.

Let the spider x be 2 (it is impossible to know 1 according to the number of wings) and 6 dragonflies according to the number of wings, then the number of legs is 8x+6*2+6*1=110 (no solution).

If the cicada is 4, and the number of dragonflies is 5 according to the number of wings, then the number of legs is.

8x+6*5+6*4=110, the solution is x is 7, spider is 7, dragonfly is 5, and cicada is 4

If the cicada is 6, and the dragonfly can only have 4 according to the number of wings, then the number of legs is.

8x+6*4+6*6=110, (no solution).

If the cicada is 8, and the dragonfly can only have 3 according to the number of wings, then the number of legs is.

8x+6*3+6*8=110, (no solution).

If the cicada is 10, and only the dragonfly has 2 according to the number of wings, then the number of legs is.

8x+6*2+6*10=110, (no solution).

Zongshang can get 7 for spiders, 5 for dragonflies, and 4 for cicadas

9-4=5 dragonflies.

2.If there are x cicadas, there will be dragonflies (14-x) 2 and only spiders (110-6x-6(14-x) 2) 8.

Then add them up to 16 and you're done.

Let the spider x be 2 (it is impossible to know 1 according to the number of wings) and 6 dragonflies according to the number of wings, then the number of legs is 8x+6*2+6*1=110 (no solution).

If the cicada is 4, and the number of dragonflies is 5 according to the number of wings, then the number of legs is.

8x+6*5+6*4=110, the solution is x is 7, spider is 7, dragonfly is 5, cicada is 4 If the cicada is 6, according to the number of wings, only dragonflies have 4 know, then the number of legs is.

8x+6*4+6*6=110, (no solution).

If the cicada is 8, and the dragonfly can only have 3 according to the number of wings, then the number of legs is.

8x+6*3+6*8=110, (no solution).

If the cicada is 10, and only the dragonfly has 2 according to the number of wings, then the number of legs is.

8x+6*2+6*10=110, (no solution).

Zongshang can get 7 for spiders, 5 for dragonflies, and 4 for cicadas

Example 6] Spiders have 8 legs, dragonflies have 6 legs and 2 pairs of wings, cicadas have 6 legs and 1 pair of wings, and now there are 18 of these three kinds of insects, with 118 legs and 18 pairs of wings, how many spiders, dragonflies, and cicadas are there? ( a
b
c
d

This question you can have dragonflies x only. This way we can conclude that the cicadas have (14-x) only. At this point, we can't count their legs.

Dragonflies are 6*x legs. The cicada has 6*(14-x) legs, here we assume that x is a known number, and we can calculate the number of spiders (13+3x 4) from the sum of the legs of the three insectsThe sum of the three insects is 16

That is, x+(14-2x)+(13+3x)=16 The unary equation brother will always calculate x=5 14-2x=4 13+3x 4=7 That is, there are 7 spiders, 5 dragonflies, and 4 cicadas. Brother give me more points, hehe.

Set a spider, a dragonfly b, and a cicada.

8a+6b+2c=110……（1）

2b+c=14……（2）

8a+2b=82,4a+b=41

b=41-4a

a, b, and c are all positive integers.

So a = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10b=37,33,29,25,21,17,13,9,5,1c=14-2b

So b = 5,1

That is, the three insects have 9, 5, 4 or 10, 1, 12 (only) and then they can be counted to have several legs.

1.16*8=128 128-110=18 18 (8-6)=9 16-9=7 spiders.

9*2=18 18-14=4 4 (2-1)=4 cicadas.

9-4=5 dragonflies.

2.If there are x cicadas, there will be dragonflies (14-x) 2 and only spiders (110-6x-6(14-x) 2) 8.

Then add them up to 16 and you're done.

Set x spiders, dragonflies, and cicadas.

8x+6y+6z=128

2y+z=22

x+y+z=20

The solution is x=4y=6 z=10

"Chickens and rabbits in the same cage"Issue.

Hypothesis: all spiders, 16 heads, 16x8 = 128 feet, 18 more feet, 18 divided by (8-6) = 9 (only) 16-9 = 7 spiders, 9 cicadas and dragonflies. Cicadas must be odd numbered, can be 5 cicadas and 4 dragonflies.

You do the math, it's absolutely OK. It's too much trouble with 3 unknowns.

Think of dragonflies and cicadas as one kind, because they are both 6 legs, assuming that they are all spiders: with legs: 8 20 = more than 160:

160-132 = 28 dragonflies and cicadas have a total of: 28 (8-6) = 14 spiders: 20-14 = 6 and then analyze from the wings:

Suppose all dragonflies have wings: 14 2 = 28 pairs more: 28-18 = 10 pairs then the cicada has:

Sunny has 6 legs, 2 pairs of wings; Spiders have 8 legs and no wings; The cicada has 6 legs and 1 pair of wings. There are now some dragonflies, spiders and cicadas, and their total number is known to be 18 with a total of 118 legs and 20 pairs of wings. How many of each insect are there?

The format of this question is much like the chicken-rabbit cage problem, but it is more complicated, with three animals mixed together. You can try to separate the easiest of them first.

Dragonflies and cicadas both have 6 legs, and only spiders have 8 legs. So the first step can consider 6-legged insects and 8-legged insects, so that there are only two types left, first find the number of 8-legged insects, you can know how many spiders there are.

Assuming that the 18 insects are all 6-legged dragonflies and cicadas, then the total number of legs will be.

6 18 = 108 (strips).

There are actually 118 legs, a difference.

118-108=10 (strips).

Take an 8-legged spider in and change it for a 6-legged dragonfly or cicada out, and for each spider you change, you add 2 legs, so the spiders that change in are shared.

10 2 = 5 (only).

So it was found that there were 5 spiders.

Now you can proceed to the second step and find the number of the other two insects. Subtract the number of spiders from the total number of insects to get the dragonfly and cicada in common.

18-5=13 (only).

Assuming that all 13 of them are cicadas, then they have a total of 13 pairs of wings. There are actually 20 pairs, which is not bad.

20-13=7 (right).

Take one dragonfly in and change it to a cicada and add a pair of wings, so you have to change in 7 dragonflies and leave 6 cicadas.

In the end, there were 7 dragonflies, 5 spiders, and 6 cicadas.

There are x spiders, y dragonflies, and z cicadas.

8x+6y+6z=110

2y+z=14

x+y+z=16

The solution is x=7, y=5, z=4

The math master is already on it, give him points.

If you don't learn to repeat.

Elementary school students who have solved equations can solve them in this way.

Let's assume that all bugs have only 6 legs, and the number of legs is.

The extra feet are spider's.

There are 5 spiders.

In the same way, it is assumed that dragonflies and cicadas only have one pair of wings.

The extra wings are dragonfly's.

The number of dragonflies is 7

Cicada 18-5-7 6

A: 5 spiders, 7 dragonflies, and 6 cicadas.

There are x spiders, y dragonflies, and z cicadas.

8x+6y+6z=118

2y+z=20

x+y+z=18

lz can solve the equation by himself

Set x spiders, dragonflies, and cicadas.

8x+6y+6z=118

2y+z=20

x+y+z=18

x=5y=7z=6

Think of dragonflies and cicadas as one kind, because they are both 6 retreats.

Let's say it's all spiders:

With legs: 8 30 = 240 strips.

Many: 240-194 = 46.

Dragonflies and cicadas total: 46 (8-6) = 23.

Spiders: 30-23 = 17 pcs.

Again from the wing analysis:

Let's say it's all dragonflies.

Winged: 23 2 = 46 pairs.

Many: 46-37 = 9 pairs.

Then the cicadas have: 9 (2-1)=9.

There are 23-9=14 dragonflies.

A: 17 spiders, 14 dragonflies, and 9 cicadas.

If you are a junior high school student, you can use equations to solve them

Spiders, dragonflies, and cicadas have x, y, and z respectively, and they have the following according to the title

x+y+z=30

8x+6y+6z=194

2y+z=37

Just solve this system of equations.

Leg cutting method: After cutting off the 6 legs of each spider, dragonfly and cicada, there are only 2 legs left of the spider, then the spider has:

194-30x6) 2=7 (only).

In this way, dragonflies and cicadas have a total of 30-7=23 (only), after cutting off 1 pair of wings of each dragonfly and cicada, there will be 1 pair of wings of dragonflies left, then dragonflies have:

37-23x1=14 (only).

The number of cicadas is: 30-7-14 = 9 (only).

Assuming that 18 are all spiders, it should be a total of 18*8=144 legs, but now it is 118 legs, and the less 26 legs are because.

Dragonflies and cicadas both have 6 legs, which is 2 fewer than spiders, so using 26 2=13 is the total number of dragonflies and cicadas.

So the spider is 18-13 = 5 (only), which is the number of spiders, 13 dragonflies and cicadas.

Assuming all 13 of them are dragonflies, there should be 13*2=26 pairs of wings but now there are 20

To the wings is because.

Cicadas have 1 pair of wings less than dragonflies, so.

The number of cicadas is.

Dragonfly population. 18-5-6 = 7.

I hope the landlord adopts thank you! If you don't understand, please ask. . .

There are x cicadas.

According to the number of wings, the number of dragonflies is (14-x) and the number of dragonflies is 16-(14-x) 2-x.

Spider*8 + Dragonfly*6 + Cicada*6=110

Solution. x=4, so there are 4 cicadas.

5 dragonflies. 7 spiders.

From the fact that spiders only have legs, and the number of legs of dragonflies and cicadas is 6, the number of spiders can be found: (112-18 6) (8-6) = 2 Now the problem becomes: dragonflies have 2 pairs of wings, cicadas have 1 pair of wings, and there are 18-2 = 16 and 22 pairs of wings for these 2 kinds of insects. How many dragonflies are there?

Number of dragonflies: (22-16 1) (2-1) = 6 cicadas number: (16 2-22) (2-1) = 10 Of course, 16-6 = 10 can be counted.

For the specific solution, you can refer to the chickens and rabbits in the same cage.

There are x spiders, y dragonflies, and z cicadas.

8x+6y+6z=112 ①

x+y+z=18 ②

2y+z=22 ③

Derived from the equation.

z=22-2y ④

Bring in. x+y+22-2y=18

x-y=-4

y=x+4 ⑤

Bring in. 8x+6y+132-12y=112

8x-6y= -20

4x-3y=-10 ⑥

Bring in. 4x-3(x+4)=-10

x-12=-10

x = 2 substitutions.

y=2+4y =6

Substitution. z=22-2*6

z=10 In summary, there were 2 spiders, 6 dragonflies, and 10 cicadas.

If you need to use matrix, C, assembly language, matlab, etc. to solve the problem, please @me.