i need HELP solving a math problem?

Question

There are 6 chairs and stools. The chairs have 4 legs and the stool have 3 legs. All together there are 20 legs. How many chairs and how many stools are there?

thanks :)

Answer 1

x = number of chairs
y = number of stools

4x + 3y = 20
x + y = 6

Got me so far?

x + y = 6
y = 6 - x

4x + 3y = 20
4x + 3(6 - x) = 20
4x + 18 - 3x = 20
x + 18 = 20
x = 20 - 18
x = 2

plug in the answers....

x + y = 6
2 + y = 6
y = 4

4x + 3y = 20

Do the values match?

4*2 + 3*4 = 20
8 + 12 = 20

All you have to do is solve one equation for either x or y, then plug that solution into the other equation to solve for the other one.

Don't try to learn the 'answer', learn the process.
Give a man a fish, he will eat for a day - teach a man to fish and he can feed himself for life.

Wouldn't you rather 'learn to fish'? I would so no more homework help.

Have fun with Calculus. :)

Answer 2

In math talk: this is a problem with two equations and two variables. We will write the two equations and combine them to solve for one variable first, then the other.

The hardest part is writing the equations, so let's do that first. Since we want to find the number of chairs and stools we will choose our two variables (letters) as C for the number of chairs and S for the number of stools. Always pick letters that make it easy to remember what it is you're trying to find.

"There are six chairs and stools" means: C + S = 6 (equation 1)
This is our first equation.

"The chairs have 4 legs and the stool have 3 legs. All together there are 20 legs" means (chairs) x 4 = chair legs and (stools) x 3 = stool legs, with a total of 20 legs:
4 C + 3 S = 20 (equation 2)

Now we have our two equations. Two methods can be used: Substitution and Addition. Either way, we will combine them to make one of our variables go away for a bit.

Substitution: solve equation (1) for one of the variables. I choose S:

S = 6 - C

Now in equation (2) wherever I see and S I'm going to plug in 6 - C since they're equal:

4 C + 3 ( 6 - C) = 20 and now we have only C to worry about.

Distribute: 4 C + 18 - 3 C = 20 Simplify to get C = 2.

Now plug in 2 for C back into equation (1):

2 + S = 6 so S = 4

Check: 4 stools = 12 legs and 2 chairs = 8 legs. 6 seats. 20 legs.

Answer 3

Let chairs be y, and stools be x.
4 legs on a chair, 3 legs on a stool

Make two equations.
(1). y + x = 6
(2). 4y + 3x = 20

Use The Elimination Method.
y + x = 6 Multiplying by -3 on both sides of equation (1).
4y + 3x = 20

-3y - 3x = -18
y + 3x = 20
--------------------
y + 0x = 2 Adding

y = 2 Solving for y.

Then
2 + x = 6 Substituting 2 for y in equation (1).

x = 6 - 2 Subtract 2 from both sides.

x = 4

We obtain (4, 2), or x = 4, y = 2. This checks, so it is a solution.

(1). y + x = 6----- 2 + 4 = 6 True
(2). 4y + 3x = 20---4(2) + 3(4) = 20 True

So, There are 2 chairs and 4 stools.

Answer 4

2 chairs 4 stools

Answer 5

4 stools, 2 chairs
4 x 3 = 12 legs
2 x 4 = 8 legs
12 + 8 = 20 legs

Answer 6

4 stools
2 chairs

4x3=12
2x4=8
8+12=20
There ya Go!

Answer 7

Set up the problem like this.
c+s=6
4c+3s=20

So, say multiply the top problem by -3.

-3c+-3s=-18
4c+3s=20

Add these together. You'll find the s values cancel out.

1c=2

2/1=2 which is your c value.

4(2)+3s=20 is your new problem

You can reduce this to:

8+3s=20

20-8=12 so you have now:
3s=12
divide 12 by 3 now

12/3= 4 which is your s value

Now to check it.

2+4=6

4(2)+3(4)= ?
8+12=20

The basic idea behind this is to get both sides as close to zero as possible.

Hope this helps!

Answer 8

2x + 2y = 20 using the simultaneous method 2x - 2y = 4 ____________ 4x = 24 ( divide by four) x = 6 substitute x in one of the two equations 2x + 2y = 20 2 (6) + 2y = 20 12 +2y = 20 2y = 20 - 12 2y = 8 (divide by 2) y = 4 Answer : X = 6 , Y = 4

Answer 9

2 chairs and 4 stools

wow, I did mine without the long, drawn out equations!

Answer 10

nothing because you are getting nothing