the cost of 3 makers and 2 pencils is $1.80. The cost of 4 markers and 6 pencils is $2.90. What is the cost of each item?

The cost of 3 markers and 2 pencils is $1.80, which is $.36 cent each, and The cost of 4 markers and 6 pencils is $2.90 which is $.29 cent each.

To find the cost of each item, we can set up a system of equations based on the given information.

Let's assume the cost of each marker is "m" dollars and the cost of each pencil is "p" dollars.

From the first statement, we know that:
3m + 2p = 1.80 ----(Equation 1)

From the second statement, we know that:
4m + 6p = 2.90 ----(Equation 2)

We can solve this system of equations using the method of substitution or elimination.

Let's solve the system of equations using the method of substitution:

From Equation 1, we can rewrite it as:
3m = 1.80 - 2p ----(Equation 1.1)

Substituting Equation 1.1 into Equation 2, we get:
4(1.80 - 2p) + 6p = 2.90

Expanding the equation:
7.20 - 8p + 6p = 2.90

Combining like terms:
-2p + 7.20 = 2.90

Moving the constant term to the other side:
-2p = 2.90 - 7.20
-2p = -4.30

Dividing both sides by -2:
p = -4.30 / -2
p = 2.15

Now that we know the cost of each pencil (p = $2.15), we can substitute this value into Equation 1.1 to find the cost of each marker:

3m = 1.80 - 2(2.15)
3m = 1.80 - 4.30
3m = -2.50

Dividing both sides by 3:
m = -2.50 / 3
m ≈ 0.83

So, the cost of each marker is approximately $0.83, and the cost of each pencil is $2.15.

To solve this problem, we can set up a system of equations. Let's first assign variables to the costs of the markers and pencils.

Let's assume the cost of each marker is 'M' and the cost of each pencil is 'P'.

According to the given information:
1) The cost of 3 markers and 2 pencils is $1.80.
This can be expressed as: 3M + 2P = 1.80.

2) The cost of 4 markers and 6 pencils is $2.90.
This can be expressed as: 4M + 6P = 2.90.

Now, we have a system of two equations with two unknowns, M and P. We can solve these equations to find the values of M and P.

To eliminate one of the variables, let's multiply the first equation by 2 and the second equation by 3. This will result in the coefficients of the 'P' terms being equal and opposite, allowing us to eliminate 'P'.

Multiplying the first equation by 2: 2(3M + 2P) = 2(1.80)
Simplifying: 6M + 4P = 3.60

Multiplying the second equation by 3: 3(4M + 6P) = 3(2.90)
Simplifying: 12M + 18P = 8.70

Now, let's subtract the equation obtained from multiplying the first equation by 2 from the equation obtained from multiplying the second equation by 3, as this will eliminate 'P':

(12M + 18P) - (6M + 4P) = 8.70 - 3.60
Simplifying: 6M + 14P = 5.10

Now we have the equation 6M + 14P = 5.10.

Next, we can solve this equation simultaneously with the other equation: 3M + 2P = 1.80.

We can use various methods like substitution or elimination to solve this system of equations. Let's proceed with the substitution method.

From the equation 3M + 2P = 1.80, we can express M in terms of P:
3M = 1.80 - 2P
M = (1.80 - 2P)/3

Substituting this value of M into the equation 6M + 14P = 5.10:

6((1.80 - 2P)/3) + 14P = 5.10

Simplifying and solving for P:

(10.80 - 12P)/3 + 14P = 5.10
10.80 - 12P + 42P = 15.30
30P = 4.50
P = 4.50/30
P = 0.15

Now that we have found the value of P, we can substitute it back into either of the original equations to find the value of M. Let's use the first equation:

3M + 2(0.15) = 1.80
3M + 0.30 = 1.80
3M = 1.80 - 0.30
3M = 1.50
M = 1.50/3
M = 0.50

Therefore, the cost of each marker (M) is $0.50, and the cost of each pencil (P) is $0.15.

3m+2p=1.8 (qeuation A)

4m+6p=2.9 (equation B)
"equation A" times "3":
9m+6p=5.4 (equation A')
"equation A'" - "equation B":
5m=2.5
[m=0.5]
substitude "m=0.5" into "equation A":
3(0.5)+2p=1.8
1.5+2p=1.8
2p=0.3
[p=0.15]