Four pens cost $2 less than two markers. Three pens and nine markers cost $54. Find the cost of each.

4 p = 2 m -2 so m = 2 p + 1

3 p + 9 m = 54
3 p + 9 (2 p + 1) = 54 etc

4p - 2m = 2 ---> 2p - m = 1 or m = 2p-1

3p + 9m = 54

sub in
m=2p-1 into the other equation and solve

To solve this problem, let's assign variables to the unknowns. Let's call the cost of a pen "p" and the cost of a marker "m."

According to the problem, four pens cost $2 less than two markers. This can be expressed as:

4p = 2m - 2 ... (Equation 1)

Also, three pens and nine markers together cost $54. This can be expressed as:

3p + 9m = 54 ... (Equation 2)

Now we have a system of two equations with two variables. We can solve this system using either substitution or elimination method.

Let's solve it by the substitution method:

From Equation 1, we can rewrite it as:

2m = 4p + 2 ... (Equation 1a)

Now we can substitute this expression for "2m" into Equation 2:

3p + 9(4p + 2) = 54

Let's simplify this equation:

3p + 36p + 18 = 54

Combine like terms:

39p + 18 = 54

Subtract 18 from both sides:

39p = 36

Divide both sides by 39:

p = 36/39

Now let's simplify this fraction:

p = 4/13

So the cost of a pen is $4/13.

To find the cost of a marker, substitute the value of "p" into either Equation 1 or Equation 2. Let's substitute it into Equation 1:

4(4/13) = 2m - 2

16/13 = 2m - 2

Add 2 to both sides:

16/13 + 2 = 2m

Multiplying both sides by 13/2 to isolate "m":

m = (16/13 + 2) * 13/2

m = (16 + 26) / 13

m = 42/13

So the cost of a marker is $42/13.

To summarize:

The cost of each pen is $4/13, and the cost of each marker is $42/13.