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.