A clothing store is having a sale on shirts and jeans.Five shirts and 3 pairs jeans cost $220. Six shirts and 2 pairs of jeans cost $200. How much is the cost of one shirt? How much is the cost of pair of jeans?
5s + 3j = 220
6s + 2j = 200
so now just solve for s and j
To find the cost of one shirt and one pair of jeans, we can use a system of equations.
Let's assign some variables:
Let x be the cost of one shirt.
Let y be the cost of one pair of jeans.
From the given information, we can set up two equations:
Equation 1: 5x + 3y = 220 (Five shirts and three pairs of jeans cost $220)
Equation 2: 6x + 2y = 200 (Six shirts and two pairs of jeans cost $200)
Now we can solve this system of equations to find the values of x and y.
Multiplying Equation 1 by 2, we get:
10x + 6y = 440
Subtracting Equation 2 from Equation 3, we get:
(10x + 6y) - (6x + 2y) = 440 - 200
4x + 4y = 240
Dividing both sides of Equation 4 by 4, we get:
x + y = 60
Subtracting Equation 5 from Equation 6, we get:
(x + y) - (x + y) = 60 - 40
0x = 20
Since 0x = 20 is not a valid equation, it means there is no unique solution for this system of equations. This indicates that there might be an error or inconsistency in the given information.
To find the cost of one shirt and the cost of a pair of jeans, we need to set up a system of equations based on the given information.
Let's assume the cost of one shirt is S and the cost of one pair of jeans is J.
From the first piece of information, we know that 5 shirts and 3 pairs of jeans cost $220. This can be represented as the equation:
5S + 3J = 220 ... (Equation 1)
From the second piece of information, we know that 6 shirts and 2 pairs of jeans cost $200. This can be represented as the equation:
6S + 2J = 200 ... (Equation 2)
Now we have a system of equations with two variables (S and J). We can solve this system to find the values of S and J.
Multiplying Equation 2 by 3, we get:
18S + 6J = 600 ... (Equation 3)
Now we can subtract Equation 1 from Equation 3 to eliminate the J variable:
(18S + 6J) - (5S + 3J) = 600 - 220
18S + 6J - 5S - 3J = 380
This simplifies to:
13S + 3J = 380 ... (Equation 4)
Now we have a new equation with only the S variable remaining.
Next, we need to eliminate the J variable between Equation 1 and Equation 4.
Multiplying Equation 1 by 3, we get:
15S + 9J = 660 ... (Equation 5)
Now we can subtract Equation 4 from Equation 5:
(15S + 9J) - (13S + 3J) = 660 - 380
15S + 9J - 13S - 3J = 280
This simplifies to:
2S + 6J = 280 ... (Equation 6)
We now have another equation with only the S variable remaining.
Now we have a system of equations:
13S + 3J = 380 ... (Equation 4)
2S + 6J = 280 ... (Equation 6)
We can solve this system using any method we prefer, such as substitution or elimination.
Let's solve this system using the elimination method:
Multiply Equation 4 by 2 and Equation 6 by 13, we get:
26S + 6J = 760 ... (Equation 7)
26S + 78J = 3640 ... (Equation 8)
Subtract Equation 7 from Equation 8:
(26S + 78J) - (26S + 6J) = 3640 - 760
26S + 78J - 26S - 6J = 2880
This simplifies to:
72J = 2880
Divide both sides of the equation by 72:
J = 40
Now we know that the cost of one pair of jeans is $40.
Substitute J = 40 back into Equation 4:
13S + 3(40) = 380
13S + 120 = 380
Subtract 120 from both sides of the equation:
13S = 380 - 120
13S = 260
Divide both sides of the equation by 13:
S = 20
Therefore, the cost of one shirt is $20.
In conclusion, the cost of one shirt is $20, and the cost of one pair of jeans is $40.