A box of chalk and 2 staplers cost $10
Three boxes of chalk and 2 staplers cost $18. Find the total cost of 1 box of chalk and 1 stapler?
staples = s
chalk = c
c + 2s = 10
3c + 2s = 18
-c-2s = -10
3c + 2s = 18
2c = 8
c = 4
s = 3
c + s
4 + 3 = 7
The total cost $7.
Set this up as a system of equation then solve:
x = chalk
y = stapler
x+2y = 10 <--- equation 1
3x+2y = 18 <---- equation 2
Now, there are many methods when solving systems of equation, but I would use substitution method for this problem.
Look at the first equation:
x+2y=10
step 1) We should pick a variable to isolate, it doesn't matter which one. I'll choose x.
x=10-2y
step 2) substitute what we got for the first equation after isolating x, into the second equation:
3(10-2y)+2y = 18
now you can see that this equation only has 1 variable to solve. so solving for y:
30-6y+2y = 18
30-4y = 18
-4y = -12
y = 3
Great, now we know what y is.
next step is to substitute what we know for y into the first equation:
x+2y=10
x+2(3)=10
x+6=10
x=4
So x= 4 and y = 3
the total cost of 1 box of chalk is 4
the total cost of 1 staler is 3
Let's assume the cost of one box of chalk is 'C' dollars and the cost of one stapler is 'S' dollars.
According to the given information:
1 box of chalk + 2 staplers = $10 ...(1)
3 boxes of chalk + 2 staplers = $18 ...(2)
To find the total cost of 1 box of chalk and 1 stapler, we need to solve these two equations.
From equation (1), we can write:
C + 2S = 10 ...(3)
From equation (2), we can write:
3C + 2S = 18 ...(4)
Now, let's solve equations (3) and (4) to find the values of C and S.
Multiplying equation (3) by 3, we get:
3C + 6S = 30 ...(5)
Subtracting equation (4) from equation (5), we get:
(3C + 6S) - (3C + 2S) = 30 - 18
3C - 3C + 6S - 2S = 12
4S = 12
S = 12/4
S = 3
Substituting the value of S in equation (1), we get:
C + 2(3) = 10
C + 6 = 10
C = 10 - 6
C = 4
Therefore, the cost of 1 box of chalk is $4, and the cost of 1 stapler is $3.
The total cost of 1 box of chalk and 1 stapler is $4 + $3 = $7.
To find the total cost of one box of chalk and one stapler, we can set up a system of equations based on the given information.
Let's assume the cost of one box of chalk is C, and the cost of one stapler is S.
From the first statement, we know that one box of chalk and two staplers cost $10. So we can write the equation as:
C + 2S = 10 ---------------(Equation 1)
From the second statement, we know that three boxes of chalk and two staplers cost $18. So we can write the equation as:
3C + 2S = 18 ---------------(Equation 2)
Now, we need to solve this system of equations to find the values of C and S. There are several methods to solve a system of equations, such as substitution, elimination, or matrices. We will use the substitution method in this case.
From Equation 1, we can rewrite it as C = 10 - 2S.
Now, substitute this value of C in Equation 2:
3(10 - 2S) + 2S = 18
30 - 6S + 2S = 18
Combine like terms:
-4S = 18 - 30
-4S = -12
Divide both sides by -4:
S = 3
Now substitute the value of S back into Equation 1 to find C:
C + 2(3) = 10
C + 6 = 10
C = 10 - 6
C = 4
So, one box of chalk costs $4 and one stapler costs $3. The total cost of 1 box of chalk and 1 stapler is $7.