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.
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?
2 answers
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
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