Martin buys the pencil as well as the cover for 80 cents. At the same store, Gloria buys the cover as well as an eraser for $1.20, as well as Zachary buys the pencil as well as an eraser for 70 cents. How most would it price to buy 3 pencils, 3 notebooks, as well as 3 erasers?
can u additionally discuss it me how to compromise it?
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i believe 2.50
3 pens .45
3 notebooks 1.95
3 erasers 1.65
Together they bought 2pencils, 2notebooks, 2erasers for $2.70
So another of each would cost half as much again
$2.70 + $1.35 = $4.05
There are various ways to solve this problem. The first definite step is to write systems of equations:
p+n=.8
n+e=1.2
p+e=.7
You can find the individual costs of each item by using substitution or elimination. For example, you can move around the first equation to get p=.8-n
Then you can plug p into the third equation to get e. Then you can plug in e into the second equation to get n. Then plug each of those into this equation you’re trying to solve the major equation:
3p+3n+3e=x
It majorly depends on you for which way you wish to solve this equation though.
erasers cost 55 cents, notebooks cost 65 cents, and pencils cost 15 cents
total cost is
1.65+1.95+.45=4.05$
here you go:
p = pencil
n = notebook
e = eraser
❶p + n = 0.80
❷n + e = 1.20
❸p + e = 0.70
❶ p = 0.80 – n
❷ e = 1.20 – n
substitute:
❸ (0.80 – n) + (1.20 – n ) = 0.70
0.80 – n + 1.20 – n = 0.70
-2n = 0.70 – 0.80 – 1.20
-2n = -1.30
n = -1.30 / -2
n = 0.65
❶p + n = 0.80
p + (0.65) = 0.80
p = 0.80 – 0.65
p = 0.15
❷n + e = 1.20
(0.65) + e = 1.20
e = 1.20 – 0.65
e = 0.55
3p = (0.15)*3 = 0.45
3n = (0.65)*3 = 1.95
3e = (0.55)*3 = 1.65
———————————
…………………..4.05 <- sum
Good Luck
p = pencil, n = notebook, e = eraser.
What is given (numbers are in cents):
(1) p + n = 80
(2) n + e = 120
(3) p + e = 70
From (3) we know that p = 70 – e. Sub this into (1)
70 – e +n = 80
this becomes:
n = 10 + e
Sub this into (2):
10 +e + e = 120
2 * e = 110
Now we know that e = 110/2 = 55. Using this, (3) becomes:
p +55 = 70
p = 15
Now plug that into (1)
15 + n = 80
n = 65
So now we have n = 65, e = 55, and p = 15.
So to buy one of each it would cost 65 + 55 +15 = 135.
Then, to buy 3 of each would cost 3 x 135 = 405.
Remember, that is in cents, so the answer is $4.05.
p+n=0.80
n+e=1.20
p+e=0.70
p+n+1.20=n+e+0.80
p+1.20=e+0.80
p+0.40=e
p+(p+0.40)=0.70
2p=0.30
p=0.15
0.15+n=0.80
n=0.65
0.15+e=0.70
e=0.55
Check with n+e=1.20
0.55+0.65
Damn I had a phone call. I’m sure someone else has finished and you can plug the rest in anyhow,