SOLUTION: Alice stopped by a coffee shop three days in a row at a conference to buy drinks and pastries. On the first day, she bought a cup of coffee, a muffin and a scone for which she pa

Question 1181999: Alice stopped by a coffee shop three days in a row at a conference to buy drinks and
pastries. On the first day, she bought a cup of coffee, a muffin and a scone for which
she paid |6.15. The next day she bought two cups of coffee, three muffins and a
scone (for herself and friends). Her bill was |12.20. The last day she bought a cup
of coffee, two muffins and two scones, and paid |10.35. Determine the price of a cup
of coffee, the price of a muffin and the price of a scone. Clearly explain your set-up
for the problem

Found 5 solutions by mananth, josgarithmetic, ikleyn, greenestamps, MathTherapy:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
Let price of coffee be $x, muffin be $y, and scone be $z
On the first day, she bought a cup of coffee, a muffin and a scone for which
she paid |6.15.
x +y+z = 6.15
The next day she bought two cups of coffee, three muffins and a
scone. Her bill was |12.20.
2x+3y+z = 12.20
The last day she bought a cup of coffee, two muffins and two scones, and paid |10.35.
x+2y + 2z= 10.35
x + y + z = 6.15 --------------1
2 x + 3 y z = 12.2 --------------2

x + 2 y + 2 z 10.35 --------------3

consider equation 1 &2 Eliminate y
Multiply 1 by -3 1
Multiply 2 by 1 1
we get
-3 x + -3 y + -3 z = -18.45
2 x + 3 y + 1 z = 12.2
Add the two
-1 x + 0 y + -2 z = -6.25 -------------4 4
consider equation 2 & 3 Eliminate y
Multiply 2 by -2
Multiply 3 by 3
we get
-4 x + -6 y + -2 z = -24.4
3 x + 6 y + 6 z = 31.05
Add the two
-1 x + 0 y + 4 z = 6.65 -------------5
Consider (4) & (5) Eliminate x
Multiply 4 by -1
Multiply (5) by 1
we get 2
1 x + 2 z = 6.25
-1 x + 4 z = 6.65
Add the two
0 x + 6 z = 12.9
/
z = 2.15

Plug the value of z in (5)
-1 x + 8.60 = 6.65
-1.00 x = -1.95
x = 1.95
plug value of x & z in 1.00
1.95 + 1.00 y + 2.15 = 6.15
1.00 y = 6.15 + -1.95 + -2.15
y= 2.05

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
c coffee
m muffin
x scone

Change E2 to E2-2*E1.
Change E3 to E3-E1.

Look carefully at the E2 and E3 equations; and see adding corresponding members eliminates x, and m is found very quickly.
-------------Muffin price.

You should find a path to solve the other two prices now knowing m.
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

Writing equations for such problem is traditionally considered as very ROUTINE step,
which does not require any explanations - - - so OBVIOUS it is.

The best way to describe a setup in such problems is to say

            "write equations as you read the problem".

Do not try to present this routine procedure as some high-intellectual activity.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

The two solutions you have seen so far are both valid but very different algebraic methods for solving the problem (even if the calculations shown are not always correct)....

There are an endless number of different ways for solving a system of three equations (using algebra -- not Cramer's rule or matrices). You shouldn't take ANY response you get to your post as saying that a particular way of solving the problem is what you should do.

Here is what I did; it's not a formal, well-defined process, but it turned out to make the work of solving this example easy.

Allow me to use "$" -- it's much easier for me to type then "|"....

x = coffee price
y = muffin price
z = scone price

x+y+z=$6.15 [1]
2x+3y+=$12.20 [2]
x+2y+2z=$10.35 [3]

I suppose because the first equation has x, y, and z once each, my first thought was to compare each of the other equations to the first. That gave me...

x+2y=$6.05 [4]
y+z=$4.20 [5]

Then I noted that, by chance, the "x+2y" in [4] was part of [3]. So substituting [4] in [3] gives
$6.05+2z=$10.35
2z=$4.30
z=$2.15 [6]

Then substituting [6] in [5] gives
y+$2.15=$4.20
y=$2.05 [7]

And finally substituting [6] and [7] in [1] gives
x+$2.05+$2.15=$6.15
x=$6.15-$4.20=$1.95

ANSWER:
coffee x = $1.95
muffin y = $2.05
scone z = $2.15

That was not a very good formal mathematical method, because it didn't follow any well-defined strategy... but it worked out nicely for this problem.

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Alice stopped by a coffee shop three days in a row at a conference to buy drinks and
pastries. On the first day, she bought a cup of coffee, a muffin and a scone for which
she paid |6.15. The next day she bought two cups of coffee, three muffins and a
scone (for herself and friends). Her bill was |12.20. The last day she bought a cup
of coffee, two muffins and two scones, and paid |10.35. Determine the price of a cup
of coffee, the price of a muffin and the price of a scone. Clearly explain your set-up
for the problem

Don't ever look at any problems those 2 people try to help you with. They will CONFUSE you, and have been WRONG, probably 99% of the times. One of them has this RIDICULOUS and STUPID habit of multiplying equations by 1, as if that'll change the equation. How QUAINT, SENSELESS and UTTERLY STUPID! Correct approach: Let a cup of coffee, a muffin, and a scone cost C, M, and S, respectively Then we get: = C + M + S = 6.15 ----- eq (i) 2C + 3M + S = 12.20 ----- eq (ii) C + 2M + 2S = 10.35 ----- eq (iii) 2C + 2M + 2S = 12.30 ----- Multiplying eq (i) by 2 ------ eq (iv) Cost of a cup of coffee, or ------ Subtracting eq (iii) from eq (iv) C + 2M = 6.05 ----- Subtracting eq (i) from eq (ii) ------- eq (v) 1.95 + 2M = 6.05 ----- Substituting 1.95 for C in eq (v) 2M = 4.10 Cost of a muffin, or 1.95 + 2.05 + S = 6.15 ------ Substituting 1.95 and 2.05 for C and M, respectively, in eq (i) 4 + S = 6.15 Cost of a scoce, or

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