Question 1011774: The owner of men's clothing store bought six belts and eight hats for $140. A week later, at the same prices, he bought nine belts and six hats for $132. Find the price of a belt and the price of a hat.
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WHAT I KNOW
The equation for the first time he bought his items wod be 6b+8h=140
The b would stand for belts while the h would stand for hats
The other equation would be 9b+6h=132. I know that i have to multiply a number so o e of the variables would be the same but i do not understand the rest. Thank you for looking through my question. Have a nice day
Found 3 solutions by stanbon, mathmate, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The owner of men's clothing store bought six belts and eight hats for $140. A week later, at the same prices, he bought nine belts and six hats for $132. Find the price of a belt and the price of a hat.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
WHAT I KNOW
The equation for the first time he bought his items wod be 6b+8h=140
The b would stand for belts while the h would stand for hats
The other equation would be 9b+6h=132. I know that i have to multiply a number so o e of the variables would be the same but i do not understand the rest. Thank you for looking through my question. Have a nice day
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6b + 8h = 140 dollars
9b + 6h = 132 dollars
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Modify for elimination:
54b + 72h = 9*140
54b + 36h = 6*132
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Subtract and solve for "h"::
36h = 468
h = 13 (# of hats he hought)
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6b = 140-8*13 = 36
b = 6 (# of belts sold)
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Cheers,
Stan H.
Answer by mathmate(429)  (Show Source):
You can put this solution on YOUR website!
Question:
The owner of men's clothing store bought six belts and eight hats for $140. A week later, at the same prices, he bought nine belts and six hats for $132. Find the price of a belt and the price of a hat.
Solution:
Let
B=cost of a belt
H=cost of a hat
Set up system of equations according to the total cost:
6B+8H=140...........(1)
9B+6H=132...........(2)
Solve for B and H by elimination
9B+12H=210..........1.5*(1)
9B+6H=132...............(2)
Subtract
6H=(210-132)=78
H=13
back-substitute in (1) to find H
6B+8*13=140.............(1)
6B=140-104=36
B=6
Check
6(6)+8(13)=140, 9*6+6*13=132, ok.
Answer:
Belts cost $6 each, hats cost $13 each.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! you have two equations that have to be solved simultaneously.
the two equations are:
6b + 8h = 140
9b + 6h = 132
multiply both sides of the first equation by 3 and multiply both sides of the second equation by 2.
you will get:
18b + 24h = 420
18b + 12h = 264
subtract the second equation from the first to get:
0b + 12h = 156
simplify to get 12h = 156
divide both sides of this equation by 12 to get:
h = 78
the cost of a hat is 13 dollars.
go back to one of the original equations and replace h with 13 and solve for b.
6b + 8h = 140 becomes 6b + 8*13 = 140 which becomes 6b + 104 = 140.
subtract 104 from both sides of this equation to get 6b = 140 - 104.
simplify to get 6b = 36.
divide both sides of this equation by 6 to get b = 6.
the cost of a hat is 13 dollars.
the cost of a belt is 6 dollars.
replace both original equations with 13 dollars for h and 6 dollars for b to get:
6b + 8h = 140 becomes 6*6 + 8*13 = 140 which becomes 36 + 104 = 140 which becomes 140 = 140 which is true.
9b + 6h = 132 becomes 9*6 + 6*13 = 132 which becomes 54 + 78 = 132 which becomes 132 = 132 which is true.
the numbers check out ok.
your solution is the cost of a belt is 6 dollars and the cost of a hat is 13 dollars.
here's a reference on how to solve equations simultaneously.
http://www.purplemath.com/modules/systlin1.htm
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