SOLUTION: A postal clerk sold some $.34 stamps and some $.23 stamps. Altogether, 15 stamps were sold for the total cost of $4.44. How many of each type of stamp were sold?

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Question 180802This question is from textbook Beginning Algebra with Applications
: A postal clerk sold some $.34 stamps and some $.23 stamps. Altogether, 15 stamps were sold for the total cost of $4.44. How many of each type of stamp were sold? This question is from textbook Beginning Algebra with Applications

Found 2 solutions by josmiceli, mgmoeab:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the number of $.34 stamps = a
Let the number of $.23 stamps = b
given:
(1) a+%2B+b+=+15
(2) .34a+%2B+.23b+=+4.44
Multiply both sides of (1) by .23 and
subtract from (2)
(2) .34a+%2B+.23b+=+4.44
(1) -.23a+-+.23b+=+-.23%2A15
------------------------------
(2) .34a+%2B+.23b+=+4.44
(1) -.23a+-+.23b+=+-3.45
.11a+=+.99
a+=+9
And, since
(1) a+%2B+b+=+15
b+=+15+-+a
b+=+6
9 of the $.34 stamps were sold and
6 of the $.23 stamps were sold
check:
(2) .34a+%2B+.23b+=+4.44
.34%2A9+%2B+.23%2A6+=+4.44
3.06+%2B+1.38+=+4.44
4.44+=+4.44
OK

Answer by mgmoeab(37) About Me  (Show Source):
You can put this solution on YOUR website!
LET X BE THE NUMBER OF $0.34 STAMPS
LET Y BE THE NUMBER OF $0.23 STAMPS
SYSTEM OF LINEAR EQUATIONS
a) ($0.34)X + ($0.23)Y = $4.44
b) X + Y = 15

SOLVING
a) (0.34)X + (0.23)Y = 4.44
b) (-0.34)X - (0.34)Y = -5.10 <---- MULTIPLY BY -0.34 TO ELIMINATE X
YOU WILL GET A NEW EQUATION
c) -0.11Y = -0.66
IF YOU SOLVE FOR Y, THEN YOU WILL GET Y= 6
TO FIND X YOU TAKE THE VALUE OF Y AND SUSTITUTE IN THE ORIGINAL EQUATION
b) X + Y = 15, ---> X + 6 = 15; THEN X = 9

THE ANSWER WILL BE:
9 STAMPS OF $0.34, AND 6 STAMPS OF $0.23 WERE SOLD