help with a couple of algebra ques?

Question

1. MRS BOYD HAS A DESK FULL OF QUATERS AND NICKELS. IF SHE HAS A TOTAL OF 20 COINS WITH A TOTAL FACE VALUE OF $4.00, HOW MANY OF THE COINS ARE DIMES, AND HOW MANY ARE NICKELS?




2. 30,000 PEOPLE ATTENDED A BALLGAME AT A STADIUM THAT OFFERS TWO KINDS OF SEATS:GENERAL ADMISSION AND RESERVED. THE DAY'S RECEIPTS WERE $228,000. HOW MANY PEOPLE PAID $12 FOR RESERVED SEATS, AND HOW MANY PAID $6 FOR GENERAL ADMISSION.

THANKS

Answer 1

1. Let x = numer of nickels and y = number of quarters
(1) x + y = 20
(2) 5x + 25y =400 (multiplied by 100 to get rid of decimal)
-5x - 5y = -100 ((1) multiplied by -5)
20y = 300
y = 15 quarters
x = 20 - 15 = 5 nickels

2. Same method as 1.
let x = no. of $12 seats, y = no.of $6 eats
x + y = 30,000
12x + 6y = 228,000
. 6x + 6y = 180,000
6x = 48,000
x = 8,000
y = 30,000 - 8,000
y = 22,000

Answer 2

Let

x = Nickels

y = Quarters

.05x = Nickels value

.25 = Quarters value

20 = total coins

4.00 = Total value of all coins

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x + y = 20 - - - - - - - - - -Equation 1
.05x + .25y = 4.00 - - - - Equation 2
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Multiply equation 2 by 100

100(.05x) + 100(.25y = 100(4.00)

5x + 25y = 400 - -New equation 2

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x + y = 20

5x + 25y = 400

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Substitution Method Equation 1

x + y = 20

x + y - x = 20 - x

y = 20 - x

Insert the y value into equation 2

- - - - - - -- --

5x + 25y = 400

5x + 25(20 - x) = 400

5x + 500 - 25x = 400

- 20x + 500 = 400

-20x + 500 - 500 = 400 - 500

- 20x = - 100

- 20x/-20 = - 100/ - 20

x = 5

The answer is x = 5

Insert the x value intop equation 1

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x + y = 20

5 + y = 20

5 + y - 5 = 20 - 5

y = 15

The answer is y = 15

Insert the y value into equation 1

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Check

x + y = 10

5 + 15 = 20

20 = 20

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There are five (5) nickels

There are fifteen (15) quarters

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2 nd Problem

x + y = 30000 - - - - - - - - Equation 1

6x + 12y = 228000- - - - - -Equation 2

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Substitution Method equation 1

x + y = 30000

x + y - x = 30000 - x

y = 30000 - x

Insert the y value into equation 2

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6x + 12y = 228000

6x + 12(30000-x) = 228000

6x + 360000 - 12x = 228000

- 6x + 360000 -= 228000

-6x + 360000- 360000 = 228000 - 360000

- 6x = - 132000

- 6x/-6 = - 132000 / - 6

x = 22000

The answer is x = 22000

Insert the x value into equation 1

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x + y = 30000

22000 + y = 30000

22000 + y - 22000 = 30000 - 22000

y = 8000

The answer is y = 8000

Insert the y value into equation 1

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x + y = 30000

22000 + 8000 = 30000

30000 = 30000

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Check for equation 2

6x + 12y = 228000

6(22000) + 12(8000) = 228000

132000 + 96000 = 228000

228000 = 228000

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The solution set is { 22000, 8000 }

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Answer 3

1. The question is wrong! The two closest answers are 13 quarters with 7 dimes equals $3.95, and 14 quarters with six dimes equals $4.10 ?

2. 22,000 at $6 and 8000 at $12

Answer 4

let quarters be q
and nickels be n

q+n=20

and

25q+5n=400 (convert everything to cents)

q=20-n
substitute for q in the second equation

25(20-n)+5n=400

500-20n=400

-20n=-100

n=5

substitute into the first equation to find the value of q

q=20-n

q=20-5

q=15

so you have 15quarters and 5nickels

You can do the second problem in the exact same way. by making 2 equations and then using this substitution method.

Answer 5

1.the sum is wron.please check it

2.x+y=30000
12x+6y=228000
eqn (2)-eqn.(1)*6
6x=48000
x=8000substituting
y=22000
check
96000+132000=228000
checks