What are the equations for these word problems?

Question

My sister needs help with her homework and I haven't done this in a looooong time...

1) April buys 8 books for $44. Paperbacks cost $4 each and hardbacks cost $8 each. How many of each book did she buy?

2) Suppose you have a coin collection of dimes and quarters containing 46 coins. If you have $6.70, how many of each type of coin do you have?

Answer 1

Hi,


1) April buys 8 books for $44. Paperbacks cost $4 each and hardbacks cost $8 each. How many of each book did she buy?

Let x = # of hardbacks and y = # of paperbacks

Then x + y = 8

If paperbacks are $4 each and hardbacks are $8 each then the prices give you the equation:

8x + 4y = 44

If we solve the first equation for x, it becomes:

x + y = 8
x = 8 - y we can substitute this expression into the second equation in place of x. then we can solve for y.

8x + 4y = 44
8(8 - y) + 4y = 44
64 - 8y + 4y = 44
64 - 4y = 44
-4y = -20
y = 5, so she bought 5 paperbacks. Since x + y = 8, then x = 3 for 3 hardbacks.

2) Suppose you have a coin collection of dimes and quarters containing 46 coins. If you have $6.70, how many of each type of coin do you have?

Let d = # of dimes and q = # of quarters so like the first problem we start with:

d + q = 46

Then since each dime is worth 10 cents and each quarter is worth 25 cents, this gives the equation 10d + 25q = 670 (Notice all money is entered in cents, not dollars)

d + q = 46 <== solve for d
d = 46 - q <==== substitute this in the other equation for "d".
10d + 25q = 670
10(46 - q) + 25q = 670
460 - 10q + 25q = 670
460 + 15q = 670
15q = 210
q = 14, so there are 14 quarters. 46 - 14 = 32, so there are 32 dimes.

I hope that helps!! (And it's nice you are helping your sister)

Answer 2

Ok, so let's say X amount of books cost 4$ and Y amount of books cost 8$
So 4X+8Y=44$
And X+Y= 8 Books

First solve for one Variable, Y=8-X
Then plug that in into the first equation for Y
4X+8(8-X)=44

Then Multiply it out
4X+64-8X=44

Add Like Numbers and Move the Same Variables to one side
4X-8X=44-64

-4X=-20
(You can remove the negative side since both sides are negative)
So 4X=20

divide both sides by 4 and ...
X=5

Then Plug in X for either of the equations .....
So either 5+Y=8
or 5(4)+8Y=44

and you should end up with Y=3


2. Do the same thing with #2 , just plug in the different numbers
X + Y = 46
.25X + .1Y = $6.70

Solve for 1 Variable
Y=46-X

Then plug it into the other equation
.25X + .1(46-X)=6.70

Multiply it out
.25X+4.6-.1X = 6.70

Add Like Numbers,
.15X + 4.6 = 6.70

Separate the Variables
.15X = (6.70-4.6)
.15X = 2.1

Divide
2.1/.15 = 14

Solve for Y
14 + Y =46

So Y = 32

Answer 3

1. p = paperbacks; 8-p = hardbacks

4p + 8(8-p) = 44


2. d = number of dimes; 46-d = number of quarters

0.10 d + 0.25 (46-d)=6.70

These are just the equations.

Answer 4

1) P+H = 8....and..... 4*P+8*H = 44......... Solve these 2 simultaneously.

2) D+Q = 46......and........(0.10)*D + (0.25)*Q = 6.70