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Default Algebra - 02-20-2013, 05:00 PM

Help me with these questions, I honestly didn't pay attention at all like past 3 days. Idk how to solve. And I'm lazy.

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Last edited by Matt; 02-20-2013 at 05:02 PM.
  
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Default 02-20-2013, 05:11 PM

First one is pretty clear.

X Hot dogs purchased
Y Hamburgers purchased
3(AMOUNT/X) * 4(AMOUNT/Y)= $10
(2,1) - (x,y)
3(2 hot dogs) * 4(1 hamburger)=$10
The order pair represents 2 hot dogs ($6) and 1 burger ($4) purchased.

For the second one, you want to solve for X, but we have two variables. Your best approach is substitution.
15x + 10y = 700
x + y = 60
Solve for y in the second equation.
y = 60 - x
Plug into first equation. Solve for x.
15x + 10 (60 - x) = 700
Distribute 10.
15x - 10x + 600 =700
Subtract 600 on both sides and collect like terms.
5x = 100
Finish.
x = 20

Tell me if you understand so far.



  
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Default 02-20-2013, 05:12 PM

Quote:
Originally Posted by Mr 223 View Post
First one is pretty clear.

X Hot dogs purchased
Y Hamburgers purchased
3(AMOUNT/X) * 4(AMOUNT/Y)= $10
(2,1) - (x,y)
3(2 hot dogs) * 4(1 hamburger)=$10
The order pair represents 2 hot dogs ($6) and 1 burger ($4) purchased.

For the second one, you want to solve for X, but we have two variables. Your best approach is substitution.
15x + 10y = 700
x + y = 60
Solve for y in the second equation.
y = 60 - x
Plug into first equation. Solve for x.
15x + 10 (60 - x) = 700
Distribute 10.
15x - 10x + 600 =700
Subtract 600 on both sides and collect like terms.
5x = 100
Finish.
x = 20

Tell me if you understand so far.
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Default 02-20-2013, 05:13 PM

1. A
2. B 20
3. kinda weird still doin it.


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Default 02-20-2013, 05:16 PM

X = 2, Y = 1

3(2) + 4(1) = 10
6 + 4 = 10
10 = 10

Hotdogs = X Hamburgers = Y

Since we know that the X and the Y are dealing with hotdogs and hamburgers. And we also know that X=2 and Y=1. Therefore Hotdogs = 2, Hamburgers = 1.

The 2 and the 1 are not the amount of money spent on the burgers and hotdogs but the actual amount of burgers and hotdogs. The 3 and 4 Represent the amount each costs.


Your answer is A.


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Default 02-20-2013, 05:26 PM

We'll be doing the same algebraically for the third. We want to understand everything about our values. (Price [x,y], what 5x and 4y equals, etc.)
5x + 4y = 32
x + 6y = 22
Solve for x to get an equation fit for substitution.
x = 22 - 6y
Plug it in.
5 (22 - 6y) + 4y = 32
Distribute the 5.
110 - 30y + 4y = 32
Subtract 110 on both sides and collect like terms.
-26y = -78
Finish.
y = 3
We now know the price for 1 daisy because the problem says they are y dollars each.
Now, we'll find the price of the rose by plugging in our value and solve for x. You can use any given equation for this step.
x + 6(3) = 22
x + 18 = 22
Subtract 18.
x = 4
Plug your two values to test your values.
5(4) + 4(3) = 32?
20 + 12 = 32?
32 = 32 CORRECT

Seeing that one of your choices is "A rose costs $1 more than a daisy.", we can choose that answer immediately. We found that a rose was $4 each and a daisy is $3 each.




Last edited by Mr 223; 02-20-2013 at 05:28 PM.
  
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Default 02-21-2013, 08:48 AM

Quote:
Originally Posted by Mr 223 View Post
We'll be doing the same algebraically for the third. We want to understand everything about our values. (Price [x,y], what 5x and 4y equals, etc.)
5x + 4y = 32
x + 6y = 22
Solve for x to get an equation fit for substitution.
x = 22 - 6y
Plug it in.
5 (22 - 6y) + 4y = 32
Distribute the 5.
110 - 30y + 4y = 32
Subtract 110 on both sides and collect like terms.
-26y = -78
Finish.
y = 3
We now know the price for 1 daisy because the problem says they are y dollars each.
Now, we'll find the price of the rose by plugging in our value and solve for x. You can use any given equation for this step.
x + 6(3) = 22
x + 18 = 22
Subtract 18.
x = 4
Plug your two values to test your values.
5(4) + 4(3) = 32?
20 + 12 = 32?
32 = 32 CORRECT

Seeing that one of your choices is "A rose costs $1 more than a daisy.", we can choose that answer immediately. We found that a rose was $4 each and a daisy is $3 each.
Read this^ explains it perfectly.


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Default 02-21-2013, 08:51 AM

You got to do simultaneous equation for 2 and 3.It is all very easy, paying attention in class is half the battle won. If you can practice often, it will be a piece of cake.


  
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