Algebra Homework Help

[Halo 3 Seller]

Ok so with this information i need to know the cost of 1 two by six and the cost of one two by fours.

30 Two by fours and 50 two by sixes had a cost of 341$.
70 two by fours and 10 two by sixes had a cost of 273$

If someone could tell me the cost of 1 two by four and 1 two by six with that info that would be great!

 

[Tckr]

A Two by Four would be 3.2
A Two By Six would be 4.9

30x+50y=341
(70x+10y=273)-5

30x+50y=341
-250x -50y= -1365
-320x=-1024
x=3.2

30(3.2) + 50y = 341
96+50y=341
50y=245
y=4.9

Hopefully you have a basic understanding and know what I just did. If you have any other questions about this just pm me.

 

[malyce]

Ok so with this information i need to know the cost of 1 two by six and the cost of one two by fours.

30 Two by fours and 50 two by sixes had a cost of 341$.
70 two by fours and 10 two by sixes had a cost of 273$

If someone could tell me the cost of 1 two by four and 1 two by six with that info that would be great!

This is a class "2 problems, 2 unknowns" type problem. Here's how to solve it:

Step 1: Lets first rewrite it. Let a 2x4 = x, and a 2x6 = y.

30x + 50y = 341
70x + 10y = 273.

Step 2: Lets solve one equation in terms of one variable. Lets solve equation 2 to give us y.

First subtract 70x from both sides. 10y = 273 - 70x
Then divide by 10. y = 27.3 - 7x.

Step 3: Now, lets plug that y value into the y from the first equation.

30x + 50(27.3 +7x) = 341 (Do you see what I did here? We solved for y using the second equation and plugged that y value into the first to solve for x).

30x - 350x + 1365 = 341.
-320x = -1024 (subtract 1365 from both sides)
x = 3.2. One 2x4 = $3.2

Now that we have x = 3.2, plug this number into either the first OR the second equation to solve for y. I'll use the first.

30(3.2) + 50y = 341
96 + 50y = 341
50y = 245
y = $4.9

2x4 = $3.2
2x6 = $4.9


If you have any questions at all, feel free to ask. Be sure to work out this problem for yourself using my steps first, to see if you can better understand it!

 

[malakadang]

The above both answered your question.

If your still unfamiliar, then you can google 'Simultaneous Equations'.

Heres a youtube video showing you how to solve problems of that nature - http://www.youtube.com/watch?v=XM7Q4Oj5OTc