SOLUTION: How do I set this problem up to solve and check? A company produces 2 model bikes; Model A and Model B. Model A takes 2 hours to assemble, where Model B takes 3 hours to assemble

Algebra ->  Systems-of-equations -> SOLUTION: How do I set this problem up to solve and check? A company produces 2 model bikes; Model A and Model B. Model A takes 2 hours to assemble, where Model B takes 3 hours to assemble      Log On


   

Question 236144: How do I set this problem up to solve and check?
A company produces 2 model bikes; Model A and Model B. Model A takes 2 hours to assemble, where Model B takes 3 hours to assemble. Model A cost $25.00 to make per bike, where model B takes $30.00 to make per bike. If the company has a total of 34 hours and $350.00 per day for these two models, how many of each models can be made in a day?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = number of model A.
Let y = number of model B.

First equation:

2x + 3y = 34

2 times x are the total number of hours required to make model A.
3 times y are the total number of hour required to make model B.

Second equation:

25x + 30y = 350

25 times x is the total cost to make model A.
30 times y is the total cost to make model B.

You need to solve these two equations simultaneously which means that the same value of x and the same value of y apply to both equations.

The two equations are:

2x + 3y = 34 (equation 1)
25x + 30y = 350 (equation 2)

You can solve by substitution or elimination.

You can also graph these equations to see where they intersect, if anywhere.

We'll solve first and graph later.

We'll solve by elimination.

The equations are (again):

2x + 3y = 34 (equation 1)
25x + 30y = 350 (equation 2)

Multiply equation 1 by 10 to get:

20x + 30y = 340 (equation 1 multiplied by 10)
25x + 30y = 350 (equation 2)

Subtract equation 2 from equation 1 to get:

5x = 10

Divide both sides by 5 to get:

x = 2

Substitute in original equation 1 and solve for y.

2x + 3y = 34 (equation 1) becomes

2*2 + 3y = 34 which becomes:
4 + 3y = 34
Subtract 4 from each side to get:
3y = 30
Divide both sides by 3 to get:
y = 10

You have:

x = 2
y = 10

Substitute in both original equations to confirm the answers can solve both equations simultaneously.

2x + 3y = 34 (equation 1)
25x + 30y = 350 (equation 2)

become:

2*2 + 3*10 = 34 (equation 1)
25*2 + 30*10 = 350 (equation 2)

Simplify and combine like terms to get:

35 = 35 (equation 1)
350 = 350 (equation 2)

To graph these equations, you need to solve for y.

solving for y in each equation, we get:

y = (34-2x)/3 (equation 1)

y = (350x-25)/30 (equation 2)

The graph is shown below

graph%28300%2C300%2C-1%2C4%2C-1%2C11%2C%2834-2x%29%2F3%2C%28350-25x%29%2F30%29

It's a little difficult to see, but the intersection of these two lines is at (x,y) = (2,10) meaning these lines intersect when the value of x = 2 and the value of y = 10.