i have a huge project can someone help me create a table for the word problem below

Question

At the Apple Pan, 4 burgers and 3 fries cost $26.50. 5 burgers and 5 fries cost $36.25. What is the cost for each item?

 

Here are the 2 equations i came up with. .

4b + 3f = 26.50
and
5b + 5f = 36.25

Answer 1

4b + 3f = 26.50

 

5b + 5f = 36.25

 

5b=36.25-5f

b=7.25-f

 

4(7.25-f)+3f=26.50

29-4f+3f=26.5

29-f=26.5

f=2.5

 

4b+3(2.5) = 26.50

4b+7.5 = 26.50

4b=19

b=4.75

Answer 2

We can use the elimination method to cancel out one of the variables.

 

Let's multiply -5 to both sides of the first equation.

 

-5 (4b + 3f ) = -5(26.50)

 

=

 

-20b - 2f = -132.50

 

 

Now let's multiply 4 to both sides of the second equation.

 

 

4(5b + 5f) = 4(36.25)

 

=

 

20b +20f = 145

 

 

Now let's add these new equations together, and hopefully one of the variables will be eliminated, allowing us to solve for the other variable more easily.

 

 

-20b - 2f = -132.50

 

+

20b +20f = 145

-------------------------

0b + 18f = 12.5

 

 

18f = 12.5

 

f = 12.5/18

 

  = 0.69

 

 

So the cost of a pack of fries is f = $0.69. We can plug this value into the original first equation and solve for the cost of a burger b.

 

4b + 3f = 26.50

 

4b + 3(0.69) = 26.50

 

4b + 2.08 = 26.50

 

4b = 26.50 - 2.08

 

4b = 24.42

 

b = 24.42/4

 

   = 6.10

 

So the cost of a burger is b = $6.10 . (A little pricey if you ask me...)

 

We can double-check these values by plugging them into the original equations. If the statements are true, then the values are correct.

 

4b + 3f = 26.50

 

4(6.10) + 3(0.69) = 26.50

 

24.40 + 2.07 = 26.50

 

26.47 ≈ 26.50                      TRUE!

 

 

5b + 5f = 36.25

 

5(6.10) + 5(0.69) = 36.25

 

30.50 +  3.45 = 36.25

 

33.95 ≈ 36.25                  Close, but a bit of a stretch

 

 

Miscellaneous: 

I chose to multiply the two equations in the beginning by 5 and -4 because I knew that would give me a 20b in the first equation and a -20b in the second equation. Those two terms would cancel when I added the two equations, leaving me with only one variable f to solve for.