Algebra 1 word problem

Question

Macy bought 3 hamburgers and 6 small fries for $27. Kirk bought 7 hamburgers and 4 fries for $35.50. How much would it cost to buy 1 hamburger and 1 small fry?

Answer 1

Macy bought 3 hamburgers and 6 small fries for $27. Kirk bought 7 hamburgers and 4 fries for $35.50. How much would it cost to buy 1 hamburger and 1 small fry?

This is called a system of equations. We have two variables, the price of hamburgers, and the price of small fries. I will represent these variables with B for Burgers and F for fries.

Macy bought 3 hamburgers and 6 small fries for $27.

From here we know that 3B (3 burgers) + 6F (6 small fries) costs $27. In equation form, we see that:

27 = 3B + 6F

Kirk bought 7 hamburgers and 4 fries for $35.50.

From this, we know that 7B (7 burgers) + 4F (4 fries) costs $35.50. In equation form, we see that:

35.50 = 7B + 4F

So we now have our system of equations:

27 = 3B + 6F

35.50 = 7B + 4F

If we can make one of our variables go away, then we can solve for the other one. We can do this by making a term be equal in both equations. For example, we see in our first equation that we have 6 Fries, and in our second equation, we have 4 fries. If we make those have the same number of fries, we can solve for the price of a burger.

To do this, we will multiply our first equation by 2 and our second equation by 3, like so:

First equation * 2:

(27 = 3B + 6F) * 2

>>> 27*2 = 3B*2 + 6F*2

>>> 54 = 6B + 12F

Second equation * 3:

(35.50 = 7B + 4F) * 3

>>> 35.50*3 = 7B*3 + 4F*3

>>> 106.50 = 21B + 12F

So now our two equations are

54 = 6B + 12F

106.50 = 21B + 12F

Do you see how we have the same number of fries in each? Now if we subtract one equation from the other, our fries factor will disappear entirely and we can solve for B (the price of a burger).

(106.50 = 21B + 12F)

-(54 = 6B + 12F)

>>> 106.50-54 = 21B-6B + 12F-12F

>>> 52.50 = 15B + 0F

>>> 52.50 = 15B

So, now we know that buying 15 burgers would cost $52.50, which means it only takes simple division to solve for the price of one burger.

52.50 = 15B

>>> 52.50 / 15 = B

>>> 3.50 = B

One burger costs $3.50

Now that we know how much a burger costs, we can substitute that value into one of our original equations to find the cost of a small fry.

We know that

27 = 3B + 6F

and that

B = 3.5

so,

27 = 3(3.5) + 6F

>>> 27 = 10.5 + 6F

>>> 27 - 10.5 = 6F

>>> 16.5 = 6F

>>> 16.5/6 = F

>>> 2.75 = F

And now we know that the price of one small fry is $2.75.

So, to answer the final question:

How much would it cost to buy 1 hamburger and 1 small fry?

One burger costs $3.50 and one small fry costs $2.75, so

3.50 + 2.75 = 6.25

It would cost $6.25.

Answer 2

it is a system, burgers fries money

system equation

3x+6y=27

7x+4y=35.5

divide first equation by 3

x+2y=9

7x+4y=35.5

multiply first equation by 2

2x+4y=18

7x+4y=35.5

subtract first equation form second equation

2x+4y=18

5x+0y=17.5

yay,, use second equation to conclude x=3.5

now can say 2*3.5+4y=18 from first equation

y=11/4

check 3*3.5+6*11/4=10.5+16.5

7*3.5+4*11/4=24.5+11

in more advanced algebra this problem will be put in matrix form

and immediately one has x and y if there is a solution