Word problem in algebra

Question

Luis picked up five scones and four large coffees for the office. He paid 21.50. Rachel picked up the next day 6 scones and 7 large coffees for the office and paid 30.09. What is the cost of one scone? What is the cost of one large coffee?

Answer 1

Question: Luis picked up five scones and four large coffees for the office. He paid 21.50. Rachel picked up the next day 6 scones and 7 large coffees for the office and paid 30.09. What is the cost of one scone? What is the cost of one large coffee?

Answer: I now answer this question with a nice, warm cup of coffee at my desk.

Let's set up a system of equations. We'll need one equation for Luis, and one equation for Rachel.

Let's refer to scones by the variable s, and to large coffee by the variable c.

LUIS: 5s + 4c = 21.50

RACHEL: 6s + 7c = 30.09

This is a system that can be solved by means of elimination. It doesn't matter which variable you solve for first, but just watch your step at all times. I will solve for s first, so that means I will be eliminating c. I need to force 4c and 7c somehow. Maybe by using a least common multiple and making one of the numbers negative? The LCM of 4 and 7 is 28, so check this out:

(-7) [5s + 4c = 21.50]

(4) [6s + 7c = 30.09]

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That system becomes:

-35s - 28c = -150.50

24s + 28c = 120.36

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The -28c and 28c will add to 0c, or 0. It's been eliminated. Add up what's leftover:

-11s = -30.14

s = 2.74 ----> One scone costs $2.74.

Now that I know what s is, I can plug it into one of the equations in the original system. Pick the easier-looking one. (Either one will work.) I'll use Luis' equation. Plug in s where you see it and solve for c.

LUIS: 5s + 4c = 21.50

5(2.74) + 4c = 21.50

13.70 + 4c = 21.50

4c = 7.80

c = 1.95 -----> One large cup of coffee costs $1.95.

Let's check our work:

LUIS: 5s + 4c = 21.50

RACHEL: 6s + 7c = 30.09

LUIS: 5(2.74) + 4(1.95) = 21.50

RACHEL: 6(2.74) + 7(1.95) = 30.09

LUIS: 13.70 + 7.80 = 21.50

RACHEL: 16.44 + 13.65 = 30.09

LUIS: 21.50 = 21.50

RACHEL: 30.09 = 30.09

We get both true statements, so our answer is correct.

One scone costs $2.74, and one large cup of coffee costs $1.95.

Answer 2

Heidi

If the price of scones is s, and coffee c

5s + 4c = 21.50 (1)

6s + 7c = 30.09 (2)

We need to find s and c.

Lets make the coefficients of s both 30. So we multiply (1) by 6, and (2) by 5

30s + 24c = 129 (3)

30s + 35c = 150.45 (4)

Take (3) from (4)

11c = 21.45

c = 1.95 (Starbucks ??)

So coffee price is 1.95

You can substitute this back into (3) to find s.

Mike

Subtract

Answer 3

cost of scones = x

cost of coffees = y

cost of 5 scones is 5x, cost of 4 coffees is 4y

cost of 6 scones is 6x, cost of 7 coffees is 7y

5x + 4y = 21.50

6x + 7y = 30.09

Solve by elimination. Multiply top row by 6 and bottom row by 5

30x + 24y = 129

30x + 35y + 150.45

Subtract bottom row from top row

-11y = -21.45

y = 1.95, cost of one coffee = $1.95

Substitute value of y back into one of the original equations to solve for x

5x + 4(1.95) = 21.50

5x + 7.8 = 21.50

5x = 13.70

x = 2.74, cost of one scone = $2.74