algebra 1 word problems

Question

Kyle works at a donut factory where a 10-oz cup of coffe costs 95 cents. A 14-oz cup costs $1.15 and a 20-oz cup costs $1.50. during one busy period kyle served 24 cups of coffee using 384 ounces of coffee while collecting a total of $30.60. how many cups of each side did kyle fill?

Answer 1

This is a system of equations problem. The way we know this is because there are multiple variables/unknowns in the question.

1. You should identify what it is that you don't know. We seem to not know how many of each cup type was sold/filled.
2. Let x = number of 10-oz cups
3. Let y = number of 14-oz cups
4. Let z = number of 20-oz cups
5. We need to group/organize the information. **Look for key words that repeat such as “ounces”, “$”, etc.*** Usually, information with the same key words will be the same equation.
6. The 95 cents, the $1.15, and the $1.50 will all be related to the $30.60.
7. We should also notice that there is information about ounces, too, so the 10 oz, 14 oz, and 20 oz will form an equation with the 384 ounces.
8. The number of cups (our variables) will also form an equation with the total number of cups.
9. Make equations out of the information
10. Cups info: x+ y + z = 24
11. Prices info: 0.95x + 1.15y + 1.50z = $30.60
12. Ounces info: 10x + 14y + 20z = 384
13. Solve using substitution, elimination, or graphing.
14. Eq 1: x+ y + z = 24
15. Eq 2: 0.95x + 1.15y + 1.50z = $30.60
16. Eq 3: 10x + 14y + 20z = 384

I chose to eliminate z and then use substitution to solve for variable y.

(x+y+z = 24) * 1.5 = -1.50x - 1.50y - 1.50z = -36

Eq 1: -1.50x -1.50y - 1.50z = -36

+ Eq 2: 0.95x + 1.15y + 1.50z = 30.60

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

Eq 4: -0.55x - 0.35y = -5.4

Now I will eliminate the same variable (z) from equations two and three.

Eq 2: 0.95x + 1.15y + 1.50z = $30.60

Eq 3: 10x + 14y + 20z = 384

Eq 2: (0.95x+1.15y + 1.50z = 30.60)* 20 = 19x + 23y + 30z = 612

Eq 3: (10x + 14y + 20z =384) *-1.5 = -15x - 21y - 30z = -576

Eq 2: 19x + 23y + 30z = 612

+ Eq 3: -15x - 21y - 30z = -576

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

Eq 5: 4x + 2y = 36

Substitution for equations 4 and 5 by solving for y.

1. TIP 1: If there is a variable with a coefficient of 1, it will be easy to solve for that variable and use substitution.
2. TIP 2: If you can divide all of the coefficients in the equation by a number (such as 2), then substitution is a good choice.

Eq 4: -0.55x - 0.35y = -5.4

Eq 5: 4x + 2y = 36

4x + 2y = 36

-4x    - 4x

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

2y = 36 -4x Divide by 2

Y = 18 - 2x

Now I substitute what y equals (18-2x) into equation 4 for “y”.

-0.55x - 0.35(18-2x) = -5.4 Distribute the -0.35

-0.55x - 6.3 + 0.7x = -5.4  Combine like terms

0.15x-6.3= -5.4    Add 6.3 to both sides

 0.15x = 0.9     Divide both sides by 0.15

 X = 6       

       

Plug x = 6 back into equation 4 or 5 to find y.

Eq 1: 4(6) + 2y = 36 —> 24 + 2y = 36 Subtract 24 from both sides

2y = 12  Divide both sides by 2

 Y = 6

Now that we have x and y, we can plug them *both* back into one of the original equations to find “z”. The easiest equation would be equation 1.

Eq 1: x+ y + z = 24

6 + 6 + z = 24

12 + z = 24

Z = 12

The answer can be written as (x,y,z) so it would be (6,6,12).

So he filled 6 10-oz cups, he filled 6 14-ounce cups, and filled 12, 20-oz cups.

Answer 2

For this problem, you will need to write a system of equations and then solve for the variables.

SET UP

1) Define your variables

x = 10 oz cup of coffee

y = 14 oz cup of coffee

z = 20 oz cup of coffee

2) Write an equation for the number of cups of coffee sold (total cups sold = 24 cups)

x + y + z = 24

3) Write an equation for the number of ounces of coffee sold (total ounces = 384 ounces)

10x + 14y + 20z = 384

3) Write an equation for the total amount of money made (total money = $30.60)

0.95x + 1.15y + 1.50z = 30.60

Now you have a system of 3 equations to solve.

SOLVE

1) Choose 2 equations and choose a variable to eliminate.

Let's use the first 2 equations and eliminate x:

x + y + z = 24 (multiply this equation by -10)

10x + 14y + 20z = 384

-10x - 10y - 10z = -240

10x + 14y + 20z = 384

The x values cancel out and you are left with:

4y + 10z = 144

2) Choose another set of 2 equations and eliminate the same variable (in this case, x)

Let's use the first and third equation:

x + y + z = 24 (multiply this equation by -95)

95x + 115y + 150z = 3060 (multiplied everything by 100 to eliminate the decimals because I prefer whole #s)

-95x -95y - 95z = -2280

95x + 115y + 150z = 3060

The x values cancel out and you are left with:

20y + 55z = 780

3) Solve the system of equations for the two equations you just made

4y + 10z = 144 (multiply this equation by -5)

20y + 55z = 780

-20y - 50z = -720

20y + 55z = 780

The y values cancel out and you are left with:

5z = 60

z = 12

4) Plug in the value of z into the equation 4y + 10z = 144 to find y

4y + 10(12) = 144

4y + 120 = 144

4y = 24

y = 6

5) Plug in the values for y and z into the equation x + y + z = 24 to find x

x + 6 + 12 = 24

x + 18 = 24

x = 6

6) Check your work by plugging in the values of x, y, and z into each equation

Let me know if you need clarification or have further questions :)

Answer 3

x+y+z = 24

,95x + 1.15y + 1.5z = 30.6

10x + 14y + 20z = 384

3 equations 3 unknowns

if they're linearly independent equations, not inconsistent, there's one unique solution

solve by substitution, elimination or matrix algebra. the calculations get a little tedious

but the solution is x=6=y and z=12

6 10 oz.

6 14 oz

12 20 oz

6+6+12 = 24

.95(6)+1.15(6)+1.5(12) = 5.7+6.9+18=30.6

10(6)+14(6) +20(12)= 60+84+240 = 384