Please help with word problem. Thank you

Question

At the track meet, Ms. Nesland bought 3 bags of popcorn, 4 candy bars, and 2 drinks for a total of $ 15.75. Mrs. Wick spent $14.00 on 5 bags of popcorn, 2 candy bars, and 1 drink. Mrs. Zinn bought 3 bags of popcorn, 3 candy bars, and 3 drinks for $16.50. (Yes, they were sharing with other people.). Write and solve a system of equations to find the purchase price for each item. You may solve this problem using matrices, but clearly show the steps taken to arrive at the final solution.

Answer 1

Popcorn is $1.75

Candy bars are $1.50

Drinks are $2.25

While several approaches will work, I used a system of equations in 3 unknowns.

Assign each item to a variable

x = popcorn

y = candy bars

z = drinks

Set up equations for each purchase

Equation 1

3x + 4y + 2z = 15.75

Equation 2

5x + 2y + z = 14.00

Equation 3

3x + 3y + 3z = 16.50

Review you equations carefully for the use of Elimination where possible

Notice you can eliminate two variables y and z then solve directly for x as follows;

Multiply Equation 2 by negative 2 and add it to Equation 1

-2(5x + 2y + z = 14.00)

-10x - 4y -2z = -28.00

Combine this equation with Equation 1

-10x - 4y -2z = -28.00

3x + 4y + 2z = 15.75

Both y and z are eliminated

-7x = -12.25

Divide both sides by -7

x = 1.75

Now combine the original Equation 1 and Equation 2 together to give

8x + 6y + 3z = 29.75

Subtract Equation 3 from the combination above to yield an equation in x and y

8x + 6y + 3z = 29.75

-(3x + 3y + 3z = 16.50) to give

8x + 6y + 3z = 29.75

-3x - 3y - 3z = -16.50

This eliminates z to give an equation in x and y only.

5x + 3y = 13.25

You already have a value for x and you can use it to solve for y

5(1.75) + 3y = 13.25

8.75 + 3y = 13.25

Subtract 8.75 from both sides

3y = 13.25 - 8.75

3y = 4.50

y = 1.50

Now plug x and y into Equation 2, to solve for the single z

Equation 2

5x + 2y + z = 14.00

5(1.75) + 2(1.50) + z = 14.00

8.75 + 3.00 + z = 14.00

11.75 + z = 14.00

Subtract 11.75 from both sides

z = 14.00 - 11.75

z = 2.25

You can check the values by substituting them in all the original equations. I'll leave that to you

Answer 2

Let's start with what we want to solve for. "The price of the items", so since there are three items, we need to solve for three things, X, Y, and Z.

X is the price of popcorn

Y is the price of candy bars

Z is the price of a drink

Let's add up how much of each item they bought. When we do this, let's add it all up to equal how much they spent.

Ms. Nesland's purchase can be shown as 3x+4y+2z=$15.75

Mrs. Wick's purchase is similar and is shown as 5x+2y+1z=$14.00

Mrs. Zinn's purchase is similar and is shown as 3x+3y+3z=$16.50

Now to solve this problem, we need to start solving for each variable. However, when we get an equation for each variable, we want it to only have one other variable in the equation. This will take some extra steps.

I'll randomly decide to solve for X in the first equation (Ms. Nesland's).

3x+4y=15.75-2z

3x=15.75-2z-4y

x=(15.75-2z-4y)/3

*see how Z and Y are both in this equation for X? This is good for now, but we will need to keep going. We need to get only Z or Y in the equation for X. But we will get to that in a minute.

Next Step, plug this new X equation into another equation and solve for Y. Let's plug it into Mrs. Wick's.

5((15.75-2z-4y)/3)+2y+1z=14.00

Solve for Y.

26.25-(10z/3)-(20y/3)+2y+1z=14

-(10z/3)-(20y/3)+2y+1z=14-26.25=-12.25

Combine z's and y's

-(7z/3)-(14y/3)=-12.25

Solve for Y

-14y/3=-12.25+(7z/3)

-14y=-36.75+7z

y=(-36.75+7z)/-14

y=2.625-0.5z

We only have Z in the equation for Y. Great! Let's also get only Z in the equation for X too. Let's take that equation we found in the beginning. x=(15.75-2z-4y)/3 And let's plug in the Y equation we just got into it: y=2.625-0.5z

x=(15.75-2z-4(2.625-0.5z))/3

x=(15.75-2z-10.5+2z)/3

x=(5.25-0z)/3 *The z's cancelled out*

x=5.25/3

x=1.75

Let's take an original equation and plug in what we know. x=1.75 and y=2.625-0.5z

I'll randomly select this one, but you could have picked the other two.

3x+3y+3z=$16.50

3(1.75)+3(2.625-0.5z)+3z=16.5

Solve for Z

5.25+7.875-1.5z+3z=16.5

13.125+1.5z=16.5

1.5z=3.375

z=2.25

Now we know X and Y. Let's plug these into ANY equation we have been given or found.

I'll again randomly choose the same equation.

3x+3y+3z=$16.50

3(1.75)+3y+3(2.25)=16.5

5.25+3y+6.75=16.5

12+3y=16.5

3y=4.5

y=1.5

Now we have our answers!

x=1.75

y=1.5

z=2.25

Remember we defined these in the beginning.

X is the price of popcorn

Y is the price of candy bars

Z is the price of a drink

Popcorn is $1.75

Candy Bars are $1.50

Drinks are $2.25

You can double check this. Plug these values into ANY equation.

Let's pick Mrs. Wick's purchase equation: 5x+2y+1z=$14.00

5(1.75)+2(1.5)+1(2.25)=14

8.75+3+2.25=14

14=14

Since this worked, we know we are right.

Answer 3

Your system looks like this, where p is for popcorn, c for candy, and d for drinks. And the right hand sides f the equations are in dollars:

3p + 4c + 2d = 15.75

5p + 2c + 1d = 14.00

3p + 3c + 3d = 16.50

Set up a matrix equation where, a matrix A is a 3 x 3 matrix comprised of the coefficients of the left hand sides of the three equations, a vector c is the transpose of (15.75, 14, 16.5), the right hand sides of the equations, and a vector b is the transpose of (p, c, d), the unknown product prices.

Then we can represent the linear system as:

Ab = c

Now if we want b, the list of purchase prices for each item, we can left multiply both sides of the matrix equation by A^-1 , the inverse of A

A^-1 Ab= A^-1 c which, if there is a solution, produces b, by itself, a list of your purchase prices in p, c, d order.