Question

Part C $8 for every medium-sized collar and $12 for every large-sized collar He has $1.020 to spend and Demarco gives away dog collars and ID tags during a local event The cost for making dog collars is wants to make a total of 100 dog collars. Write a system of linear equations to model the situation. Let x be the number of medium- sized collars, and let y be the number of large-sized collars. Equation 1: Equation 2: 7b How many of each size collar can Demarco make? Show your work. Demarco wants to order the design plus $9 for each tag How many ID tags c​

Answer 1

`\textsf{Equation 1} :\sf 8x + 12y = 1020`

`\textsf{Equation 2} :\sf x + y = 100`

Demarco can make 45 medium-sized collars and 55 large-sized collars.

Explanation:

Let's use the following variables:

• x = number of medium-sized collars
• y = number of large-sized collars
• c = number of ID tags

We know that the cost for making dog collars is $8 for every medium-sized collar and $12 for every large-sized collar.

We also know that Demarco wants to make a total of 100 dog collars. This gives us the following system of linear equations:

`\textsf{Equation 1} :\sf 8x + 12y = 1020`

`\textsf{Equation 2} :\sf x + y = 100`

To find how many of each size collar Demarco can make, we can solve the system of equations.

We can do this by subtracting the second equation from the first equation by substitution or elimination. Let's use the substitution method.

From Equation 2,

solve for x:

`\sf x = 100 - y`

Substitute this value of x into Equation 1:

`\sf 8(100 - y) + 12y = 1020`

Open bracket on left side

`\sf 800 - 8y + 12y = 1020`

Solve like terms,

`\sf 800+4y = 1020`

Subtract both sides by 800,

`\sf 800+4y -800= 1020 -800`

`\sf 4y = 220`

Divide both side by 4.

`\sf (4y)/(4)=(220)/(4)`

`\sf y = 55`

Now that we have the value of y (number of large-sized collars), substitute it back into the expression for x:

`\sf x = 100 - y`

`\sf x = 100 - 55`

`\sf x = 45`

So, Demarco can make 45 medium-sized collars and 55 large-sized collars.