You sell small and large candles at a craft fair. Small candles cost $4 each and large candles cost $6 each. You collect $144 after selling a total of 28 candles. How many of each type of candle did you sell? Write a linear system to represent the situation. Be sure to define your variables.

Answer: Small Candle = 12, Large Candle = 16

Step-by-step explanation:

Equation is this, s+l=28 2nd equation is 4s+6l=144

now you multiply every number in the second equation by -6 since positive 6 is the second equations second constant. Now you got -6s-6l=-168 And 4s+6l=144. The -6l and positive cancel so now you got -6s + 4s = -168+144. Now you got -2s = -24. Divide them both by negative two then you get s = 12. 28 candles total - 12 = 16 so 16 large candles and 12 small. L = large candles, S = small candles.

Answer:

8 small candles

16 large candles

Step-by-step explanation:

1. Approach

Set up a system of equations to solve this problem. Call the number of small candles sold (x), and the number of large candles sold (y). Have one expression model the amount of money earned, the other model the number of candles sold. Then, solve this system of equations using the process of elimination. In the process of elimination, one multiplies one of the equations by a term such that one of the coefficients of a variable in one of the equations is the additive inverse of the corresponding variable in the other equation. This means that when one adds the two equations, one of the variables will eliminate. Then one can solve for the other variable, and finally, substitute the value of the variable that one found back into one of the equations, and solve for the value of the other variable.

2. Set up a system of equations

Use the given information to set up a system of equations to describe the amount earned, and the amount sold

x + y = 28 -> amount sold

4x + 6y = 144 -> amount earned

3. Solve the system

As explained above, the process of elimination is a method of solving a system of equations. In the process of elimination, one multiplies one of the equations by a term such that one of the coefficients of a variable in one of the equations is the additive inverse of the corresponding variable in the other equation. This means that when one adds the two equations, one of the variables will eliminate.

x + y = 28 (*-4)

4x + 6y = 144

-4x - 4y = -112

4x + 6y = 144

___________

2y = 32

/2     /2

y = 16

4. Solve to find out how much of the other candle was sold

Now back solve, substitute the found value into one of the equations, and solve to find the other value

x + y = 28

y = 16

x + 16 = 28

    -16     -16

x = 8

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