Mark B. answered • 03/05/19
PhD Candidate and Algebra I Tutor with 20 Years of Experience
Hello Jay,
As you are most likely aware, you have a systems of equations problem above. While you correctly pointed out on of the equations - 2t + 3e = 780, you need the other equation to solve this problem. The best part is the equation is actually given. Therefore, I am going to work the problem with both equations, showing you how I would work the problem below.
First equation is:
t + e = 300 <-------"Middle school sold 300 tacos and enchiladas."
2t + 3e = 780 <---- "Tacos sold for $2 and enchiladas for $3. They made $780."
Please note: You can treat the prices of the tacos and enchiladas as whole numbers remembering to add your correct decimal points if that was what the problem is asking. However, the question is asking for the number of tacos sold. This is why I am not converting the 2 and 3 to $2.00 and $3.00.
Step #1: First equation
t + e = 300
t = 300 - e
Step #2: Second equation
2t + 3e = 780
Substitute for t in the above equation - the second equation with the value of 300 - e for 't"
2(300 - e) + 3e = 780
600 - 2e + 3e = 780
600 + e = 780
e = 180 The number of enchiladas sold.
Now, substitute the value of 180 for "e" in the first equation.
t + e = 300
t + 180 = 300
t = 120 The number of tacos sold.
Now, let's check our work, okay?
First, we know from the problem that a total of $780 was made, correct?
Therefore, if tacos sold for $2 and 120 were sold, $240 was made from tacos, right?
And, if Enchiladas sold for $3, a total of 180 times multiplied by $3 equals, 540. right?
Adding 240 plus 540 should equal the amount earned by selling both the tacos and enchiladas, correct?
And, in fact the numbers check. Therefore, the solution is valid,
I hope I have helped and that you have a great week. Please feel free to provide feedback to my solution, seek clarity, or ask further questions related to the solution. Best!
