Using system of equations answer this word problem. 1) Scott and Paul are selling Pies for a school fundraiser. Customers can buy apple pies and lemon meringue pies. Scott sold 3 apple pies and 12 lemon meringue pies for a total of 141$. Paul sold 2 apple pies and 5 lemon meringue pies for a total of 64$ What is the cost of each of one apple pie and one lemon meringue pie? (Show your work and tell me how to do it and I'll give you brainlist!)

Step-by-step explanation:

We first need to come up with a system of equations that models this situation. Let's call apple pies "a" and lemon pies "L". Then we look to each individual involved in selling these pies and come up with an equation for each. Scott sold 3 apple and 12 lemon pies. The combination of this number of these 2 types of pies grossed him $141. In equation form, the combination of this number of these 2 types of pies is an addition thing. Namely,

3a + 12L = 141

For Paul, his combination was 2 apple and 5 lemon for a total of $64. His equation is

2a + 5L = 64

We need to solve this system either by substitution or elimination/addition to come up with the cost of both types of pies. I chose elimination/addition because I think in this case it is easier, and easier is good.

Setting up for elimination/addition:

3a + 12L = 141

2a + 5L = 64

Let's first eliminate the a terms. That means that the coefficient on each a has to be the same but with different signs. The LCM of both 3 and 2 is 6, so we will multiply the top equation by 2 and the bottom equation by - 3 (we could also have chosen to eliminate the L's first; we also could have chosen to eliminate the a's first but made the 2 negative instead of the 3. As long as you do the math correctly, you'll get the same answer regardless of how you choose to do it). Multiplying the first equation by 2 and the second by - 3 gives us a new system:

6a + 24L = 282

-6a - 15L = - 192

The reason this is called the elimination or addition method is because the goal is to eliminate, but you do that by adding the columns together. As you can see, 6a - 6a eliminates the a terms, right? What you're left with then is

9L = 90 and

L = 10

Now that we know the cost of a lemon pie, we can back sub in to solve for the cost of an apple pie. It doesn't matter which equation you back sub into, as long as it has both an L and an a in it. I'll choose the first equation we wrote when we took on this nightmare:

3a + 12L = 141. Subbing in 10 for L gives us

3a + 12 (10) = 141 and

3a + 120 = 141 and

3a = 21 so

a = 7

That means that lemon pies cost 10 each and apple pies cost 7 dollars each.