Question

At a fundraiser breakfast, the bill for three glasses of orange juice and five pancake specials is $7.60, whereas the bill for one glass of orange juice and two pancake specials is $2.90. The system of equation {(3j [email protected] 2p=2.90)┤ represents this situation, where j represents the cost of a glass of orange juice and p represents the cost of a pancake special. What would be the bill for one glass of orange juice and one pancake special?

Answer 1

We have a system of equation:
3j + 5p= $7.60 (1)
j + 2p= $2.90 (2)

Multiply equation (2) by 3:
3j + 6p = $8.70 (3)

Take (3)- (1), we have:
(3j) - (3j) + (6p - 5p) = $8.70- $7.60
⇒ p= $1.10

Now, we use equation (2) to find j:
j= $2.90- 2* $1.10
⇒ j= $0.70

j+ p= $0.70 + $1.10= $1.80

The bill for one glass of orange juice and one pancake special would be $1.80.

Hope this helps~

Answer 2

Ok, so you are looking for the price of the orange juice and the pancake special individually. So, the first step is to declare our variables: 

j = price of the orange juice. 
p = price of the pancake special. 

So the first bill, translated with our variables is 
3j + 5p = 7.60 

The second bill, translated with our variables is 
1j + 2p = 2.90 

Now, it says to use the substitution method, so we need to solve one of those equations for one variable. Since in the second equation, we have one term with a coefficient of 1, that will be the easiest to solve for. So we move everything other than that term to the other side. 

1j = 2.90 - 2p 

Now we have j solved for, we will substitute it into the other equation 

3(2.90 - 2p) + 5p = 7.60 

Now we have only one variable, so we just simplify 

8.70 - 6p + 5p = 7.60 
-1p = -1.10 
p = 1.10 

Now we know how much the pancake special costs, we can plug this into either of the two original equations and solve for the price of the orange juice. 

1j + 2(1.10) = 2.90 
j = 0.70 

Now part c asks for the cost of one orange juice and one pancake special. So, with our variables, that will be: 
j + p 

We already know the values of these variables, so we can plug them in. 

0.70 + 1.10 = $1.80 

2) Linear combination is the same as the addition/elimination method. This means your goal is to add the equations together. But before we do that, we want to multiply one row by some random number so that when we add the rows together, we can get rid of one of the variables. 

a - 2b = 2 
a - 1b = 6 

So we have two choices here. We can multiply one of the a's by -1 to make it negative. That way one of them is positive, and the other is negative, and they will cancel when we add them together. OR we can multiply the bottom equation by -2. This will make the b variable in the bottom row a +2b and it will cancel out with the -2b in the top equation. 

Both will give you the right answer, so it won't matter which way you go. Since it's a smaller number, I'm going to go with the first way, and multiply the top a by -1. 

Note: You need to multiply EVERYTHING in the equation by the number! 

-1(1a - 2b = 2) 
a - b = 6 

-a + 2b = -2 
a - b = 6 

When you add the rows together, simply add like terms. Notice the a will cancel out. The +2b will add with the -b to make 1b and the -2 will add with the 6 to make 4 

So we have: 

1b = 4 

And now we can go back up and plug this info in to solve for the other variable 

a - (4) = 6 
a = 10