How to solve linear system with three lines

Question

On Monday,  Julia's buys 2kg of apples and 4kg of bananas for $7.30. On Tuesday,  she buys 1kg of apples and 4kg of carrots for $18.25. On Wednesday,  she buys 1 kg of apples,  3kg of dates and 1 kg of carrots for $51.25. How much will it cost to by 1kg of each of the four items? 

Answer 1

Hi Andrew,

 

Let's start off by writing down equations describing how much money Julia spent on fruits each day:

 

     2a + 4b = $7.30 (on Monday)

     1a + 4c = $18.25 (on Tuesday)

     1a + 1c + 3d = $51.25 (on Wednesday)

 

We've defined a, b, c, and d to be the cost of 1kg of apples, bananas, carrots, and dates, respectively.

 

The problem asks us to find out how much our total would be if we wanted to buy 1kg of each of the four different types of fruit. In a mathematical statement, the problem wants us to find:

 

     1a + 1b + 1c + 1d = ???

 

We can solve this problem using matrices, but it turns out that matrices are really hard to write out online. Instead, let's just add multiples of the equations together. We'll call the Monday equation M, the Tuesday equation T, and the Wednesday equation W.

 

First, let's note that of the three equations, only M contains any reference to b, while only W contains any reference to d. Let's start by dividing W by 3 so that we end up with 1d in our result:

 

     (1/3)a + (1/3)c + 1d = $17.083333

 

Now, let's divide M by 4 so that we end up with 1b in our result:

 

     (1/2)a + 1b = $1.825

 

Let's add these two equations together to get:

 

     (5/6)a + 1b + (1/3)c + 1d = $18.908333

 

Great! We're halfway there. Now we just need to figure out how to get the coefficients for the a and c terms to equal 1. As it turns out, if we divide equation T by 6:

 

     (1/6)a + (2/3)c = $3.041666

 

and add this to our previous equation, we end up with:

 

     1a + 1b + 1c + 1d = $21.95.

 

Looks like Julia needs to come up with about $22 to buy 1kg of each fruit.