Respuesta :
We will use the variable r to represent the number of scones sold to Robbie. We will use the variable x to represent the increase in scones sold to each person (since the increase is the same, we can use the same variable each time).
Scones sold to Cameron = r + x.
Scones sold to Louis = the scones sold to Cameron + x = r + x + x = r + 2x.
Scones sold to Tom = scones sold to Louis + x = r + 2x + x = r + 3x.
Scones sold to Charlie = scones sold to Tom + x = r + 3x + x = r + 4x.
We add all of these together to get our total of 60:
r + (r + x) + (r + 2x) + (r + 3x) + (r + 4x) = 60
Combine our like terms and we have 5r + 10x = 60 (there are 4 r's and 10 total x's).
We have another equation involved here. We know that Robbie + Cameron, or r + r + x = 2r + x, is 3/7 of the total of the other three boys (this total given by r + 2x + r + 3x + r + 4x = 3r + 9x). We can write this as:
2r + x = 3/7(3r + 9x)
We can cancel the 7. Fraction bars are the same as division, so we cancel by multiplying: 7(2r + x) = 3(3r + 9x), which simplifies to 14r + 7x = 9r + 27x after using the distributive property.
In order to work with a system of equations (two equations at the same time) it's easier if they look the same. Let's get r and x on one side of the equation. We will start by cancelling 9r. Since it's positive, we can cancel or move it by subtracting: 14r + 7x - 9r = 9r - 9r + 27x, which gives us 5r + 7x = 27x. Now we want to move the 27x. Since it's also positive we will subtract it:
5r + 7x - 27x = 27x - 27x, which gives us 5r -20x = 0.
Let's align them vertically:
[tex]5r+10x=60 \\ 5r-20x=0[/tex]
The coefficients of our r's are the same, so we will eliminate those by subtracting:
[tex] \left \{ {{5r+10x=60} \atop {-(5r-20x=0)}} \right. [/tex]
Now that our r's are gone, we have 30x (10x - -20x = 30x) and 60 on the other side:
30x = 60
Divide both sides by 30:
[tex] \frac{30x}{30}= \frac{60}{30} \\ x=2[/tex]
Plug this into our first equation, 5r + 10x = 60:
[tex]5r+10(2)=60 \\ 5r+20=60[/tex]
Subtract 20 from both sides:
[tex]5r+20-20=60-20 \\ 5r=40[/tex]
Divide both sides by 5:
[tex] \frac{5r}{5}= \frac{40}{5} \\r=8 [/tex]
This tells us that r, the scones sold to Robbie, were 8 and x, the change between the number of scones sold to each person, was 2. That means she sold 8 to Robbie, 10 to Cameron, 12 to Louis, 14 to Tom and 16 to Charlie.
Scones sold to Cameron = r + x.
Scones sold to Louis = the scones sold to Cameron + x = r + x + x = r + 2x.
Scones sold to Tom = scones sold to Louis + x = r + 2x + x = r + 3x.
Scones sold to Charlie = scones sold to Tom + x = r + 3x + x = r + 4x.
We add all of these together to get our total of 60:
r + (r + x) + (r + 2x) + (r + 3x) + (r + 4x) = 60
Combine our like terms and we have 5r + 10x = 60 (there are 4 r's and 10 total x's).
We have another equation involved here. We know that Robbie + Cameron, or r + r + x = 2r + x, is 3/7 of the total of the other three boys (this total given by r + 2x + r + 3x + r + 4x = 3r + 9x). We can write this as:
2r + x = 3/7(3r + 9x)
We can cancel the 7. Fraction bars are the same as division, so we cancel by multiplying: 7(2r + x) = 3(3r + 9x), which simplifies to 14r + 7x = 9r + 27x after using the distributive property.
In order to work with a system of equations (two equations at the same time) it's easier if they look the same. Let's get r and x on one side of the equation. We will start by cancelling 9r. Since it's positive, we can cancel or move it by subtracting: 14r + 7x - 9r = 9r - 9r + 27x, which gives us 5r + 7x = 27x. Now we want to move the 27x. Since it's also positive we will subtract it:
5r + 7x - 27x = 27x - 27x, which gives us 5r -20x = 0.
Let's align them vertically:
[tex]5r+10x=60 \\ 5r-20x=0[/tex]
The coefficients of our r's are the same, so we will eliminate those by subtracting:
[tex] \left \{ {{5r+10x=60} \atop {-(5r-20x=0)}} \right. [/tex]
Now that our r's are gone, we have 30x (10x - -20x = 30x) and 60 on the other side:
30x = 60
Divide both sides by 30:
[tex] \frac{30x}{30}= \frac{60}{30} \\ x=2[/tex]
Plug this into our first equation, 5r + 10x = 60:
[tex]5r+10(2)=60 \\ 5r+20=60[/tex]
Subtract 20 from both sides:
[tex]5r+20-20=60-20 \\ 5r=40[/tex]
Divide both sides by 5:
[tex] \frac{5r}{5}= \frac{40}{5} \\r=8 [/tex]
This tells us that r, the scones sold to Robbie, were 8 and x, the change between the number of scones sold to each person, was 2. That means she sold 8 to Robbie, 10 to Cameron, 12 to Louis, 14 to Tom and 16 to Charlie.
