Four times the sum of the number of reds and blue exceeded 3 times the number of whites by 3. Five times the sum of the number of blues and whites exceeded 8 times the number of reds by 13. If there were 5 more whites than blues, how many of each color were there?
just put the clues into symbols:
4(r+b) = 3w+3
5(b+w) = 8r+13
w = b+5
now substitute for w in the 1st two equations:
4(r+b) = 3(b+5)+3
5(b+b+5) = 8r+13
or,
4r + b = 18
8r - 10b = 12
8r - 10(18-4r) = 12
48r = 192
r = 4
so, b=2 and w=7
To solve this problem, we need to set up a system of equations based on the given information.
Let's assign variables to each of the colors:
Let's say the number of reds is R.
The number of blues is B.
The number of whites is W.
Now, let's translate the given information into equations.
1) "Four times the sum of the number of reds and blues exceeded 3 times the number of whites by 3."
This can be written as:
4(R + B) = 3W + 3
Simplifying this equation, we get:
4R + 4B = 3W + 3
2) "Five times the sum of the number of blues and whites exceeded 8 times the number of reds by 13."
This can be written as:
5(B + W) = 8R + 13
Simplifying this equation, we get:
5B + 5W = 8R + 13
3) "If there were 5 more whites than blues."
This can be written as:
W = B + 5
Now we have a system of three equations with three variables. We can solve this system to find the values of R, B, and W.
Let's substitute the value of W from equation 3 into equations 1 and 2.
Substituting W = B + 5 into equation 1:
4R + 4B = 3(B + 5) + 3
4R + 4B = 3B + 15 + 3
4R + 4B = 3B + 18
Substituting W = B + 5 into equation 2:
5B + 5(B + 5) = 8R + 13
5B + 5B + 25 = 8R + 13
10B + 25 = 8R + 13
Now we have two equations in terms of R and B:
4R + 4B = 3B + 18 ----(4)
10B + 25 = 8R + 13 ----(5)
We can solve this system of equations by substitution or elimination method.
Using the elimination method, let's multiply equation 4 by 2:
8R + 8B = 6B + 36 ----(6)
Now, let's subtract equation 6 from equation 5:
(8R + 8B) - (8R + 6B) = (6B + 36) - (8R + 13)
8R - 8R + 8B - 6B = 6B + 36 - 8R - 13
2B = -8R + 23
Now, let's rearrange the equation:
8R - 2B = 23 ----(7)
Now we have two equations:
2B = -8R + 23 ----(7)
10B + 25 = 8R + 13 ----(5)
We can solve this system of equations to find the values of R and B.
Now, let's find the value of R by multiplying equation 7 by 5 and adding it to equation 5:
(8R - 2B) * 5 + (10B + 25) = (23) * 5 + (13)
40R - 10B + 10B + 25 = 115 + 13
40R = 128
R = 128/40
R = 16/5
R = 3.2
Since we are looking for whole numbers, we know that the number of reds cannot be 3.2. Hence, there is no solution for this set of equations.