susan is buying three different colors of tiles for her kicthen floor. she is buying 25 more red tiles than beige tiles , and three times as many navy-blue tiles as beige titles. if susan buys 435 tiles altogher, how many tiles of each color does she buy?
R=B+25
NB=3B
R+B+NB=435
you can solve this a number of ways. Post your work, and I will check.
bobpursley, I am confused to the equations that you have just posted. However, this is what I figured out but I'm not if it's correct.
((25+R)-B)+3N=435
To solve this problem, let's assign variables to represent the number of beige, red, and navy-blue tiles.
Let's say:
- Beige tiles = B
- Red tiles = R
- Navy-blue tiles = N
According to the information given, there are three conditions we can form equations from:
1. Susan is buying 25 more red tiles than beige tiles, which can be written as: R = B + 25.
2. Susan is buying three times as many navy-blue tiles as beige tiles, which can be written as: N = 3B.
3. Susan is buying a total of 435 tiles, which can be written as: B + R + N = 435.
Now, we have a system of three equations that we can solve simultaneously.
Using equation 1 and equation 3, substitute the value of R from equation 1 into equation 3:
B + (B + 25) + N = 435
2B + N + 25 = 435
Using equation 2, substitute the value of N from equation 2 into the simplified equation:
2B + 3B + 25 = 435
5B + 25 = 435
Next, isolate the variable B:
5B = 435 - 25
5B = 410
B = 410 / 5
B = 82
Now that we know the value of B, we can substitute it back into the other equations to find the values of R and N.
Using equation 1, substitute the value of B:
R = B + 25
R = 82 + 25
R = 107
Using equation 2, substitute the value of B:
N = 3B
N = 3(82)
N = 246
Therefore, Susan buys 82 beige tiles, 107 red tiles, and 246 navy-blue tiles for her kitchen floor.