alice folded a piece of paper into 12 equal squares and colored them. She had 3 times as many red parts as green parts and 2 more yellow parts than blue parts. How many parts did she have of eeach color?
There are 3 yellow parts, 1 blue parts, 2 green parts and 6 red parts.
She could have had 1 green, 3 red, 3 blue, and 5 yellow.
Or -- 2 green, 6 red, 1 blue, and 3 yellow.
go ms. sue!
do you know anything about calculus?
Sorry, Nicole -- I never took calculus! My math knowledge stops about 7th grade.
yeah mine too lol
To solve this problem, let's break it down step by step.
First, let's assume that the number of green squares is "x". Since Alice had 3 times as many red parts as green parts, the number of red squares would be 3 times "x", or 3x.
Next, we know that Alice had 2 more yellow parts than blue parts. Let's assume the number of blue squares is "y". Therefore, the number of yellow squares would be "y + 2".
Now, we can set up an equation based on the total number of squares. Since Alice folded the piece of paper into 12 equal squares in total, we have:
x + 3x + y + (y + 2) = 12
Combining like terms, we can simplify the equation to:
4x + 2y + 2 = 12
Subtracting 2 from both sides of the equation, we have:
4x + 2y = 10
To solve for x and y, we can use trial and error or substitution. Let's try substitution:
From the previous equation, we can rewrite "y" in terms of "x" as follows:
2y = 10 - 4x
Simplifying further, we get:
y = (10 - 4x) / 2
Now, we can substitute this expression for "y" back into the equation 4x + 2y = 10:
4x + 2((10 - 4x) / 2) = 10
Simplifying, we get:
4x + 10 - 4x = 10
The x terms cancel out, and we're left with:
10 = 10
This equation is true, which means that "x" can take any value. Therefore, there are infinitely many solutions to this problem, depending on the value of "x".
In summary, we cannot determine the exact number of squares Alice had for each color without additional information.