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Algebra

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There is 1 gallon of red paint, 1 gallon of blue paint, and 1 gallon of yellow paint.
Ann used 3/8 of the red paint, 1/4th of the blue paint, and 1/2 of the yellow.
Margie used 1/2 of the red paint, 5/8 of the blue paint, and 1/8 of the yellow.
Thus, we have have the equations
1=3/8+1/2+ r (r is the left over red paint)
1=1/4+5/8+b (b is the left over blue paint )
1=1/2+1/8+y (y is the left over yellow paint).
Now, we just have to solve for r, b, and y.
Step 1: Solve for r
1=3/8+1/2+r (let's multiply the whole equation by 8 to get rid of the denominators)
8=3+4+8r (add 3+4)
8=7+8r (subtract 7 from both sides)
1=8r (divide by 8 from both sides)
r=1/8
Therefore there is 1/8 gallon of red paint left.
Step 2: Solve for b
1=1/4+5/8+b (multiply the whole equation by 8)
8=2+5+8b (add 2+5)
8=7+8b (subtract seven from both sides)
1=8b (divide by 8 from both sides)
b=1/8
There there is 1/8 gallon of red paint left.
Step 3: Solve for y
1=1/2+1/8+y (multiply by 8 to the whole equation)
8=4+1+8y (add 4+1)
8=5+8y (subtract five from both sides)
3=8y (divide by 8 from both sides)
y=3/8
Therefore there is 3/8 gallons of yellow paint left.
Problem 2: Just add what's left of the yellow and blue paints together
1/8+3/8=4/8=1/2 gallons of paint.
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