##### ............................................

label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

Oct 3rd, 2015

Let's write down what we know.

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=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=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=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.

Please let me know if you need any clarification. I'm always happy to answer your questions. Leave a positive review if you are happy!
Sep 15th, 2015

...
Oct 3rd, 2015
...
Oct 3rd, 2015
Sep 22nd, 2017
check_circle