Question

Two boxes contained 155 lb of flour. If you take 20 lb from the first and add it to the second, the first box will contain 12/19 of what is now in the second. What amount of flour was originally in each box?

Answer 1

it is 80 and 75

Explanation:

I go to RSM and when I answered this answer it said that it was correct

Answer 2

There were originally 80lb in the first box and 75lb in the second box.

The first equation describes the original weight in each box.
x+y=155
The second equation describes the weight after the 20 pounds is removed from the first box and added to the second:

`x-20=(12)/(19)(y+20)`, because 20 pounds is taken from the first box, x, and added to the second box, y; and that makes the first box equal to 12/19 of the second one.
We now have this system of equations:

`\left \{ {{x+y=155} \atop {x-20=(12)/(19)(y+20)}} \right.`
We will simplify the bottom equation first by multiplying both sides by 19 to cancel the fraction:

`19(x-20)=19((12)/(19))(y+20)`
We use the distributive property on the left, and cancel the 19 on the right:

`19*x-19*20=12(y+20) \\19x-380=12*y+12*20 \\19x-380=12y+240`
We will move the 12y to the left side of the equation by subtracting:
19x-380-12y=12y+240-12y
19x-12y-380=240
Now we will cancel the 380 by adding:
19x-12y-380+380=240+380
19x-12y=620
Now our system looks like this:

`\left \{ {{x+y=155} \atop {19x-12y=620}} \right.`
To eliminate a variable, we want the coefficients to be the same. We will multiply the top equation by 19 to achieve this:

`\left \{ {{19(x+y=155)} \atop {19x-12y=620}} \right. \\ \\ \left \{ {{19x+19y=2945} \atop {19x-12y=620}} \right.`
Now we will subtract the bottom equation to cancel x:

`\left \{ {{19x+19y=2945} \atop {-(19x-12y=620)}} \right. \\ \\19y--12y=2945-620 \\19y+12y=2325 \\31y=2325`
Divide both sides by 31:
31y/31=2325/31
y=75

Substituting this back into our original first equation:
x+75=155
Subtract both sides by 75:
x+75-75=155-75
x=80