# John had two bowls of peanuts. If he transfers 17 peanuts from Bowl A to Bowl B, both bowls will have the same number of peanuts. If instead, he transfers 13 peanuts from Bowl B to Bowl A, Bowl A...

John had two bowls of peanuts. If he transfers 17 peanuts from Bowl A to Bowl B, both bowls will have the same number of peanuts. If instead, he transfers 13 peanuts from Bowl B to Bowl A, Bowl A will have 6 times as many peanuts as Bowl B. How many peanuts were there in Bowl A at first?

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To solve, assign variables that represent the number of peanuts in each bowl.

Let x be the original number of peanuts in Bowl A. And let y be the original number of peanuts in Bowl B.

If 17 peanuts are transferred from Bowl A to Bowl B, the number of peanuts in each bowl now are:

# of peanuts in Bowl A = x - 17

# of peanuts in Bowl B = y + 17

And the bowls now have same amount of peanuts. So the first equation is:

`x - 17 = y + 17`

If 13 peanuts are transferred from Bowl B to Bowl A, the number of peanuts in each bowl now are:

# of peanuts in Bowl A = x + 13

# of peanuts in Bowl B = y - 13

And the number of peanuts in Bowl A now is six times as many peanuts in Bowl B. So, the second equation is:

`x+13 = 6(y-13) `

Use these two equations

EQ1: `x-17 = y+17`

EQ2: `x +13= 6(y-13)`

to determine the original number of peanuts in Bowl A.

To solve for the value of x, consider the first equation. Isolate the y.

`x - 17 = y + 17`

`x - 17 - 17 = y`

`x - 34 = y`

Then, plug-in this to the second equation.

`x+ 13= 6(y - 13)`

`x +13 =6(x - 34 - 13)`

`x + 13 = 6(x -47)`

`x + 13 = 6x -282`

Bring together the terms with x on one side of the equation. And the terms without x on the other side.

`282+13 = 6x - x`

`295=5x`

`295/5=x`

`59=x`

**Therefore, there were originally 59 peanuts in Bowl A.**