Sales for a business were 3 million dollars more the second year than the first, and sales for the third year were double the sales for the second year.

Question

If sales for the third year were 38 million dollars, what were sales, in million of dollars, for the first year?  Explain your process.

Answer 1

Let x be the first year's sales

Let x+ 3 be the second year's sales.. since they sold 3 million more in

              the second than on the first.

 

Let 2(x+3) be the third year's sales... since they are double the second

                 year's sales.

 

Using the Distributive Property, this becomes 2x+6.. 

 

Now, 2x+6= 38

              -6   -6   Subtract 6 from both sides

          2x  = 32    Divide both sides by 2

             x= 16

So the sales for the first year are 16 millions, 19 on the second, and 38 on the third.

 

I hope this helps.

Answer 2

This problem is easier if we work backwards. Let's set sales for the third year equal  to S(3) _{From the problem we know S(3) =38 (units in millions).}

 

Now let's set sales for the second year as S(2). We are told that the third year sales are twice as much as the second year sales or S(3)=2*S(2). Combining these two equations we get 38=2*S(2). Solving for S(2) we get (38/2)=(2*S(2))/2 or S(2)=19.

 

 

Finally, let's set the sales for the first year as S(1). We are told that the second year sales were 3 million more than the first year or S(2)=S(1)+3. Combining this with S(2)=19 we get 19=S(1)+3. Solving for S(1) we get 19-3=(S(1)+3)-3 or 16=S(1).

 

So the sales for the first year are 16 million.