Question

One positive integer is 8 more than twice another. If their product is 384, find the numbers.

Answer 1

Answer:  The numbers are 32 and 12

 

Step-by-step explanation:

Let   say those two positve integers are x and y.

x has to be 8 more than twice y  , and that could be represent by the equation  

   x = 2y + 8  

Their product is 384 meaning that  x times y has to equal to 384 , and that can also be represented by the equation  xy = 384

x = 2y + 8

xy = 384  

Now using both equations, substitute the value for x into the second equation and solve for y.

(2y + 8)(y ) = 384  

  2y^2 + 8y = 384     Subtract 348  from both sides  

          -384      -384

2y^2 + 8y - 384 =  0   Factor  the left sides by the GCF

2(y^2 + 4y - 192)  = 0        

 

    x = -b ± $\sqrt{b^2 -4ac } /2$      

The variables  a= 1   , b=4  and c is -192

   

x = -4 ± $\sqrt{784} /2$

x = -4 + 14     or x = -4 - 14

 x = 12    or x = -16

Since we are dealing with length   the value of y has to be  12.  

Now that we know that  y is 12 input 12 into one of the equations and solve for x.

x = 2(12) + 8

x = 24 + 8

x = 32