Question 1007274
<pre>
Let the number of kgs. of Alloy1 be x
Let the number of kgs. of Alloy2 be y


                                 Number
Type       Number   Percent      of kg
 of        of kg    which is    of copper
coin      of metal   copper     in each
-------------------------------------------
Alloy1       x       0.30        0.30x
Alloy2       y       0.60        0.60y
-------------------------------------------
mixture     80      0.48        0.48(80)

 The first equation comes from the second column.

  {{{(matrix(3,1,Kilograms,of,Alloy1))}}}{{{""+""}}}{{{(matrix(3,1,kilograms,of,Alloy2))}}}{{{""=""}}}{{{(matrix(4,1,total,kilograms,of,alloy))}}}

                 x + y = 80

 The second equation comes from the last column.
  {{{(matrix(6,1,Kilograms,of,PURE,COPPER,in,Alloy1))}}}{{{""+""}}}{{{(matrix(6,1,Kilograms,of,PURE,COPPER,in,Alloy2))}}}{{{""=""}}}{{{(matrix(6,1,Kilograms,of,PURE,COPPER,in,mixture))}}}

           0.3x + 0.6y = 0.48(80)

           0.3x + 0.6y = 38.4


Get rid of decimals by multiplying every term by 10:

               3x + 6y = 384

 So we have the system of equations:
           {{{system(x + y = 80,3x + 6y = 384)}}}.

We solve by substitution.  Solve the first equation for y:

           x + y = 80
               y = 80 - x

Substitute (80 - x) for y in 30x + 60y = 3840

      3x + 6(80 - x) = 384
       3x + 480 - 6x = 384
           -3x + 480 = 384
                 -3x = -96
                   x = 32 = the number of kg. of Alloy1.

 Substitute in y = 80 - x
               y = 80 - 32 = 48 kg. of Alloy2. 
Edwin</pre>