Achemical company makes two brands of antifreeze. the first brand is 65% pure antifreeze, and the second brand is 90% pure antifreeze. in order to obtain 50 gallons of a mixture that contains 70% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Let x = amount of 65% antifreeze
Let y = amount of 90% antifreeze    
EQUATION 1:   x + y = 50    (total of 50 gallons mixed)    
EQUATION 2:  .65x + .90y = .70(x + y) 
Simplify and solve the system of equations    
Multiply second equation by 100 on both sides to remove the decimals        
  65x + 90y = 70(x + y)    
Combine like terms        
  65x + 90y = 70x + 70y      
     65x -70x+ 90y-70y = 0    
         -5x + 20y = 0      
Now we have the following system of equations:      
  x  +    y = 50       
 -5x + 20y = 0    
 Multiply the first equation by 5 to get opposite coefficients for x;  
add the equations to eliminate x        
   5x + 5y = 250       
  -5x + 20y = 0     
         
          25y = 250    
 Solve for y            
y = 10               
Since the total mixed gallons is 50,
x = 50 - 10 = 40 
So we need 40 gallons of the 65% antifreeze and 10 gallons of the 90% antifreeze