Achemical company makes two brands of antifreeze. the first brand is 45% pure antifreeze, and the second brand is 95% pure antifreeze. in order to obtain 90 gallons of a mixture that contains 60% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Let x = amount of 45% antifreezeLet y = amount of 70% antifreeze     EQUATION 1:   x + y = 150    (total of 150 gallons mixed)     EQUATION 2:  .45x + .75y = .55(x + y) Simplify and solve the system of equations     Multiply second equation by 100 on both sides to remove the decimals           45x + 75y = 55(x + y)     Combine like terms           45x + 75y = 55x + 55y           45x - 55x + 75y - 55y = 0             -10x + 20y = 0       Now we have the following system of equations:              x  +    y = 150        -10x + 20y =     0     Multiply the first equation by -10 to get opposite coefficients for x;  add the equations to eliminate x           10x + 10y = 1500         -10x + 20y =       0                            30y = 1500      Solve for y            30y = 1500                y = 50      Since the total mixed gallons is 150, x = 150 - 50 = 100 So we need 100 gallons of the 45% antifreeze and 50 gallons of the 70% antifreeze

Think of it as five or above give it a shove. change the 5 or above to the right to a zero and add one to the number on the left.