a chemical company makes two brands of antifreeze. the first brand is 35% pure antifreeze and the second is 85% pure antifreeze. in order to obtain 70 gallons of a mixture that contains 80% 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

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Step-by-step explanation:

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.

answer: b.36 feet

step-by-step explanation: