SOLUTION: Your assistance is greatly appreciated.
Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an app
Question 57310This question is from textbook Elementary and Intermediate algebra
: Your assistance is greatly appreciated.
Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of 40% antifreeze solution. How many quarts of each should she use? This question is from textbook Elementary and Intermediate algebra
Found 3 solutions by aaaaaaaa, CrazyMan Jr., ptaylor:Answer by aaaaaaaa(138) (Show Source): You can put this solution on YOUR website! p = quarts of pure antifreeze sol.
i = quarts of the impure one
The second equation needs some explanation, 50% of 20 is 10.
Solve: We'll use substitution. After moving 1*i to the right, we get: , or . Substitute that
into another equation: and simplify: So, we know that i=16.6666666666667. Since , p=3.3333333333333.
0.4x+1y = 0.5 0.6y = -7.5 y = -12.5
0.4x+0.4y = 8 0.6y/0.6 = -7.5/0.6
Sub In
0.4x+1(-12.5) = 0.5
0.4x(+-)12.5 = 0.5
0.4x-12.5(+12.5) = 0.5(+12.5)
0.4x = 13
0.4x/0.4 = 13/0.4
x = 32.5
Proof - Sub In Again
0.4(32.5)+1(-12.5) = 0.5
13+-12.5 = 0.5
0.5 = 0.5
N.P.
J-Man Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website! Let x=amt of pure antifreeze that is mixed in
Then 20-x=amt of 40% antifreeze solution that is mixed in
We know that the amt of pure antifreeze that is mixed in (x)(1) plus the amt of pure antifreeze in the 40% solution that is mixed in (20-x)(.40) equals the amt of pure antifreeze in the final solution (20)(.50).
Thus, our equation to solve is:
(x)(1)+(20-x)(.40)=(20)(.50) Simplifying, we get:
x+8-.4x=10 or
.6x=2; 6x=20
x=3.3333 qts of pure antifreeze
20-x=20-3.3333=16.6667 qts of the 40% antifreeze solution
Hope this helps-----ptaylor