A20% solution of fertilizer is to be mixed with a 80% solution of fertilizer in order to get 150 gallons of a 60% solution. how many gallons of the 20% solution and 80% solution should be mixed?

Let n be the amount of 80% solution; and 150-n be the amount of 20% solution. Then:
.8n+.2(150-n)=.6(150)
.8n+30-.2n=90
.6n=60
n=100
150-n=50
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Hello,

Let x represent the number of gallons of 20% solution of fertilizer to be used be.

Let y represent the number of gallons of 80% solution of fertilizer to be used be.

You want to find out how many gallons of the 20% solution and 80% solution should be mixed. So, as the solution is being mixed, the final volume (gallons) adds up.

First to write an equation, change the percent to decimal form:

20%=.20
80%=.80

We now have:

x gallons of a 20% solution contains .20x gallons of fertilizer.
y gallons of an 80% solution contains .80y gallons of fertilizer.

To find how many gallons total will result after both solutions being mixed, write:

.20x+.80y = gallons of fertilizer 

Amount of fertilizer in 150 gallons of 60% solution of fertilizer:

.60*150=90 gallons of fertilizer.

Thus, .20+.80=90 is your second equation

Now, to find how many gallons of the 20 percent solution and 80 percent solution should be mixed in order to get 150 gals solve for your system of equations:

 x+y=150
.20x+.80y=90

Use the Substitution Method (if you want):

x=50

Substitute x=50 into the first equation:

50+y=150

Subtract 50 from both sides of the equation; 50 cancels out.

We now have y=100

Thus, when x=50 y=100

Therefore, 100 gallons of the 20% solution and 80% solution should be mixed in order to get 150 gallons of a 60% solution.

Faith xoxo

Just a second zzz