Please help me with this Math Problem?

let the amount of 90% antifreeze solution required = x gallons

The amount of antifreeze in 90 % solution = 0.9 x

The amount of 15% antifreeze in 100 gallons = 100*0.15 = 15

Gallons of resultant 80 % mixture = 100 + x gallons

The amount of antifreeze in (100 + x) gallons = (100 + x)*0.8

balance amount of antifreeze

=> (100 + x)0.8 = 0.9x + 15

=> 80 + 0.8x = 0.9x + 15

=> 0,1 x = 65

x = 65/0.1 = 650 gallons

For these problems it's good to organize what you know

tables work best, but unfortunately it is difficult to get them to display correctly here

For the current antifreeze solution (15%), in gallons, you have:

antifreeze: 15

totalamnt: 100

You are adding 90% antifreeze, but we don't know how much, so let's assign the variable x to represent how much of it to add. Since it is 90% antifreeze, the amount of antifreeze would be 0.9*x:

antifreeze: 0.9*x

totalamnt: x

To represent the total amount you will have after mixing, you add the total amounts:

100 + x

To represent the total amount of antifreeze, add the antifreeze amounts in the same way:

15 + 0.9*x

So basically, if you want to get a mixture that is 80% antifreeze, this means that you want the amount of antifreeze divided by the total amount to be 0.8, meaning:

(15 + 0.9*x)/(100 + x) = 0.8

At this point solve the equation:

15 + 0.9x = 80 + 0.8x

0.1x = 65

x = 650

So you would have to add 650 gallons of 90% antifreeze solution