Question

how many quarts of pure antifreeze must be added to 6 quarts of a 10% antifreeze solution to obtain a 40% antifreeze solution

Answer 1

Let n be the amount of pure antifreeze to be added. Then:
0.1(6)+n=0.4(6+n)
.6+n=2.4+0.4n
0.6n=1.8
n=3 qts of pure antifreeze needed.
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Answer 2

First off, let's change the percentage amounts to decimal format, thus, a pure antifreeze is 100% antifreeze, so it has a 100% concentration, thus if we were to add an amount of "x", then the amount on antifreeze in that will then be (100/100) * x, or 1x, or just "x".

the mixture will be say, "y" amount of 40% antifreeze, thus, the concentrated amount in it will be (40/100) * y or 0.4y.

thus

$\bf \begin{array}{lccclll} &amount(quarts)&concentration& \begin{array}{llll} concentrated\\ amount \end{array}\\ &------&------&------\\ \textit{pure antifreeze}&x&1.00&1x\\ \textit{10\% solution}&6&0.10&0.6\\ ------&------&------&------\\ mixture&y&0.40&0.4y \end{array}$

now, bear in mind that, whatever "x" is,
x + 6 = y    <--- the amounts added must yield the amount of the mixture

and x + 0.6 = 0.4y <--- the contrated amounts sum will also add up to the mixture's

$\bf \begin{cases} x+6=\boxed{y}\\ x+0.6=0.4y\\ ----------\\ x+0.6=0.4\left( \boxed{x+6} \right) \end{cases} \\\\\\ x+0.6=0.4x+2.4\implies 0.6x=1.8\implies x=\cfrac{1.8}{0.6}$