(12,5)
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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
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
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
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