i really need help on this question!! I need to solve it using a system of two equations, the question says that an automobile radiator contains 16L of antifreeze and water. this mixture is 30%antifreeze. how much of this mixture should be drained and replaced with pure antifreeze so that there will be 50% antifreeze?
Yes, you had best do this. It is January!
the initial condition has
.3 * 16 = 4.8 L antif
.7 * 16 = 11.2 L H2O
we take away x liters of mix
that is we take away .3 x L antif
and take way .7 x L H2O
then we add the same x liters antif
the result must be 8 L of antif
and 8 L of H2O
so
4.8 L antif - .3 x L antif + x L antif = 8 L antif
4.8 - .3 x + x = 8
.7 x = 3.2
x = 4.57
Now check that carefully. I am sleepy.