Question 640207
Let {{{x}}} be the number of quarts drained and replaced.
After draining {{{x}}} quarts, you have {{{20-x}}} quarts of 30% solution in the radiator. That solution is 30% antifreeze and 70% water (100%-30%=70%).
The amount of water in the radiator at that point is
{{{0.70*(20-x)}}} quarts.
After you add {{{x}}} quarts of antifreeze, you end up with {{{20}}} quarts of solution, that is supposed to be 50% antifreeze and 50% water.
The amount of water in the radiator at that point is supposed to be 50% of 20 quarts, or
{{{0.50*20}}} quarts = {{{10}}} quarts.
Since you did not add any water, all that water is exactly the same amount you had after the draining step, or
{{{0.70*(20-x)}}} quarts.
So {{{10=0.70*(20-x)}}}
{{{10=0.70*(20-x)}}} --> {{{10/0.70=0.70*(20-x)/0.70}}} --> {{{10/0.70=20-x}}}      --> {{{10/0.70+x-10/0.70=20-x+x-10/0.70}}} --> {{{x=20-10/0.70}}} --> {{{highlight(x=5.7)}}}  quarts (rounding)
NOTE: In {{{x=20-10/0.70}}} or x=20-10/0.70, according to order of operations rules (conventions) the division is done first. Most calculators know that, and will do it that way if you enter "20-10/0.7" before pressing "=".