As per linear equation, 1.6 quarts of mixture should be drained and replace by pure antifreeze so that the resultant mixture will contain 60% antifreeze.
"A linear equation is an equation in which the highest power of the variable is always 1."
Given, the radiator contains 8 quarts of a mixture of water and antifreeze.
40% of the mixture is antifreeze.
Therefore, the amount of antifreeze in radiator is
[tex]= \frac{(8)(40)}{100}[/tex] quarts
[tex]= 3.2[/tex] quarts
If some amount of mixture is drained and replace by pure antifreeze, then the total amount mixture will be
[tex]= (8-x+x)[/tex] quarts
[tex]= 8[/tex] quarts
The new amount antifreeze in 8 quarts of mixture will be
[tex]= \frac{(8)(60)}{100}[/tex] quarts
[tex]= 4.8[/tex] quarts
Therefore, the amount of mixture is drained and replace by pure antifreeze is
[tex]= (4.8 - 3.2)[/tex] quarts
[tex]= 1.6[/tex] quarts
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