Question 1090336
In math class, "mixture problems" are solved as a set of linear equations.
Defining variables:
{{{x}}}= volume of 10% alcohol solution to be used, in gallons
{{{y}}}= volume of 60% alcohol solution to be used, in gallons
Setting up equations:
Because Jennifer wants to make 10 gallons of solution,
{{{x+y=10}}} .
Because the final solution must contain 40% alcohol,
the alcohol amounts in the mixed solutions, must add to the alcohol amount in the final solution.
The equation stating that is
{{{0.1x+0.6y=0.4*10}}} .
{{{0.4*10}}} represents the amount of alcohol in 10 gallons of 40% solution,
{{{0.1x}}} represents the amount of alcohol in x gallons of 10% solution, and
{{{0.6y}}} represents the amount of alcohol in y gallons of 60% solution.
Those amounts are expressed in the same units, so it does not matter what those units are.
Solving the system of equations:
{{{0.1x+0.6y=0.4*10}}} would be simplified to {{{0.1x+0.6y=4}}} ,
and every term would be multiplies times {{{10}}} to get the equivalent equation
{{{x+6y=40}}}
Then you would solve the system
{{{system(x+y=10,x+6y=40)}}} 
by whichever way is easier for you.
Maybe you would choose substitution,
solving {{{x+y=10}}} for {{{x}}} to find {{{x=10-y}}} ,
and then substituting the expression {{{10-y}}} for {{{x}}}
in {{{x+6y=40}}} to find {{{y}}} first.
{{{10-y+6y=40}}}
{{{10+5y=40}}}
{{{5y=40-10}}}
{{{5y=30}}}
{{{y=30/5}}}
{{{highlight(y=6)}}} .
After that, it would be easiest to substitute the value found
for {{{y}}} in equation {{{x=10-y}}} , to find
{{{x=10-6}}}
{{{highlight(x=4)}}} .
 
NOTE:
In pharmaceutics class, or in chemical engineering class,
you would ask what was meant by alcohol.
Is it ethanol (ethyl alcohol) or isopropyl alcohol?
You would ask for clarification about the units, and figure out units for the amount of alcohol.
You would know that when you mix volumes of solutions, the final volume is not exactly the sum of the volumes, but the sum may work as a good approximation for your purposes.