Question 1025670
The two questions are <i>almost</i> the same kind.  I will do just your #2 mixture percents question.


x,y,z for the low, medium, and high(er) concentration acids.  The quantities of them IN LITERS to use.


<i>...using 3 times as much of the 55% solution as the 35% solution.</i>
This means  {{{z/y=3}}}, and from this, {{{z=3y}}}.


Make the material sum  equation.
{{{x+y+z=75}}}, which you may use later.


Substituting for z, the material sum is also {{{x+y+3y=75}}},
{{{x+4y=75}}}.


Make the percentage relationship equation for the mixture to prepare.
{{{(25x+35y+55z)/(75)=45}}}, which you might need effort to understand. 
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Again substitute for z in this percents equation.
{{{(25x+35y+55*3y)/75=45}}}


{{{(25x+200y)/75=45}}}
and you should reduce this, since it can be reduced.


{{{(5(x+40y))/75=5*9}}}


{{{(x+40y)/75=9}}}


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Summarize your equations as a system, keeping in mind, you temporarily eliminated z.
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{{{highlight_green(system(x+4y=75,(x+40y)/75=9))}}}
This is your system to solve for x and y.  This IS a linear system; just multiply the members of the percents equation by 75.


{{{highlight_green(system(x+4y=75,x+40y=675))}}}


If you not remember how to use Elimination Method, then just use Substitution Method.
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{{{x=75-4y}}}
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{{{(75-4y)+40y=675}}}
{{{75-4y+40y=675}}}
{{{36y=600}}}
{{{y=600/36}}}
{{{y=300/18}}}
{{{y=150/9}}}
{{{highlight(y=50/3)}}}
or
{{{highlight(y=16&2/3)}}}----------the 35% acid, liters to use
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FIND x.
{{{x=75-4*(16&2/3)}}}
{{{highlight(x=8&1/3)}}}---------liters to use of the 25% acid