Question 993323
You are not given the amount of original acid mixture, so give this a variable for its volume.
v, the initial volume in gallons of the starting mixture of unknown percent concentration.


Let p be the percent concentration of your original acid mixture.
Additionally, you are forced to assume that your acid to add in pure form, the 100%, is a liquid.


ADDING 25 GALLONS
{{{(25*100+v*p)/(v+25)=80}}}


ADDING 25 GALLONS OF WATER
{{{(25*0+v*p)/(v+25)=60}}}


The system of equations slightly reworked is
{{{system(vp+2500=80(v+25),vp=60(v+25))}}}


STEPS SOLVING THE SYSTEM
{{{vp+2500=80v+8*25*10}}}
{{{vp+2500=80v+2000}}}
{{{vp+500=80v}}}
{{{vp-80v=-500}}}
{{{80v-vp=500}}}
{{{v(80-p)=500}}}
{{{highlight_green(v=500/(80-p))}}}
-
Next use the simpler, dilution equation.
{{{(500/(80-p))p=60((500/(80-p))+25)}}}
{{{500p/(80-p)=30000/(80-p)+1500}}}
{{{5p/(80-p)=300/(80-p)+15}}}
{{{p/(80-p)=60/(80-p)+3}}}
{{{p=60+3(80-p)}}}
{{{p=60+240-3p}}}
{{{4p=60+240}}}
{{{4p=300}}}
{{{p=300/4}}}
{{{highlight(highlight(p=75))}}}