Respuesta :
so, how much acid is there in 6 gallons? well is 20% acid or (20/100), so the amount of acid in it just (20/100) * 6 or 1.2, the rest is say water.
now, if we want a 90% solution, and say we add "y" gallons, how much acid is in it? well (90/100) * y, or 0.9y.
now let's add "x" gallons of pure acid, now, pure acid is just pure acid, so is 100% acid, how much acid is there in it? (100/100) * x, or 1x or just x.
we know whatever "x" and "y" amounts are, they -> x + 6 = y
and we also know that x + 1.2 = 0.9y
[tex]\bf \begin{array}{lccclll} &\stackrel{gallons}{acid}&\stackrel{acid~\%}{quantity}&\stackrel{acid~gallons}{quantity}\\ &------&------&------\\ \textit{pure acid}&x&1.00&x\\ \textit{20\% sol'n}&6&0.20&1.2\\ ------&------&------&------\\\ mixture&y&0.90&0.9y \end{array} \\\\\\ \begin{cases} x+6=\boxed{y}\\ x+1.2=0.9y\\ ----------\\ x+1.2=0.9\left( \boxed{x+6} \right) \end{cases} \\\\\\ x+1.2=0.9x+5.4\implies x-0.9x=5.4-1.2\implies 0.1x=4.2 \\\\\\ x=\cfrac{4.2}{0.1}\implies x=\stackrel{gallons}{42}[/tex]
now, if we want a 90% solution, and say we add "y" gallons, how much acid is in it? well (90/100) * y, or 0.9y.
now let's add "x" gallons of pure acid, now, pure acid is just pure acid, so is 100% acid, how much acid is there in it? (100/100) * x, or 1x or just x.
we know whatever "x" and "y" amounts are, they -> x + 6 = y
and we also know that x + 1.2 = 0.9y
[tex]\bf \begin{array}{lccclll} &\stackrel{gallons}{acid}&\stackrel{acid~\%}{quantity}&\stackrel{acid~gallons}{quantity}\\ &------&------&------\\ \textit{pure acid}&x&1.00&x\\ \textit{20\% sol'n}&6&0.20&1.2\\ ------&------&------&------\\\ mixture&y&0.90&0.9y \end{array} \\\\\\ \begin{cases} x+6=\boxed{y}\\ x+1.2=0.9y\\ ----------\\ x+1.2=0.9\left( \boxed{x+6} \right) \end{cases} \\\\\\ x+1.2=0.9x+5.4\implies x-0.9x=5.4-1.2\implies 0.1x=4.2 \\\\\\ x=\cfrac{4.2}{0.1}\implies x=\stackrel{gallons}{42}[/tex]
