Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
the first solution is 20% acid, and say we'll be using "x" liters, so how many liters of just acid are in it? well 20% of "x" or namely 0.2x. Likewise for the 60% acid solution, if we had "y" liters of it, the amount of only acid in it is 0.6y.
[tex]\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&x&0.20&0.2x\\ \textit{2nd solution}&y&0.60&0.6y\\ \cline{2-4}&\\ mixture&100&0.44&44 \end{array}~\hfill \begin{cases} x+y=100\\\\ 0.2x+0.6y=44 \end{cases}[/tex]
[tex]x+y=100\implies y=100-x~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.2x+0.6(100-x)=44} \\\\\\ 0.2x+60-0.6x=44\implies -0.4x+60=44\implies -0.4x=-16 \\\\\\ x=\cfrac{-16}{-0.4}\implies \boxed{x=40}~\hfill \boxed{\stackrel{100-40}{y=60}}[/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
