Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
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 stopping by. We are committed to providing the best answers for all your questions. See you again soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
