One canned juice drink is 15​% orange​ juice; another is 5​% orange juice. How many liters of each should be mixed together in order to get 10 L that is 14​% orange​ juice?

x = amount of liters of the 15% OJy = amount of liters of the 5% OJlet's recall that[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}[/tex]so then, the amount of juice in the 15% solution will be (15/100)*x, or 0.15xand the amount of juice in the 5% solution will be (5/100)*y or 0.05y.we know our mixture must be 10 liters at 14% or namely 14/100, which will give in juice (14/100)*10 or 1.4 liters or pure juice in the solution with water making the OJ.[tex]\bf \begin{array}{lcccl} &\stackrel{liters}{quantity}&\stackrel{\textit{\% of }}{juice}&\stackrel{\textit{liters of }}{juice}\\ \cline{2-4}&\\ \textit{15\% OJ}&x&0.15&0.15x\\ \textit{5\% OJ}&y&0.05&0.05y\\ \cline{2-4}&\\ mixture&10&0.14&1.4 \end{array}~\hfill \begin{cases} x+y=10\\ \boxed{y}=10-x\\ \cline{1-1} 0.15x+0.05y=1.4 \end{cases} \\\\[-0.35em] ~\dotfill[/tex][tex]\bf \stackrel{\textit{substituting on the 2nd equation}}{0.15x+0.05\left( \boxed{10-x} \right)}=1.4\implies 0.15x+0.5-0.05x=1.4 \\\\\\ 0.10x+0.5=1.4\implies 0.10x=0.9\implies x=\cfrac{0.9}{0.10}\implies \blacktriangleright x = 9 \blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{y=10-x}\implies y=10-9\implies \blacktriangleright y=1 \blacktriangleleft[/tex]