x = amount of liters of the 15% OJ
y = amount of liters of the 5% OJ
let's recall that

so then, the amount of juice in the 15% solution will be (15/100)*x, or 0.15x
and 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.
![\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](https://img.qammunity.org/2020/formulas/mathematics/high-school/metdsi3akipy22j6n5j4b3kcm2tngpscn0.png)
