Diluting the solution

If there is solution with the concentration \(x=\frac b{b+a}\) and we need to get the solution with concentration \(y=\frac b{b+c}\) then from \(x\left({b+a}\right)=b=y\left({b+c}\right)\) we get the ratio of masses \(\displaystyle{\frac xy=\frac{b+c}{b+a}=\frac{m_{FINAL}}{m_{INITIAL}}}\) . . . so - if the \(\displaystyle{m_{FINAL}}\) is given we need the \(\displaystyle{m_{INITIAL}=\frac yx·m_{FINAL}}\) of the initial solution , the amount of water needed to be added is \(c-a=\mathbf{d}\) , from \(\displaystyle{\frac xy=\frac{m_F}{m_I}=1+\frac d{m_I}=\frac1{1-\frac d{m_F}}}\) we see that \(\displaystyle{d=\left({\frac xy-1}\right)·m_I=\left({1-\frac yx}\right)·m_F}\) . . . checking
\(\displaystyle{d=\left({\frac xy-1}\right)·m_I=\left({\frac xy-1}\right)·\frac yx·m_F}\) and
\(\displaystyle{d=\left({1-\frac yx}\right)·m_F=\left({1-\frac yx}\right)·\frac xy·m_I}\)

---

PS! -- the below DEMO works and "in reverse" for the initial percentage being less than the final ... the result is given as a negative amount of the water to be "added" --e.g.-- describing the amount of water to be (evaporated/)extracted from the solution ... Now updated! to match the possible concentrating option ...

---

▼ ▼ ▼
► ► ► ◄ ◄ ◄
▲ ▲ ▲

---

. . . a Q.C. (in MSO Excel) for some non-intuitive cases ::

---