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
$\displaystyle{d=\left({1-\frac yx}\right)·m_F=\left({1-\frac yx}\right)·\frac xy·m_I}$

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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 ...

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

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