by Guest » Tue Oct 09, 2018 3:06 am
[tex]\begin{array}{|l}a+c=s\\m(a+c)=n(a-c)\end{array}[/tex]
[tex]\begin{array}{|l}c=s-a\\ms=n(a-(s-a))=n(2a-s)\end{array}[/tex]
[tex]2a-s=\frac{ms}{n}[/tex]
[tex]2a=s+\frac{ms}{n}=\frac{m+n}{n}\cdot s[/tex]
[tex]a=\frac{m+n}{2n}\cdot s[/tex]
[tex]c=s-a=\left(1-\frac{m+n}{2n}\right)\cdot s=\frac{n-m}{2n}\cdot s[/tex]
Verification:
[tex]a+c=\left(\frac{m+n}{2n}+\frac{n-m}{2n}\right)\cdot s=\frac{2n}{2n}\cdot s=s[/tex]
[tex]m(a+c)-n(a-c)=m.s-\cancel{n}\cdot\frac{\cancel{2}m}{\cancel{2}\cancel{n}}\cdot s=m.s-m.s=0[/tex]
[tex]\Rightarrow m(a+c)=n(a-c)[/tex]
Answer: [tex]a=\frac{m+n}{2n}\cdot s;\ c=\frac{n-m}{2n}\cdot s[/tex]