সমাধান কর: $\frac{3}{2z+1}+\frac{4}{5z-1}=2$

Question

সমাধান কর: $\dfrac{3}{2z+1}+\dfrac{4}{5z-1}=2$

Answer 1

$\dfrac{3}{2z+1}+\dfrac{4}{5z-1}=2$

বা, $\dfrac{3}{2z+1}+\dfrac{4}{5z-1}=1+1$

বা, $\dfrac{3}{2z+1}-1=1-\dfrac{4}{5z-1}$

বা, $\dfrac{3-1\left(2z+1\right)}{2z+1}=\dfrac{1\left(5z-1\right)-4}{5z-1}$

বা, $\dfrac{3-2z-1}{2z+1}=\dfrac{5z-1-4}{5z-1}$

বা, $\dfrac{2-2z}{2z+1}=\dfrac{5z-5}{5z-1}$

বা, $\dfrac{-2\left(z-1\right)}{2z+1}=\dfrac{5\left(z-1\right)}{5z-1}$

বা, $5\left(z-1\right)\left(2z+1\right)=-2\left(z-1\right)\left(5z-1\right)$

বা, $5\left(z-1\right)\left(2z+1\right)+2\left(z-1\right)\left(5z-1\right)=0$

বা, $\left(z-1\right)\left\lbrace5\left(2z+1\right)+2\left(5z-1\right)\right\rbrace=0$

বা, $\left(z-1\right)\left(10z+5+10z-2\right)=0$

বা, $\left(z-1\right)\left(20z+3\right)=0$

হয়,
$\left(z-1\right)=0$

$\therefore z=1$

অথবা,
$\left(20z+3\right)=0$

বা, $20z=-3$

$\therefore z=-\frac{3}{20}$

সুতরাং নির্ণেয় সমাধান $z=1$ অথবা $-\frac{3}{20}$ [Answer]