Question

$solve\:for\:a,\:a+b+z+x=3a$
solve it...

Answer 1

$solve\:for\:a,\:a+b+z+x=3a\quad :\quad a=-\frac{-b-z-x}{2}$

$\mathrm{Subtract\:}b+z+x\mathrm{\:from\:both\:sides}$

$a+b+z+x-\left(b+z+x\right)=3a-\left(b+z+x\right)$

$a=3a-b-z-x$

$\mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}$

$a-3a=3a-b-z-x-3a$

$-2a=-b-z-x$

$\frac{-2a}{-2}=-\frac{b}{-2}-\frac{z}{-2}-\frac{x}{-2}$

$a=-\frac{-b-z-x}{2}$

Answer 2

$solvefora,a+b+z+x=3a:a=−2−b−z−x</p><p></p><p>\mathrm{Subtract\:}b+z+x\mathrm{\:from\:both\:sides}Subtractb+z+xfrombothsides</p><p></p><p>a+b+z+x-\left(b+z+x\right)=3a-\left(b+z+x\right)a+b+z+x−(b+z+x)=3a−(b+z+x)</p><p></p><p>a=3a-b-z-xa=3a−b−z−x</p><p></p><p>\mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}Subtract3afrombothsides</p><p></p><p>a-3a=3a-b-z-x-3aa−3a=3a−b−z−x−3a</p><p></p><p>-2a=-b-z-x−2a=−b−z−x</p><p></p><p>\frac{-2a}{-2}=-\frac{b}{-2}-\frac{z}{-2}-\frac{x}{-2}−2−2a=−−2b−−2z−−2x</p><p></p><p>a=-\frac{-b-z-x}{2}a=−2−b−z−x</p><p></p><p>$