Question

verify that (x+y)+z=x+(y+z), if x=4/5 ,y= -3/6, z=-2/15

Answer 1

$\Huge\mathrm{Question:}$

Verify that (x+y)+z=x+(y+z) ,if x = 4/5 ,y= -3/6 , z = -2/15 .

$\Huge\mathrm{Answer:}$

$\Large\bf{Left-hand-side:}$

$\sf{(x + y) + z}$

$\sf{ (\frac{4}{5} + - \frac{3}{6}) + \frac{ - 2}{15} }$

$\sf{( \frac{4}{5} - \frac{1}{2}) + \frac{ - 2}{15} }$

$\sf{( \frac{8 - 5}{10}) + \frac{ -2 }{15} }$

$\sf{ \frac{3}{10} - \frac{2}{15} }$

$\sf{ \frac{9 - 4}{30} }$

$\sf{ \cancel\frac{5}{30} }$

$\sf{ \frac{1}{6} }$

$\Large\bf{Right-hand-side:}$

$\sf{x + (y + z)}$

$\sf{ \frac{4}{5} + ( \frac{ - 3}{6} + \frac{ - 2}{15} ) }$

$\sf{ \frac{4}{5} + ( \frac{ - 15 - 4}{30} )}$

$\sf{ \frac{4}{5} + ( \frac{ - 19}{30} )}$

$\sf{ \frac{4}{5} - \frac{19}{30} }$

$\sf{ \frac{ 24 - 19}{ 30} }$

$\sf{ \cancel\frac{5}{30} }$

$\sf{ \frac{1}{6} }$

$\large\bf{Thus,}$

$\Large{\boxed{\bf{ \frac{1}{6} = \frac{1}{6} }}}$

$\Large{\boxed{\bf{LHS=RHS}}}$

$\Large{\boxed{\bf{Hence, \: Verified. }}}$

Answer 2

Answer:

Form a quadratic equation whose Roots are $x,y,z$

Now, let the equation be $ax^3 + bx^2 + cx + d= 0$

So, Sum$x+y+z = -b/a$

$xy + yz + xz = c/a$

$xyz = -d/a$

If $a = 1$, equation will be $x^3 -6x^2+15x-14=0$

So,$x=2$ is a root of equation : the equation can be written as $(x-2)(x^2 - 4x + 7) =0$

Since $x^2 - 4x + 7=0$ has imaginary roots which are $2+i sqrt{3}$ and $2-i sqrt{3}$

So, the only real solution for this equation is $x = 2$

Hope this was helpful.