Question

Question If and ,$x+y=1,y+z=10$and$x+z=3$what is$x+y+z$

$\left(A\right)7\left(B\right)8\left(C\right)13\left(D\right)14$

Short answer

The option a is true, the other options are false, because the value of$x+y+z$is 7

Step-by-step solution

– Do the addition.

Add the equation$x+y=1$to the equation $y+z=10$:

$\begin{array}{l}\underset{\_}{\begin{array}{l}x+y=1\\ +y+z=10\end{array}}\\ x+2y+z=11\end{array}$

So, the resultant equation is $x+2y+z=11$.

Add the equation $x+2y+z=11$to the equation $x+z=3$

$\begin{array}{l}\underset{\_}{\begin{array}{l}x+2y+z=11\\ x+z=3\end{array}}\\ 2x+2y+2z=14\end{array}$

So, the resultant equation is $2x+2y+2z=14$.

Use the distributive property and simplify:

$\begin{array}{l}2x+2y+2z=14\\ 2(x+y+z)=14\\ x+y+z=7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividebothsidesby2}\end{array}$

– Check whether the option is correct or incorrect.

Here, in the option $\left(A\right)$it is given that, 7.

The result in step 1, one get $x+y+z=7$.

So, option $\left(A\right)$is the correct option

– Check whether the option is true or false.

Here, in the option $\left(B\right)$it is given that, 8.

The result in step 1, one get $x+y+z=7$.

So, option $\left(B\right)$is incorrect.

– Check whether the option is true or false.

Here, in the option $\left(C\right)$it is given that, 13.

The result in step 1, one get $x+y+z=7$.

So, option $\left(C\right)$is incorrect.

– Check whether the option is true or false.

Here, in the option$\left(D\right)$it is given that, 14.

The result in step 1, one get $x+y+z=7$.

So, option $\left(D\right)$is incorrect.

Hence, the option $\left(A\right)$ is true, the other options are false, because the value of $x+y+z$ is 7.