Q34. - Algebra 2

Question

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

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

Step-by-Step Solution

Answer

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

Step 1 – 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:

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

Step 2 – 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

Step 3 – 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.

Step 4 – 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.

Step 5 – 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.