Chapter 6.3, Problem 8PPS

To determine

To solve the given system of equations using elimination method

Answer to Problem 8PPS

  $y=6$ and $z=-2$

Explanation of Solution

Given:

Equations:

  $y+z=4$

  $y-z=8$

Calculation:

Let,

  $\begin{array}{l}y+z=\mathrm{4.......}(i)\\ y-z=\mathrm{8.......}(ii)\end{array}$

Now, add equation ( i ) and equation ( ii ),

  $\begin{array}{l}\Rightarrow (y+z)+(y-z)=4+8\\ \Rightarrow y+z+y-z=12\\ \Rightarrow 2y=12\\ \Rightarrow y=6\end{array}$

Now, substitute the value of y in equation ( i ),

  $\begin{array}{l}\Rightarrow 6+z=4\\ \Rightarrow z=-2\end{array}$

Conclusion:

Therefore, the required solution for the given system of equation is $y=6$ and $z=-2$ .