Chapter 2.4A, Problem 5MCQ

To determine

To find: solutions for the system of equations.

Answer to Problem 5MCQ

  $x=-\frac{1}{3},y=\frac{4}{3}\text{ and }z=8$

Explanation of Solution

Given information:

The system of equations is as follows:

  $x+y=1$....................(1)

  $2x-y=-2$....................(2)

  $4x+y+z=8$....................(3)

Calculation:

Use elimination method in first two equations to eliminate y as follows:

  $\begin{array}{l}x+y=1\\ 2x-y=-2\\ \overline{\text{ 3}x=-1}\\ \therefore x=\frac{-1}{3}\end{array}$

Now put the value of x in equation (1) and evaluate as follows:

  $\begin{array}{c}-\frac{1}{3}+y=1\\ y=1-\left(-\frac{1}{3}\right)\\ =1+\frac{1}{3}\\ =\frac{4}{3}\end{array}$

Put the value of x and y in equation (3) and solve as follows:

  $\begin{array}{c}4\left(-\frac{1}{3}\right)+y+\frac{4}{3}=8\\ -\frac{4}{3}+y+\frac{4}{3}=8\\ y+0=8\\ y=8\end{array}$

Thus, the solution $\left(x,y,z\right)$ for the given system is $\left(-\frac{1}{3},\frac{4}{3},8\right)$ .