Chapter 11, Problem 3CT

To determine

To find: the equation is $\left\{\begin{array}{l}x-y+2z=5\\ 3x+4y-z=-2\\ 5x+2y+3z=8\end{array}\right\}$ .

Answer to Problem 3CT

The equation of the solutions is $\left\{\left(x,y,z\right)x=-5z+\frac{18}{7},y=z-\frac{17}{7}\right\}$

Explanation of Solution

Given information:

The given equation is $\left\{\begin{array}{l}x-y+2z=5\\ 3x+4y-z=-2\\ 5x+2y+3z=8\end{array}\right\}$

Calculation:

Calculate the expression,

Multiply both sides of the equation $x-y+2z=5$ by 4 so as to make coefficients of $y$ negatives of one another.

So, we get

  $\left\{\begin{array}{c}4x-4y+8z=20\\ 3x+4y-z=-2\end{array}\right\}$

Add these two equation in order to eliminate $y$

  $\begin{array}{c}4x-4y+8z=20\\ 3x+4y-z=-2\\ 7x+7z=18\end{array}$

So, we get equation containing two variables, $x$ and $z$

Multiply both sides $x-y+2z=5$ by 2 so as to make the coefficients of $y$ negatives of the another.

  $\begin{array}{c}2\left(x-y+2z\right)=5.2\\ 2x-25y+4z=10\end{array}$

Add the equations,

  $\begin{array}{c}5x+2y+3z=8\\ 2x-2y+4z=10\\ 7x+7z=18\end{array}$

Since, we get the same equation in both cases,

So, express the value of $x$ in terms of $z$

Subtract $7z$ from both sides of the equation.

  $\begin{array}{c}7x+7z-7z=18-7z\\ 7x=18-7z\end{array}$

Divide both sides of the equation by 7

  $x=-z+\frac{18}{7}$

Substitute $-z+\frac{18}{7}$ for $x$ in the equation $x-y+2z=5$ ,

  $\begin{array}{l}-z+\frac{18}{7}-y+2z=5\\ \frac{18}{7}+z-y=5\end{array}$

Add $-\frac{18}{7}-z$ for $x$ in the equation,

  $\begin{array}{c}\frac{18}{7}+z-y-\frac{18}{7}-z=5-\frac{18}{7}-z\\ -y=\frac{17}{7}-z\end{array}$

Multiply both sides of the equation by -1

  $\begin{array}{c}-1-y=-1\left(\frac{17}{7}-z\right)\\ y=z-\frac{17}{7}\end{array}$

Therefore, the solution in order triplet from is $\left\{\left(x,y,z\right)x=-5z+\frac{18}{7},y=z-\frac{17}{7}\right\}$ .

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