Chapter 4.3, Problem 51E

To determine

To find: the solution of the system of the equation.

Answer to Problem 51E

The solution of the system is $x=-\frac{91}{11},y=\frac{68}{11}\text{ and }z=\frac{98}{11}$ .

Explanation of Solution

Given:

  $\begin{array}{l}4x+2y+3z=6.\\ 2x+7y=3z.\\ -3x-9y+13=-2z.\end{array}$

Concept used:

In the elimination method or substitution method:

Either add or subtract equation to get an equation in one variable.

When the coefficients in one variable are opposites then add the equations to eliminate the variable and when the coefficients of one variable are equal and subtract the equation to eliminate a variable.

Calculation:

Consider the system of equations:

  $\begin{array}{l}4x+2y+3z=6.\\ 2x+7y=3z.\\ -3x-9y+13=-2z.\end{array}$

Writing the system of equations in standard form that obtain:

  $\begin{array}{l}4x+2y+3z=6.\\ 2x+7y-3z=0.\\ -3x-9y+2z=-13.\end{array}$

One way to solve a system of three equations is to choose pairs of equations and then eliminate one of the variables. Because, the coefficient of $z\text{ is }3$ in the first equation, and the coefficient of $z=-3$ , it can eliminate $z$ to get a equation in variable $x\text{ and }y$ .

To eliminate $z$ using the first and second equations, add first and second equations.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}4x+2y+3z=6.\\ 2x+7y-3z=0.\end{array}}\\ 6x+3y=6.\\ \text{dividing by }3:\\ 2x+3y=2.\end{array}$

To eliminate $z$ using the second and third equations multiply each side of the second equation by $2$ :

  $\begin{array}{l}2\left(2x+7y-3z\right)=2\left(0\right).\\ 4x+14y-6z=0.\end{array}$

Now multiply each side of third equation by $3$ :

  $\begin{array}{l}3\left(-3x-9y+2z\right)=3\left(-13\right).\\ -9x-27y+6z=-39.\end{array}$

Then add both the results:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}4x+14y-6z=0\\ -9x-27y+6z=-39\end{array}}\\ -5x+13y=39\end{array}$

Multiplying with $-1$ :

  $5x+13y=39$ .

Now it has two linear equations in two variables.

Multiplying first equation by $5$ :

  $\begin{array}{l}5\left(2x+3y\right)=5\left(2\right).\\ \Rightarrow 10x+15y=10.\end{array}$

Multiply second equation by $-2$ :

  $\begin{array}{l}-2\left(5x+13y\right)=-2\left(39\right).\\ -10x-26y=-78.\end{array}$

Adding both the equations it is obtain as:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}10x+15y=10.\\ -10x-26y=-78.\end{array}}\\ -11y=-68.\\ y=\frac{68}{11}.\end{array}$

Now substituting the value of $y=\frac{68}{11}$ in $2x+3y=2$ it is obtained as:

  $\begin{array}{l}2x+3\left(\frac{68}{11}\right)=2.\\ 22x=-182.\\ x=-\frac{91}{11}.\end{array}$

Now substituting the value of $x=-\frac{91}{11}\text{ and }y=\frac{68}{11}$ in equation $2x+7y-3z=0$ it is obtained as:

  $\begin{array}{l}2\left(-\frac{91}{11}\right)+7\left(\frac{68}{11}\right)-3z=0.\\ \Rightarrow z=\frac{98}{11}.\end{array}$

Hence, the solution of the system is $x=-\frac{91}{11},y=\frac{68}{11}\text{ and }z=\frac{98}{11}$ .