Chapter 7, Problem 11PS

a.

To determine

To Solve: system of two equations that consists of equation 1 and equation 2.

Answer to Problem 11PS

The solution is $x=-a,y=a$ and $z=\frac{16+6a}{5}$

Explanation of Solution

Given:

The following system has one solution $x=1,y=-1,z=2$ .

  $\{\begin{array}{c}4x-2y+5z=16\to \left(1\right)\\ x+y=0\to \left(2\right)\\ -x-3y+2z=6\to \left(3\right)\end{array}$

Concept Used:

These are the linear equations and they can be solved by using substitution method.

Calculation:

Let us consider equation 1 and 2 as asked and solve them for solutions.

From equation 2 we get,

  $\begin{array}{l}x+y=0\\ x=-y\end{array}$

Let us substitute $y=a$ then $x=-a$

Where a is any arbitrary constant

By substituting x and y in equation 1 we get,

  $\begin{array}{l}4x-2y+5z=16\\ 4\left(-a\right)-2\left(a\right)+5z=16\\ -6a+5z=16\\ z=\frac{16+6a}{5}\end{array}$

Hence the solution is $x=-a,y=a$ and $z=\frac{16+6a}{5}$

Conclusion:

The solution is $x=-a,y=a$ and $z=\frac{16+6a}{5}$

b.

To determine

Solve system of two equations that consists of equation 1 and equation 3.

Answer to Problem 11PS

The solutions of 1 and 3 equations is $x=\frac{-11a+36}{14},y=\frac{13a-40}{14}$ and $z=a$

Explanation of Solution

Given:

The following system has one solution $x=1,y=-1,z=2$ .

  $\{\begin{array}{c}4x-2y+5z=16\to \left(1\right)\\ x+y=0\to \left(2\right)\\ -x-3y+2z=6\to \left(3\right)\end{array}$

Concept Used:

These are the linear equations and they can be solved by using elimination method.

Calculation:

Let us consider equation 1 and 3 as and add them

  $\begin{array}{l}4x-2y+5z-x-3y+2z=16+6\\ -14y+13z=40\end{array}$

Consider $z=a$

Where a is any arbitrary constant

Then

  $\begin{array}{l}-14y+13\left(a\right)=40\\ \Rightarrow -14y=40-13a\\ \Rightarrow y=\frac{13a-40}{14}\end{array}$

Then the value of x will be

  $\begin{array}{l}-x-3y+2z=6\\ \Rightarrow -x-3\left(\frac{13a-40}{14}\right)+2a=6\\ \Rightarrow x=\frac{-11a+36}{14}\end{array}$

Hence $x=\frac{-11a+36}{14},y=\frac{13a-40}{14}$ and $z=a$

Conclusion:

The solutions of 1 and 3 equations is $x=\frac{-11a+36}{14},y=\frac{13a-40}{14}$ and $z=a$

c.

To determine

To find: That the following system has one solution $x=1,y=-1,z=2$ .

Answer to Problem 11PS

The solutions of 2 and 3 equations is $x=a,y=a$ and $z=3+a$

Explanation of Solution

Given:

  $\{\begin{array}{c}4x-2y+5z=16\to \left(1\right)\\ x+y=0\to \left(2\right)\\ -x-3y+2z=6\to \left(3\right)\end{array}$

Concept Used:

These are the linear equations and they can be solved by using elimination method

Calculation:

Let us consider equation 2 and 3 as asked and solve them for solutions.

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

From equation (2)

  $x=-y$

Substitute $x=-y$ in equation (3)

  $\begin{array}{l}y-3y+2z=6\\ \Rightarrow -2y+2z=6\\ \Rightarrow -y+z=3\end{array}$

Let $y=a$

Then $x=-a$

Put these values in $-y+z=3\Rightarrow -a+z=3\Rightarrow z=3+a$

Hence the solutions of 2 and 3 equations is $x=a,y=a$ and $z=3+a$

Conclusion:

The solutions of 2 and 3 equations is $x=1,y=-1,z=2$

  $x=a,y=a$ and $z=3+a$

d.

To determine

To find: That the following system has one solution $x=1,y=-1,z=2$ .

Answer to Problem 11PS

Hence only one solution is $x=1,y=-1$ and $z=2$

Explanation of Solution

Given:

  $\begin{array}{l}4x-2y+5z=16\to \left(1\right)\\ x+y=0\to \left(2\right)\\ -x-3y+2z=6\to \left(3\right)\end{array}$

Concept Used:

These are the linear equations and they can be solved by using elimination method or graphical method.

Calculation:

  $\begin{array}{l}4x-2y+5z=16\to \left(1\right)\\ x+y=0\to \left(2\right)\\ -x-3y+2z=6\to \left(3\right)\end{array}$

From equation (2)

  $x=-y$

Substitute $x=-y$ in equation (1) and (3)

  $\begin{array}{l}4\left(-y\right)-2y+5z=16\Rightarrow -6y+5z=16\cdots \left(4\right)\\ -\left(-y\right)-3y+2z=6\Rightarrow -2y+2z=6\cdots \left(5\right)\end{array}$

Subtract three times of equation (5) from equation (4)

  $\begin{array}{l}-6y+5z-\left(-6y\right)-6z=16-18\\ -z=-2\\ z=2\end{array}$

Put $z=2$ in equation (5) and find the value of y

  $\begin{array}{l}-2y+2\left(2\right)=6\\ \Rightarrow -2y=6-4\\ \Rightarrow -2y=2\\ \Rightarrow y=-1\end{array}$

So, $x=1$

Conclusion:

They have only one solution $x=1,y=-1$ and $z=2$