Chapter 7.3, Problem 26E

To determine

To find : the solution to the given system of linear equations

Answer to Problem 26E

The solutions to the given system of equation are $\left(5/3,1/3,3\right)$

Explanation of Solution

Given information : The system of equation is $\{\begin{array}{l}x+y+z=7\\ x-2y+4z=13\\ 3y+4z=13\end{array}$

Concept Involved:

Solution of a system of equation is the point which makes both the equation TRUE.

Graphically the solution to the system of equation is the point where the two lines meet.

Method of Substitution:

1. Solve one of the equations for one variable in terms of the other.

2. Substitute the expression found in Step 1 into the other equation to obtain an equation in one variable.

3. Solve the equation obtained in Step 2.

4. Back-substitute the value obtained in Step 3 into the expression obtained in Step 1 to find the value of the other variable.

5. Check that the solution satisfies each of the original equations.

Method of Elimination:

To use the method of elimination to solve a system of two linear equations in x and y, perform the following steps.

1. Obtain coefficients for x (or y) that differ only in sign by multiplying all

terms of one or both equations by suitably chosen constants.

2. Add the equations to eliminate one variable.

3. Solve the equation obtained in Step 2.

4. Back-substitute the value obtained in Step 3 into either of the original

equations and solve for the other variable.

5. Check that the solution satisfies each of the original equations.

Calculation:

Description	Steps
Label the given equations	$x+y+z=5$	▶ 1st equation
	$x-2y+4z=13$	▶ 2nd equation
	$3y+4z=13$	▶ 3rd equation
In order to eliminate variable z we need to multiply -4 with 1st equation and add the result with 2nd equation	$\begin{array}{l}-4x-4y-4z=-20\\ \underset{\_}{x-2y+4z=13}\\ -3x-6y=-7\end{array}$
Label the new equation as 4th equation	$-3x-6y=-7$	▶ 4thequation
In order to eliminate variable z we need to multiply -1 with the 2nd equation and add the result with 3rd equation	$\begin{array}{l}-x+2y-4z=-13\\ \underset{\_}{3y+4z=13}\\ -x+5y=0\end{array}$
Label the new equation as 5th equation	$-x+5y=0$	▶ 5th equation
In order to eliminate the variable x we need to multiply -3 with 5th equation and add the result with 4th equation	$\begin{array}{l}3x-15y=0\\ \underset{\_}{-3x-6y=-7}\\ -21y=-7\end{array}$
Solve the resulting equation $-21y=-7$ By dividing -21 on both sides of the equation Simplifying fraction on both sides	$\begin{array}{l}\frac{-21y}{-21}=\frac{-7}{-21}\\ y=1/3\end{array}$
Substituting 1/3 for x in the 4th equation and solve for x Simplify the fraction in left side of the equation Adding 2 on both sides of the equation Combine like terms in both sides of the equation Dividing -3 on both sides and simplifying Combining like terms	$\begin{array}{l}-3x-6\left(1/3\right)=-7\\ -3x-2=-7\\ -3x-2+2=-7+2\\ -3x=-5\\ -3x/-3=-5/-3\\ x=5/3\end{array}$
Substituting $5/3$ for x, $1/3$ for y in 1st equation	$\begin{array}{l}5/3+1/3+z=5\\ \left(5+1\right)/3+z=5\\ 6/3+z=5\\ 2+z=5\\ 2+z-2=5-2\\ z=3\end{array}$

Calculation (Continued):

Description	Steps
Checking the solution $\left(5/3,1/3,3\right)$ by substituting in 1st equation $x+y+z=5$	$\begin{array}{l}\frac{5}{3}+\frac{1}{3}+3=5\\ \frac{5+1}{3}+3=5\\ \frac{6}{3}+3=5\\ 2+3=5\\ 5=5\boxed{TRUE}\end{array}$
Checking the solution $\left(5/3,1/3,3\right)$ by substituting in 2nd equation $x-2y+4z=13$	$\begin{array}{l}\frac{5}{3}-2\left(\frac{1}{3}\right)+4\left(3\right)=13\\ \frac{5}{3}-\frac{2}{3}+12=13\\ \frac{5-2}{3}+12=13\\ \frac{3}{3}+12=13\\ 1+12=13\\ 13=13\boxed{TRUE}\end{array}$
Checking the solution $\left(5/3,1/3,3\right)$ by substituting in 3rdequation $3y+4z=13$	$\begin{array}{l}3\left(\frac{1}{3}\right)+4\left(3\right)=13\\ \frac{3}{3}+12=13\\ 1+12=13\\ 13=13\boxed{TRUE}\end{array}$

Conclusion:

The solution to the given system of equation is $\left(5/3,1/3,3\right)$