Solved on Nov 14, 2023

Write a system of linear equations from the given augmented matrix, then solve using back-substitution. Variables: $x, y, z$ .

STEP 1

Assumptions1. The given matrix is an augmented matrix representing a system of linear equations. . The variables used to represent the system are $x$ , $y$ , $z$ , and if necessary, $w$ .
3. We are to use back-substitution to find the solution.

STEP 2

First, we write the system of linear equations represented by the augmented matrix. The format is as follows $\begin{align} a1x + a2y + az &= a4 \\ b1x + b2y + bz &= b4 \\ c1x + c2y + cz &= c4 \\ \end{align}$ where $a1, a2, a, a4, b1, b2, b, b4, c1, c2, c, c4$ are the elements of the augmented matrix.

STEP 3

Substitute the elements of the augmented matrix into the system of linear equations.
$\begin{align} x + \frac{5}{2}y + z &= \frac{9}{2} \\ y - \frac{7}{2}z &=2 \\ z &= -8 \\ \end{align}$

STEP 4

Now, we use back-substitution to find the solution. Start with the last equation, which is already solved for $z$ .
$z = -8$

STEP 5

Substitute $z = -8$ into the second equation and solve for $y$ .
$y - \frac{7}{2}(-8) =2$

STEP 6

implify the equation and solve for $y$ .
$y +28 =2$ $y =2 -28$ $y = -26$

STEP 7

Substitute $z = -$ and $y = -26$ into the first equation and solve for $x$ .
$x + \frac{5}{2}(-26) + (-) = \frac{9}{2}$

STEP 8

implify the equation and solve for $x$ .
$x -65 -8 = \frac{}{2}$ $x -73 = \frac{}{2}$ $x = \frac{}{2} +73$ $x = \frac{ +146}{2}$ $x = \frac{155}{2}$ $x =77.5$ The solution to the system of linear equations is $x =77.5$ , $y = -26$ , and $z = -8$ .

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Write a system of linear equations from the given augmented matrix, then solve using back-substitution. Variables: $x, y, z$ .

STEP 1

Assumptions1. The given matrix is an augmented matrix representing a system of linear equations. . The variables used to represent the system are $x$ , $y$ , $z$ , and if necessary, $w$ .
3. We are to use back-substitution to find the solution.

STEP 2

STEP 3

Substitute the elements of the augmented matrix into the system of linear equations.
$\begin{align} x + \frac{5}{2}y + z &= \frac{9}{2} \\ y - \frac{7}{2}z &=2 \\ z &= -8 \\ \end{align}$

STEP 4

Now, we use back-substitution to find the solution. Start with the last equation, which is already solved for $z$ .
$z = -8$

STEP 5

Substitute $z = -8$ into the second equation and solve for $y$ .
$y - \frac{7}{2}(-8) =2$

STEP 6

implify the equation and solve for $y$ .
$y +28 =2$ $y =2 -28$ $y = -26$

STEP 7

Substitute $z = -$ and $y = -26$ into the first equation and solve for $x$ .
$x + \frac{5}{2}(-26) + (-) = \frac{9}{2}$

STEP 8

implify the equation and solve for $x$ .
$x -65 -8 = \frac{}{2}$ $x -73 = \frac{}{2}$ $x = \frac{}{2} +73$ $x = \frac{ +146}{2}$ $x = \frac{155}{2}$ $x =77.5$ The solution to the system of linear equations is $x =77.5$ , $y = -26$ , and $z = -8$ .

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Write a system of linear equations from the given augmented matrix, then solve using back-substitution. Variables: x,y,zx, y, zx,y,z.

STEP 1

Assumptions1. The given matrix is an augmented matrix representing a system of linear equations. . The variables used to represent the system are xxx, yyy, zzz, and if necessary, www.3. We are to use back-substitution to find the solution.

STEP 2

STEP 3

Substitute the elements of the augmented matrix into the system of linear equations.x+52y+z=92y−72z=2z=−8\begin{align*} x + \frac{5}{2}y + z &= \frac{9}{2} \\ y - \frac{7}{2}z &=2 \\ z &= -8 \\ \end{align*} x+25​y+zy−27​zz​=29​=2=−8​

STEP 4

Now, we use back-substitution to find the solution. Start with the last equation, which is already solved for zzz.z=−8z = -8z=−8

STEP 5

Substitute z=−8z = -8z=−8 into the second equation and solve for yyy.y−72(−8)=2y - \frac{7}{2}(-8) =2y−27​(−8)=2

STEP 6

implify the equation and solve for yyy.y+28=2y +28 =2y+28=2y=2−28y =2 -28y=2−28y=−26y = -26y=−26

STEP 7

Substitute z=−z = -z=− and y=−26y = -26y=−26 into the first equation and solve for xxx.x+52(−26)+(−)=92x + \frac{5}{2}(-26) + (-) = \frac{9}{2}x+25​(−26)+(−)=29​

STEP 8

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Write a system of linear equations from the given augmented matrix, then solve using back-substitution. Variables: $x, y, z$ .

Assumptions1. The given matrix is an augmented matrix representing a system of linear equations. . The variables used to represent the system are $x$ , $y$ , $z$ , and if necessary, $w$ .
3. We are to use back-substitution to find the solution.

Substitute the elements of the augmented matrix into the system of linear equations.
$\begin{align} x + \frac{5}{2}y + z &= \frac{9}{2} \\ y - \frac{7}{2}z &=2 \\ z &= -8 \\ \end{align}$

Now, we use back-substitution to find the solution. Start with the last equation, which is already solved for $z$ .
$z = -8$

Substitute $z = -8$ into the second equation and solve for $y$ .
$y - \frac{7}{2}(-8) =2$

implify the equation and solve for $y$ .
$y +28 =2$ $y =2 -28$ $y = -26$

Substitute $z = -$ and $y = -26$ into the first equation and solve for $x$ .
$x + \frac{5}{2}(-26) + (-) = \frac{9}{2}$