Solved on Dec 15, 2023

Solve the system of linear equations $x + y + 2z = 1$ , $x - y = 1$ , and $x - z = 2$ .

STEP 1

Assumptions
1. We have a system of three linear equations with three variables $x$ , $y$ , and $z$ .
2. The system is assumed to have a unique solution.

STEP 2

Write down the system of equations for clarity.
\begin{align} x + y + 2z &= 1 \quad \text{(Equation 1)} \\ x - y &= 1 \quad \text{(Equation 2)} \\ x - z &= 2 \quad \text{(Equation 3)} \end{align}

STEP 3

Solve Equation 2 for $x$ in terms of $y$ .
$x = y + 1$

STEP 4

Substitute the expression for $x$ from STEP_3 into Equation 3 to solve for $z$ in terms of $y$ .
$y + 1 - z = 2$

STEP 5

Simplify the equation from STEP_4 to isolate $z$ .
$z = y + 1 - 2$

STEP 6

Simplify the expression for $z$ further.
$z = y - 1$

STEP 7

Substitute the expressions for $x$ and $z$ from STEP_3 and STEP_6 into Equation 1.
$(y + 1) + y + 2(y - 1) = 1$

STEP 8

Simplify the equation from STEP_7 to solve for $y$ .
$4y - 1 = 1$

STEP 9

Add 1 to both sides of the equation from STEP_8.
$4y = 2$

STEP 10

Divide both sides of the equation from STEP_9 by 4 to solve for $y$ .
$y = \frac{2}{4}$

STEP 11

Simplify the fraction to find the value of $y$ .
$y = \frac{1}{2}$

STEP 12

Substitute the value of $y$ back into the expression for $x$ from STEP_3.
$x = \frac{1}{2} + 1$

STEP 13

Simplify the expression to find the value of $x$ .
$x = \frac{3}{2}$

STEP 14

Substitute the value of $y$ back into the expression for $z$ from STEP_6.
$z = \frac{1}{2} - 1$

STEP 15

Simplify the expression to find the value of $z$ .
$z = -\frac{1}{2}$
The solution to the system of equations is $x = \frac{3}{2}$ , $y = \frac{1}{2}$ , and $z = -\frac{1}{2}$ .

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Solve the system of linear equations $x + y + 2z = 1$ , $x - y = 1$ , and $x - z = 2$ .

STEP 1

Assumptions
1. We have a system of three linear equations with three variables $x$ , $y$ , and $z$ .
2. The system is assumed to have a unique solution.

STEP 2

Write down the system of equations for clarity.
\begin{align} x + y + 2z &= 1 \quad \text{(Equation 1)} \\ x - y &= 1 \quad \text{(Equation 2)} \\ x - z &= 2 \quad \text{(Equation 3)} \end{align}

STEP 3

Solve Equation 2 for $x$ in terms of $y$ .
$x = y + 1$

STEP 4

Substitute the expression for $x$ from STEP_3 into Equation 3 to solve for $z$ in terms of $y$ .
$y + 1 - z = 2$

STEP 5

Simplify the equation from STEP_4 to isolate $z$ .
$z = y + 1 - 2$

STEP 6

Simplify the expression for $z$ further.
$z = y - 1$

STEP 7

Substitute the expressions for $x$ and $z$ from STEP_3 and STEP_6 into Equation 1.
$(y + 1) + y + 2(y - 1) = 1$

STEP 8

Simplify the equation from STEP_7 to solve for $y$ .
$4y - 1 = 1$

STEP 9

Add 1 to both sides of the equation from STEP_8.
$4y = 2$

STEP 10

Divide both sides of the equation from STEP_9 by 4 to solve for $y$ .
$y = \frac{2}{4}$

STEP 11

Simplify the fraction to find the value of $y$ .
$y = \frac{1}{2}$

STEP 12

Substitute the value of $y$ back into the expression for $x$ from STEP_3.
$x = \frac{1}{2} + 1$

STEP 13

Simplify the expression to find the value of $x$ .
$x = \frac{3}{2}$

STEP 14

Substitute the value of $y$ back into the expression for $z$ from STEP_6.
$z = \frac{1}{2} - 1$

STEP 15

Simplify the expression to find the value of $z$ .
$z = -\frac{1}{2}$
The solution to the system of equations is $x = \frac{3}{2}$ , $y = \frac{1}{2}$ , and $z = -\frac{1}{2}$ .

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Solve the system of linear equations x+y+2z=1x + y + 2z = 1x+y+2z=1, x−y=1x - y = 1x−y=1, and x−z=2x - z = 2x−z=2.

STEP 1

Assumptions1. We have a system of three linear equations with three variables xxx, yyy, and zzz.2. The system is assumed to have a unique solution.

STEP 2

Write down the system of equations for clarity.\begin{align*} x + y + 2z &= 1 \quad \text{(Equation 1)} \\ x - y &= 1 \quad \text{(Equation 2)} \\ x - z &= 2 \quad \text{(Equation 3)} \end{align*}

STEP 3

Solve Equation 2 for xxx in terms of yyy.x=y+1x = y + 1x=y+1

STEP 4

Substitute the expression for xxx from STEP_3 into Equation 3 to solve for zzz in terms of yyy.y+1−z=2y + 1 - z = 2y+1−z=2

STEP 5

Simplify the equation from STEP_4 to isolate zzz.z=y+1−2z = y + 1 - 2z=y+1−2

STEP 6

Simplify the expression for zzz further.z=y−1z = y - 1z=y−1

STEP 7

Substitute the expressions for xxx and zzz from STEP_3 and STEP_6 into Equation 1.(y+1)+y+2(y−1)=1(y + 1) + y + 2(y - 1) = 1(y+1)+y+2(y−1)=1

STEP 8

Simplify the equation from STEP_7 to solve for yyy.4y−1=14y - 1 = 14y−1=1

STEP 9

Add 1 to both sides of the equation from STEP_8.4y=24y = 24y=2

STEP 10

Divide both sides of the equation from STEP_9 by 4 to solve for yyy.y=24y = \frac{2}{4}y=42​

STEP 11

Simplify the fraction to find the value of yyy.y=12y = \frac{1}{2}y=21​

STEP 12

Substitute the value of yyy back into the expression for xxx from STEP_3.x=12+1x = \frac{1}{2} + 1x=21​+1

STEP 13

Simplify the expression to find the value of xxx.x=32x = \frac{3}{2}x=23​

STEP 14

Substitute the value of yyy back into the expression for zzz from STEP_6.z=12−1z = \frac{1}{2} - 1z=21​−1

STEP 15

Simplify the expression to find the value of zzz.z=−12z = -\frac{1}{2}z=−21​The solution to the system of equations is x=32x = \frac{3}{2}x=23​, y=12y = \frac{1}{2}y=21​, and z=−12z = -\frac{1}{2}z=−21​.

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Solve the system of linear equations $x + y + 2z = 1$ , $x - y = 1$ , and $x - z = 2$ .

Assumptions
1. We have a system of three linear equations with three variables $x$ , $y$ , and $z$ .
2. The system is assumed to have a unique solution.

Write down the system of equations for clarity.
\begin{align} x + y + 2z &= 1 \quad \text{(Equation 1)} \\ x - y &= 1 \quad \text{(Equation 2)} \\ x - z &= 2 \quad \text{(Equation 3)} \end{align}

Solve Equation 2 for $x$ in terms of $y$ .
$x = y + 1$

Substitute the expression for $x$ from STEP_3 into Equation 3 to solve for $z$ in terms of $y$ .
$y + 1 - z = 2$

Simplify the equation from STEP_4 to isolate $z$ .
$z = y + 1 - 2$

Simplify the expression for $z$ further.
$z = y - 1$

Substitute the expressions for $x$ and $z$ from STEP_3 and STEP_6 into Equation 1.
$(y + 1) + y + 2(y - 1) = 1$

Simplify the equation from STEP_7 to solve for $y$ .
$4y - 1 = 1$

Add 1 to both sides of the equation from STEP_8.
$4y = 2$

Divide both sides of the equation from STEP_9 by 4 to solve for $y$ .
$y = \frac{2}{4}$

Simplify the fraction to find the value of $y$ .
$y = \frac{1}{2}$

Substitute the value of $y$ back into the expression for $x$ from STEP_3.
$x = \frac{1}{2} + 1$

Simplify the expression to find the value of $x$ .
$x = \frac{3}{2}$

Substitute the value of $y$ back into the expression for $z$ from STEP_6.
$z = \frac{1}{2} - 1$

Simplify the expression to find the value of $z$ .
$z = -\frac{1}{2}$
The solution to the system of equations is $x = \frac{3}{2}$ , $y = \frac{1}{2}$ , and $z = -\frac{1}{2}$ .