Solve the system of equations by elimination. $\newline$ $2x - 3y - z = -11$ $\newline$ $-x - y - z = -11$ $\newline$ $x - 2y - 2z = -10$

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Algebra 2

Solve a system of equations in three variables using elimination

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Q. Solve the system of equations by elimination.

\newline

2x - 3y - z = -11

\newline

-x - y - z = -11

\newline

x - 2y - 2z = -10

Eliminate x: Add the first and second equations to eliminate x. $\newline$ $(2x - 3y - z) + (-x - y - z) = -11 + (-11)$ $\newline$ $2x - x - 3y - y - z - z = -22$ $\newline$ $x - 4y - 2z = -22$
Prepare for elimination: Multiply the second equation by $2$ to prepare for elimination with the third equation. $\newline$ $2(-x - y - z) = 2(-11)$ $\newline$ $-2x - 2y - 2z = -22$
Eliminate x: Add the modified second equation and the third equation to eliminate x. $\newline$ $(-2x - 2y - 2z) + (x - 2y - 2z) = -22 + (-10)$ $\newline$ $-2x + x - 2y - 2y - 2z - 2z = -32$ $\newline$ $-x - 4y - 4z = -32$
Simplify equation: Divide the last equation by $-1$ to simplify. $\frac{-x}{-1} - \frac{4y}{-1} - \frac{4z}{-1} = \frac{-32}{-1}$ $x + 4y + 4z = 32$
Eliminate $y$ and $z$ : Now we have two equations with $x$ , $y$ , and $z$ : $x - 4y - 2z = -22$ $x + 4y + 4z = 32$ Add these two equations to eliminate $y$ and $z$ . $(x - 4y - 2z) + (x + 4y + 4z) = -22 + 32$ $x + x - 4y + 4y - 2z + 4z = 10$ $2x + 2z = 10$
Solve for x: Divide the last equation by $2$ to solve for x. $\newline$ $\frac{2x}{2} + \frac{2z}{2} = \frac{10}{2}$ $\newline$ $x + z = 5$
Solve for $y$ : Substitute $x + z = 5$ into $x - 4y - 2z = -22$ to solve for $y$ .
$(5 - z) - 4y - 2z = -22$
$5 - z - 4y - 2z = -22$
$5 - 4y - 3z = -22$
Substitute $x + z = 5$ : Isolate the term with $y$ .
$-4y = -22 - 5 + 3z$
$-4y = -27 + 3z$
Solve for $z$ : Divide by $-4$ to solve for $y$ . $y = \frac{{-27 + 3z}}{{-4}}$ $y = \frac{27}{4} - \left(\frac{3}{4}\right)z$
Solve for $x$ : Substitute $x + z = 5$ into $x + 4y + 4z = 32$ to solve for $y$ .
$(5 - z) + 4y + 4z = 32$
$5 + 4y + 3z = 32$
Solve for $z$ : Isolate the term with $y$ . $4y = 32 - 5 - 3z$ $4y = 27 - 3z$
Solve for $z$ : Isolate the term with $y$ . $4y = 32 - 5 - 3z$ $4y = 27 - 3z$ Divide by $4$ to solve for $y$ . $y = \frac{27 - 3z}{4}$ $y = \frac{27}{4} - \frac{3}{4}z$
Solve for z: Isolate the term with y. $\newline$ $4y = 32 - 5 - 3z$ $\newline$ $4y = 27 - 3z$ Divide by $4$ to solve for y. $\newline$ $y = \frac{27 - 3z}{4}$ $\newline$ $y = \frac{27}{4} - \frac{3}{4}z$ We have two expressions for y that must be equal: $\newline$ $\frac{27}{4} - \frac{3}{4}z = \frac{27}{4} - \frac{3}{4}z$ $\newline$ This is always true for all $z$ , so we need to find a specific value for $z$ using another equation.
Solve for $z$ : Isolate the term with $y$ .
$4y = 32 - 5 - 3z$
$4y = 27 - 3z$ Divide by $4$ to solve for $y$ .
$y = \frac{27 - 3z}{4}$
$y = \frac{27}{4} - \frac{3}{4}z$ We have two expressions for $y$ that must be equal:
$\frac{27}{4} - \frac{3}{4}z = \frac{27}{4} - \frac{3}{4}z$
This is always true for all $z$ , so we need to find a specific value for $z$ using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $y$ $3$ to solve for $z$ .
$y$ $5$
$y$ $6$
$y$ $7$
Solve for z: Isolate the term with y. $\newline$ $4y = 32 - 5 - 3z$ $\newline$ $4y = 27 - 3z$ Divide by $4$ to solve for y. $\newline$ $y = \frac{27 - 3z}{4}$ $\newline$ $y = \frac{27}{4} - \frac{3}{4}z$ We have two expressions for y that must be equal: $\newline$ $\frac{27}{4} - \frac{3}{4}z = \frac{27}{4} - \frac{3}{4}z$ $\newline$ This is always true for all $z$ , so we need to find a specific value for $z$ using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $x - 4y - 2z = -22$ to solve for $z$ . $\newline$ $4y = 27 - 3z$ $1$ $\newline$ $4y = 27 - 3z$ $2$ $\newline$ $4y = 27 - 3z$ $3$ We already have $4y = 27 - 3z$ $3$ , so we need to find a specific value for $z$ using another equation.
Solve for z: Isolate the term with y. $\newline$ $4y = 32 - 5 - 3z$ $\newline$ $4y = 27 - 3z$ Divide by $4$ to solve for y. $\newline$ $y = \frac{27 - 3z}{4}$ $\newline$ $y = \frac{27}{4} - \frac{3}{4}z$ We have two expressions for y that must be equal: $\newline$ $\frac{27}{4} - \frac{3}{4}z = \frac{27}{4} - \frac{3}{4}z$ $\newline$ This is always true for all $z$ , so we need to find a specific value for $z$ using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $x - 4y - 2z = -22$ to solve for $z$ . $\newline$ $4y = 27 - 3z$ $0$ $\newline$ $4y = 27 - 3z$ $1$ $\newline$ $4y = 27 - 3z$ $2$ We already have $4y = 27 - 3z$ $2$ , so we need to find a specific value for $z$ using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $4y = 27 - 3z$ $6$ to solve for $z$ . $\newline$ $4y = 27 - 3z$ $8$ $\newline$ $4y = 27 - 3z$ $9$
Solve for z: Isolate the term with y. $\newline$ $4y = 32 - 5 - 3z$ $\newline$ $4y = 27 - 3z$ Divide by $4$ to solve for y. $\newline$ $y = \frac{27 - 3z}{4}$ $\newline$ $y = \frac{27}{4} - \frac{3}{4}z$ We have two expressions for y that must be equal: $\newline$ $\frac{27}{4} - \frac{3}{4}z = \frac{27}{4} - \frac{3}{4}z$ $\newline$ This is always true for all z, so we need to find a specific value for z using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $x - 4y - 2z = -22$ to solve for z. $\newline$ $x - 4\left(\frac{27}{4} - \frac{3}{4}z\right) - 2z = -22$ $\newline$ $x - 27 + 3z - 2z = -22$ $\newline$ $x + z = 5$ We already have $x + z = 5$ , so we need to find a specific value for z using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $4y = 27 - 3z$ $2$ to solve for z. $\newline$ $4y = 27 - 3z$ $3$ $\newline$ $4y = 27 - 3z$ $4$ $\newline$ $4y = 27 - 3z$ $5$
Solve for $z$ : Isolate the term with $y$ . $4y = 32 - 5 - 3z$ $4y = 27 - 3z$ Divide by $4$ to solve for $y$ . $y = \frac{27 - 3z}{4}$ $y = \frac{27}{4} - \frac{3}{4}z$ We have two expressions for $y$ that must be equal: $\frac{27}{4} - \frac{3}{4}z = \frac{27}{4} - \frac{3}{4}z$ This is always true for all $z$ , so we need to find a specific value for $z$ using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $y$ $3$ to solve for $z$ . $y$ $5$ $y$ $6$ $y$ $7$ We already have $y$ $7$ , so we need to find a specific value for $z$ using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $4y = 32 - 5 - 3z$ $1$ to solve for $z$ . $4y = 32 - 5 - 3z$ $3$ $4y = 32 - 5 - 3z$ $4$ Multiply all terms by $4$ to clear the fraction. $4y = 32 - 5 - 3z$ $6$ $4y = 32 - 5 - 3z$ $7$ Substitute $y$ $7$ into $4y = 32 - 5 - 3z$ $7$ to solve for $4y = 27 - 3z$ $0$ . $4y = 27 - 3z$ $1$ $4y = 27 - 3z$ $2$ $4y = 27 - 3z$ $3$
Solve for z: Isolate the term with y. $\newline$ $4y = 32 - 5 - 3z$ $\newline$ $4y = 27 - 3z$ Divide by $4$ to solve for y. $\newline$ $y = \frac{27 - 3z}{4}$ $\newline$ $y = \frac{27}{4} - \frac{3}{4}z$ We have two expressions for y that must be equal: $\newline$ $\frac{27}{4} - \frac{3}{4}z = \frac{27}{4} - \frac{3}{4}z$ $\newline$ This is always true for all z, so we need to find a specific value for z using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $x - 4y - 2z = -22$ to solve for z. $\newline$ $x - 4\left(\frac{27}{4} - \frac{3}{4}z\right) - 2z = -22$ $\newline$ $x - 27 + 3z - 2z = -22$ $\newline$ $4y = 27 - 3z$ $0$ We already have $4y = 27 - 3z$ $0$ , so we need to find a specific value for z using another equation.Substitute $y = \frac{27}{4} - \frac{3}{4}z$ into $4y = 27 - 3z$ $3$ to solve for z. $\newline$ $4y = 27 - 3z$ $4$ $\newline$ $4y = 27 - 3z$ $5$ Multiply all terms by $4$ to clear the fraction. $\newline$ $4y = 27 - 3z$ $7$ $\newline$ $4y = 27 - 3z$ $8$ Substitute $4y = 27 - 3z$ $0$ into $4y = 27 - 3z$ $8$ to solve for x. $\newline$ $4$ $1$ $\newline$ $4$ $2$ $\newline$ $4$ $3$ Divide by $4$ $4$ to solve for z. $\newline$ $4$ $5$ $\newline$ $4$ $6$