Objective A: Solve a System of Three Linear Equations in Three Variables $\newline$ YOUR TURN $\newline$ \#\# $1$ $\newline$ Solve the following system of equations. $\newline$ $\left\{\begin{array}{r} 2 x+2 y+z=-9 \\ -x+y+4 z=-7 \\ x+y+2 z=-9 \end{array}\right.$

View step-by-step help

Home

Math Problems

Algebra 2

Quadrants

Full solution

Q. Objective A: Solve a System of Three Linear Equations in Three Variables

\newline

YOUR TURN

\newline

\#\#

1

\newline

Solve the following system of equations.

\newline

\left\{\begin{array}{r} 2 x+2 y+z=-9 \\ -x+y+4 z=-7 \\ x+y+2 z=-9 \end{array}\right.

Write Equations: Write down the system of equations. $\newline$ We have the following system of equations: $\newline$ $1$ ) $2x + 2y + z = -9$ $\newline$ $2$ ) $-x + y + 4z = -7$ $\newline$ $3$ ) $x + y + 2z = -9$
Eliminate Variable: Use the elimination method to eliminate one variable. $\newline$ We can start by eliminating the variable $x$ . To do this, we can add equation $2$ ) and equation $3$ ) to get a new equation without $x$ . $\newline$ $(-x + y + 4z) + (x + y + 2z) = (-7) + (-9)$ $\newline$ This simplifies to: $\newline$ $2y + 6z = -16$ $\newline$ Let's call this equation $4$ ).
Eliminate x Again: Now, let's eliminate $x$ from equations $1$ ) and $3$ ). We can multiply equation $3$ ) by $2$ and subtract it from equation $1$ ) to eliminate $x$ . $(2x + 2y + z) - 2(x + y + 2z) = -9 - 2(-9)$ This simplifies to: $2x + 2y + z - 2x - 2y - 4z = -9 + 18$ Which further simplifies to: $-3z = 9$ Let's call this equation $5$ ).
Solve for z: Solve equation $5$ ) for z. $\newline$ Dividing both sides of $-3z = 9$ by $-3$ , we get: $\newline$ z = $\frac{9}{-3}$ $\newline$ z = $-3$
Substitute for y: Substitute $z = -3$ into equation $4$ ) to find $y$ . Substituting $z$ into $2y + 6z = -16$ , we get: $2y + 6(-3) = -16$ $2y - 18 = -16$ Adding $18$ to both sides gives us: $2y = 2$ Dividing both sides by $2$ , we get: $y = 1$
Find $x$ : Substitute $y = 1$ and $z = -3$ into one of the original equations to find $x$ . Let's use equation $3$ ): $x + y + 2z = -9$ . Substituting $y$ and $z$ , we get: $x + 1 + 2(-3) = -9$ $x + 1 - 6 = -9$ $x - 5 = -9$ Adding $y = 1$ $0$ to both sides gives us: $y = 1$ $1$ $y = 1$ $2$