College Preparatory Mathematics

\begin{align*} \begin{matrix} 3x + 2y - z = - 1 \\ - 2x - 2y + 3z = 5 \\ 5x + 2y - z = 3 \end{matrix} \amp \quad \text{ We will eliminate $y$ using two different pairs of equations } \\ \amp\quad \text{ } \\ 3x+2y-z=-1 \amp\quad \text{ Using the first two equations, } \\ \underline{-2x-2y+3z=5} \amp\quad \text{ Add the first two equations } \\ \text{(A)} \quad x\qquad +2z=4 \amp\quad \text{ This is equation (A), our first equation } \\ \amp\quad \\ -2x-2y+3z=5 \amp\quad \text{ Using the second two equations } \\ \underline{5x+2y-z=3} \amp\quad \text{ Add the second two equations } \\ \text{(B)} \quad 3x\qquad +2z=8 \amp\quad \text{ This is equation (B), our second equation } \\ \amp\quad \text{ } \\ (A)\quad x+2z=4 \amp\quad \text{ Using (A) and (B) we will solve this system. } \\ (B)\quad 3x+2z=8 \amp\quad \text{ We will solve by addition } \\ \amp\quad \text{ } \\ -1(x+2z)=(4)(-1) \amp\quad \text{ Multiply (A) by } -1 \\ -x-2z=-4 \amp\quad \text{ } \\ \amp\quad \text{ } \\ -x-2z=-4 \amp\quad \text{ Add to the second equation, unchanged } \\ \underline{3x+2z=8} \amp\quad \text{ } \\ 2x=4\amp\quad \text{ Solve, divide by } 2 \\ \amp\quad \text{ } \\ x=2 \amp\quad \text{ We now have $x$! Plug this into either (A) or (B) } \\ \amp\quad \text{ } \\ (2)+2z=4 \amp\quad \text{ We plug it into (A), solve this equation, subtract } 2 \\ \underline{-2\qquad -2} \amp\quad \text{ } \\ 2z=2 \amp\quad \text{ Divide by } 2 \\ \amp\quad \text{ } \\ z=1 \amp\quad \text{ We now have $z$! Plug this and x into any original equation } \\ \amp\quad \text{ } \\ 3(2)+2y-(1)=-1 \amp\quad \text{We use the first, multiply $3(2)=6$ and combine with } -1 \\ 2y+5=-1 \amp\quad \text{ Solve,subtract } 5 \\ \underline{-5\quad -5} \amp\quad \text{ } \\ 2y=-6 \amp\quad \text{ Divide by } 2 \\ \amp\quad \text{ } \\ y=-3 \amp\quad \text{ We now have $y$! } \\ \amp\quad \text{ } \\ (2,-3,1) \amp\quad \text{ Our Solution } \end{align*}

\begin{align*} 4x-3y+2z=-29 \amp\quad \text{ No variable will easily eliminate. } \\ 6x+2y-z=-16 \amp\quad \text{We could choose any variable, so we chose } x \\ -8x-y+3z=23 \amp\quad \text{ We will eliminate $x$ twice.} \\ \amp\quad \text{ } \\ 4x-3y+2z=-29 \amp\quad \text{ Start with first two equations. LCM of $4$ and $6$ is $12$. } \\ 6x+2y-z=-16 \amp\quad \text{Make the first equation have $12x$, the second } -12x \\ \amp\quad \text{ } \\ 3(4x-3y+2z)=(-29)3 \amp\quad \text{ Multiply the first equation by } 3 \\ \amp\quad \\ -2(6x+2y-z)=(-16)(-2) \amp\quad \text{ Multiply the second equation by} -2\\ -12x-4y+2z=32 \amp\quad \\ \amp\quad \\ 12x-9y+6z=-87 \amp\quad \text{ Add these two equations together } \\ \underline{-12x-4y+2z=32} \amp\quad \\ \text{(A)}\quad -13y+8z=-55\amp\quad \text{ This is our (A) equation } \\ \amp\quad \\ 6x+2y-z=-16 \amp\quad \text{ Now use the second two equations (a different pair) } \\ -8x-y+3z=23 \amp\quad \text{ The LCM of $6$ and $-8$ is $24$.} \\ \amp\quad \text{ } \\ 4(6x+2y -z)=(-16)4 \amp\quad \text{ Multiply the first equation by } 4 \\ 24x+8y-4=-64 \amp\quad \text{ } \\ \amp\quad \text{ } \\ 3(-8x-y+3z)=(23)3 \amp\quad \text{ Multiply the second equation by } 3 \\ -24x-3y+9z=69 \amp\quad \text{ } \\ \amp\quad \text{ } \\ 24x+8y-4=-64 \amp\quad \text{ Add these two equations together} \\ \underline{-24x-3y+9z=69} \amp\quad \text{ } \\ \text{(B)} \qquad 5y+5z=5\amp\quad \text{ This is our (B) equation } \\ \amp\quad \text{ } \\ \text{(A)}\quad -13y+8z=-55 \amp\quad \text{ Using (A) and (B) we will solve this system } \\ \text{(B)}\qquad \qquad 5y+5z=5 \amp\quad \text{ The second equation is solved for $z$ to use substitution} \\ \amp\quad \text{ } \\ 5y+5z=5 \amp\quad \text{ Solving for $z$, subtract } 5y \\ \underline{-5y\qquad -5y} \amp\quad \text{ } \\ 5z=5-5y \amp\quad \text{ Divide each term by } 5 \\ z=1-y \amp\quad \text{ Plug into untouched equation } \\ -13y+8(1-y)=-55 \amp\quad \text{Distribute} \\ -13y+8-8y=-55\amp\quad \text{ Combine like terms } - 13y - 8y \\ -21y+8=-55 \amp\quad \text{ Subtract } 8 \\ \underline{-8 \quad -8} \amp\quad \text{ } \\ -21y=-63 \amp\quad \text{ Divideby } -21 \\ \amp\quad \text{ } \\ y=3 \amp\quad \text{We have our $y$! Plug this into $z =$ equations} \\ z=1-(3) \amp\quad \text{ Evaluate } \\ z=-2 \amp\quad \text{ We have $z$, now find $x$ from original equation. } \\ \amp\quad \text{ } \\ 4x-3(3)+2(-2)=-29 \amp\quad \text{ Multiply and combine like terms} \\ 4x-13=-29 \amp\quad \text{ Add } 13 \\ +13 \quad +13 \amp\quad \text{ } \\ 4x=-16 \amp\quad \text{ Divideby } 4 \\ x=-4 \amp\quad \text{ We have our $x$! } \\ \amp\quad \text{ } \\ (-4,3,-2) \amp\quad \text{Our solution } \end{align*}

\begin{align*} \begin{matrix}5x-4y+3z=-4\\ - 10x + 8y - 6z = 8 \\ 15x-12y+9z=-12 \end{matrix} \amp\quad \text{ We will eliminate $x$, start with first two equations } \\ \amp\quad \text{ } \\ 5x-4y+3z=-4 \amp\quad \text{ LCM of $5$ and $-10$ is $10$.} \\ -10x+8y-6z=8 \amp\quad \text{ } \\ \amp\quad \\ 2(5x-4y+3z)=-4(2) \amp\quad \text{ Multiply the first equation by } 2 \\ 10x-8y+6z=-8 \amp\quad \text{ } \\ \amp\quad \\ 10x-8y+6z=-8 \amp\quad \text{ Add this to the second equation, unchanged} \\ \underline{-10x+8y -6=8} \amp\quad \\ 0=0 \amp\quad \text{ A true statment} \\ Infinite Solutions \amp\quad \text{ Our solution } \end{align*}

\begin{align*} \begin{matrix} 3x-4y+z=2 \\ -9x+12y-3z=-5 \\ 4x - 2y - z = 3 \end{matrix} \amp \quad \text{ We will eliminate $z$,starting with the first two equations} \\ \amp\quad \text{ } \\ 3x-4y+z=2\amp\quad \text{ The LCM of $1$ and $-3$ is $3$} \\ - 9x + 12y - 3z = - 5\amp\quad \text{ } \\ \amp\quad \text{ } \\ 3(3x-4y+z)=(2)3 \amp\quad \text{ Multiply the first equation by } 3 \\ 9x-12y+3z=6\amp\quad \text{ } \\ \amp\quad \text{ } \\ 9x-12y+3z=6 \amp\quad \text{ Add this to the second equation, unchanged} \\ \underline{- 9x + 12y - 3z = - 5} \amp\quad \text{ } \\ 0=1 \amp\quad \text{ A false statement} \\ \text{No Solution } \emptyset \amp\quad \text{ Our Solution} \end{align*}