Lesson 3: 3-D Systems of Equations

Solve $$\begin{cases}2x+y=1\\ 3x+2y=3\end{cases}$$

$$\begin{cases}2x+y=1\\ 3x+2y=3\end{cases}$$

$\bigstar$ Substitution Method

From the first equation, we obtain $y=1-2x$. We substitute $y=1-2x$ into the second equation and get

\begin{align*}3x+2(1-2x)&=3\\3x+2-4x&=3\\-x&=1\\ x&=-1\end{align*}

So $y=1-2(-1)=3$. The solution set is $\{(-1,3)\}$.

$$\begin{cases}2x+y=1\\ 3x+2y=3\end{cases}$$

We multiply both sides of the first equation by $-2$. The system becomes

$$\begin{cases}-4x-2y=-2\\ 3x+2y=3\end{cases}$$

By adding up the two equations, we obtain

$$-x=1,\text{ that is, } x=-1.$$

We substitute $x=-1$ into $2x+y=1$ to obtain $y$:

\begin{align*}2(-1)+y&=1\\-2+y&=1\\y&=3\end{align*}

The solution set is $\{(-1,3)\}$.

$$\begin{cases}2x+y=1\\ 3x+2y=3\end{cases}$$

On the graph below the red line represents $2x+y=1$ and the blue line represents $3x+2y=3$. The lines intersect at $(-1,3)$. The solution set is $\{(-1,3)\}$.

Solve $$\begin{cases}5x-y+3z=6\\ x+y+2z=3\\-x-2y+z=8\end{cases}$$

$$\begin{cases}5x-y+3z=6 \qquad\text{(A)}\\ x+y+2z=3\qquad\text{(B)}\\-x-2y+z=8\qquad\text{(C)}\end{cases}$$

Consider:

$$\begin{cases}x+y+2z=3\qquad\text{(B)}\\-x-2y+z=8\qquad\text{(C)}\end{cases}$$

Adding (B) and (C) eliminates $x$:

$$-y+3z=11\qquad\text{(D)}$$

Now consider (A) and (B):

$$\begin{cases}5x-y+3z=6\qquad\text{(A)}\\ x+y+2z=3\qquad\text{(B)}\end{cases}$$

Multiply (B) by $-5$ to eliminate $x$:

$$\begin{cases}5x-y+3z=6\qquad\text{(A)}\\ -5x-5y-10z=-15\qquad\text{(-5B)}\end{cases}$$

Adding the two equations gives:

$$-6y-7z=-9\qquad\text{(E)}$$

Consider the system consisting of (D) and (E)

$$\begin{cases}-y+3z=11\qquad\text{(D)}\\-6y-7z=-9\qquad\text{(E)}\end{cases}$$

and solve for $y$ and $z$. Multiply (D) by $-6$:

$$\begin{cases}6y-18z=-66\qquad\text{(-6D)}\\-6y-7z=-9\qquad\text{(E)}\end{cases}$$

Adding the two equations gives $-25z=-75$, so $z=3$. By (D), we have

\begin{align*} -y+3\cdot 3 &=11\\-y+9&=11\\-y&=2\\y&=-2\end{align*}

Substituting $y=-2$ and $z=3$ into (B) gives:

\begin{align*} x-2+2\cdot 3&=3\\ x-2+6&=3\\x+4&=3\\x&=-1\end{align*}

We have $x=-1$, $y=-2$ and $z=3$.

Check:

$$\begin{cases}5(-1)-(-2)+3\cdot 3\stackrel{?}{=}6 \qquad\checkmark\\ (-1)+(-2)+2\cdot 3\stackrel{?}{=}3\qquad\checkmark\\-(-1)-2(-2)+3\stackrel{?}{=}8\qquad\checkmark\end{cases}$$

So the solution set is $\{(-1,-2,3)\}$.

Solve $$\begin{cases}3x-y+z=7\\ x+y+z=-3\\4x+2z=4\end{cases}$$

$$\begin{cases}3x-y+z=7 \qquad\text{(A)}\\ x+y+z=-3 \qquad\text{(B)}\\4x+2z=4\qquad\text{(C)}\end{cases}$$

Consider:

$$\begin{cases}3x-y+z=7 \qquad\text{(A)}\\ x+y+z=-3 \qquad\text{(B)}\end{cases}$$

Adding (A) and (B) eliminates $y$:

$$4x+2z=4$$

or

$$2x+z=2\qquad\text{(D)}$$

Now consider the system consisting of (C) and (D):

$$\begin{cases}4x+2z=4\qquad\text{(C)}\\ 2x+z=2\qquad\text{(D)}\end{cases}$$

and solve for $x$ and $z$. Divide (C) by $2$:

$$\begin{cases}2x+z=2\\ 2x+z=2\end{cases}$$

Subtracting the two equations gives $0=0$. So the system has infinitely many solutions.

Solve $$\begin{cases}3x-y+z=7\\ x+y+z=-3\\4x+2z=5\end{cases}$$

$$\begin{cases}3x-y+z=7 \qquad\text{(A)}\\ x+y+z=-3 \qquad\text{(B)}\\4x+2z=5\qquad\text{(C)}\end{cases}$$

Consider:

$$\begin{cases}3x-y+z=7 \qquad\text{(A)}\\ x+y+z=-3 \qquad\text{(B)}\end{cases}$$

Adding (A) and (B) eliminates $y$:

$$4x+2z=4$$

or

$$2x+z=2\qquad\text{(D)}$$

Now consider the system consisting of (C) and (D):

$$\begin{cases}4x+2z=5\qquad\text{(C)}\\ 2x+z=2\qquad\text{(D)}\end{cases}$$

and solve for $x$ and $z$. Multiply (D) by $2$:

$$\begin{cases}4x+2z=5\\ 4x+2z=4\end{cases}$$

Subtract the two equations to get $0=1$, which is not possible. So there is no solution.

• What does it mean to solve a system of equations? 
• What is the purpose of the Method of Addition?
• What is the graphical description of the system of linear equations?

$\bigstar$ Solve the system of equations $$\begin{cases}2x+y-z=4\\x+2y+2z=1\\3x-y+2z=0\end{cases}$$

\[\begin{cases}2x+y-z=4\qquad\text{(A)}\\x+2y+2z=1 \qquad\text{(B)}\\ 3x-y+2z=0 \qquad\text{(C)}\end{cases}\]

We start by eliminating $y$ from (A)  and (B). Multiply (A) by $-2$ and keep (B).

$$\begin{cases}-4x-2y+2z=-8 \qquad\text{(-2A)}\\ x+2y+2z=1 \qquad\text{(B)}\end{cases}$$

Adding these two equations gives

$$ -3x+4z=-7\qquad\text{(D)}.$$

Now consider (A)  and (C). 

$$\begin{cases}2x+y-z=4 \qquad\text{(A)}\\ 3x-y+2z=0\qquad\text{(C)}\end{cases}$$

To eliminate $y$, we add these two equations and obtain

$$5x+z=4\qquad\text{(E)}.$$

Consider the system with equations (D) and (E):

$$\begin{cases}-3x+4z=-7\qquad\text{(D)} \\5x+z=4 \qquad\text{(E)}\end{cases}$$

and solve for $x$ and $z$. Multiply (E) by $-4$:

$$\begin{cases} -3x+4z=-7 \qquad\text{(D)}\\ -20x-4z=-16\qquad\text{(-4E)}\end{cases}$$

Addding these equations gives

\begin{align*}-23x&=-23\\x&=1\end{align*}

Substitute $x=1$ into (E):

\begin{align*}5x+z&=4\\5\cdot 1+z&=4\\5+z& = 4\\z&=-1\end{align*}

Now substitute $x=1$ and $z=-1$ into (A):

\begin{align*}2(1)+y-(-1)&=4\\2+y+1&=4\\3+y&=4 \\y&=1\end{align*}

So $x=1$, $y=1$, $z=-1$.

Check:

$$\begin{cases}2x+y-z=4\\x+2y+2z=1\\3x-y+2z=0\end{cases}$$

$$\begin{cases}2(1)+1-(-1)\stackrel{?}{=}4 \quad\checkmark\\1+2(1)+2(-1)\stackrel{?}{=}1\quad\checkmark\\3(1)-1+2(-1)\stackrel{?}{=}0\quad\checkmark \end{cases}$$

The solution set is $\{(1,1,-1)\}$.