Solving systems of linear equations by elimination  (Page 2/2)

Solve $\{\begin{array}{rrr}\hfill 6a-5b=14& \hfill & \hfill \left(1\right)\\ \hfill 2a+2b=-10& \hfill & \hfill \left(2\right)\end{array}$

Step 1: The equations are already in the proper form, $ax+by=c.$

Step 2: If we multiply equation (2) by —3, the coefficients of $a$ will be opposites and become 0 upon addition, thus eliminating $a$ .

       $$\begin{array}{lllll}\{\begin{array}{l}6a-5b=14\\ -3\left(2a+2b\right)=-3\left(10\right)\end{array}\hfill & \hfill & \to \hfill & \hfill & \{\begin{array}{l}6a-5b=14\\ -6a-6b=30\end{array}\hfill \end{array}$$

Step 3:  Add the equations.

       $\frac{\begin{array}{c}6a-5b=14\\ -6a-6b=30\end{array}}{0-11b=44}$

Step 4:  Solve the equation $-11b=44.$

       $-11b=44$
        $b=-4$

Step 5:  Substitute $b=-4$ into either of the original equations. We will use equation 2.

       $\begin{array}{rrrrr}\hfill 2a+2b& \hfill =& \hfill -10& \hfill & \hfill \\ \hfill 2a+2\left(-4\right)& \hfill =& \hfill -10& \hfill & \hfill \text{Solve\hspace{0.17em}for\hspace{0.17em}}a.\\ \hfill 2a-8& \hfill =& \hfill -10& \hfill & \hfill \\ \hfill 2a& \hfill =& \hfill -2& \hfill & \hfill \\ \hfill a& \hfill =& \hfill -1& \hfill & \hfill \end{array}$

 We now have $a=-1$ and $b=-4.$

Step 6:  Substitute $a=-1$ and $b=-4$ into both the original equations for a check.

       $\begin{array}{rrrrrrrrrrrr}\hfill (1)& \hfill & \hfill 6a-5b& \hfill =& \hfill 14& \hfill & \hfill (2)& \hfill & \hfill 2a+2b& \hfill =& \hfill -10& \hfill \\ \hfill & \hfill & \hfill 6(-1)-5(-4)& \hfill =& \hfill 14& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill & \hfill & \hfill 2(-1)+2(-4)& \hfill =& \hfill -10& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill -6+20& \hfill =& \hfill 14& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill & \hfill & \hfill & \hfill -2-8& \hfill =& \hfill -10& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill 14& \hfill =& \hfill 14& \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill & \hfill & \hfill & \hfill -10& \hfill =& \hfill -10& \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$

Step 7:  The solution is $\left(-1,-4\right).$

Solve  $\{\begin{array}{ccc}\begin{array}{l}3x+2y=-4\hfill \\ 4x=5y+10\hfill \end{array}& & \begin{array}{l}(1)\hfill \\ (2)\hfill \end{array}\end{array}$

Step 1:  Rewrite the system in the proper form.

       $\{\begin{array}{ccc}\begin{array}{l}3x+2y=-4\\ 4x-5y=10\end{array}& & \begin{array}{l}\left(1\right)\\ \left(2\right)\end{array}\end{array}$

Step 2:  Since the coefficients of $y$ already have opposite signs, we will eliminate $y$ .
     Multiply equation (1) by 5, the coefficient of $y$ in equation 2.
     Multiply equation (2) by 2, the coefficient of $y$ in equation 1.

       $\begin{array}{rrrrr}\hfill \{\begin{array}{l}5\left(3x+2y\right)=5\left(-4\right)\\ 2\left(4x-5y\right)=2\left(10\right)\end{array}& \hfill & \hfill \to & \hfill & \hfill \{\begin{array}{l}15x+10y=-20\\ 8x-10y=20\end{array}\end{array}$

Step 3:  Add the equations.

       $\frac{\begin{array}{c}15x+10y=-20\\ 8x-10y=20\end{array}}{23x+0=0}$

Step 4:  Solve the equation $23x=0$

       $23x=0$

       $x=0$

Step 5:  Substitute $x=0$ into either of the original equations. We will use equation 1.

       $\begin{array}{rrrrr}\hfill 3x+2y& \hfill =& \hfill -4& \hfill & \hfill \\ \hfill 3\left(0\right)+2y& \hfill =& \hfill -4& \hfill & \hfill \text{Solve\hspace{0.17em}for\hspace{0.17em}}y.\\ \hfill 0+2y& \hfill =& \hfill -4& \hfill & \hfill \\ \hfill y& \hfill =& \hfill -2& \hfill & \hfill \end{array}$

 We now have $x=0$ and $y=-2.$

Step 6:  Substitution will show that these values check.

Step 7:  The solution is $\left(0,-2\right).$

Solve $\{\begin{array}{rrr}\hfill 2x-y=1& \hfill & \hfill \left(1\right)\\ \hfill 4x-2y=4& \hfill & \hfill \left(2\right)\end{array}$

Step 1: The equations are in the proper form.

Step 2: We can eliminate $x$ by multiplying equation (1) by –2.

       $\begin{array}{rrrrr}\hfill \{\begin{array}{c}-2\left(2x-y\right)=-2\left(1\right)\\ 4x-2y=4\end{array}& \hfill & \hfill \to & \hfill & \hfill \{\begin{array}{c}-4x+2y=-2\\ 4x-2y=4\end{array}\end{array}$

Step 3:  Add the equations.

       $\frac{\begin{array}{c}-4x+2y=-2\\ 4x-2y=4\end{array}}{\begin{array}{c}0+0=2\\ 0=2\end{array}}$

 This is false and is therefore a contradiction. The lines of this system are parallel.  This system is inconsistent.

Solve  $\{\begin{array}{rrr}\hfill 4x+8y=8& \hfill & \hfill \left(1\right)\\ \hfill 3x+6y=6& \hfill & \hfill \left(2\right)\end{array}$

Step 1:  The equations are in the proper form.

Step 2:  We can eliminate $x$ by multiplying equation (1) by –3 and equation (2) by 4.

       $\begin{array}{rrrrr}\hfill \{\begin{array}{c}-3\left(4x+8y\right)=-3\left(8\right)\\ 4\left(3x+6y\right)=4\left(6\right)\end{array}& \hfill & \hfill \to & \hfill & \hfill \{\begin{array}{c}-12x-24y=-24\\ 12x+24y=24\end{array}\end{array}$

Step 3:  Add the equations.

       $\frac{\begin{array}{c}-12x-24y=-24\\ 12x+24y=24\end{array}}{\begin{array}{c}0+0=0\\ 0=0\end{array}}$

 This is true and is an identity. The lines of this system are coincident.

 This system is dependent.