# 11.3 Elimination by addition  (Page 2/2)

 Page 2 / 2

## Practice set a

$\left\{\begin{array}{l}x+y=6\\ 2x-y=0\end{array}$

$\left(2,4\right)$

$\left\{\begin{array}{l}x+6y=8\\ -x-2y=0\end{array}$

$\left(-4,2\right)$

## Sample set b

Solve the following systems using the addition method.

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

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 $a$ .

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

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

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.

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

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

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

Step 1:  Rewrite the system in the proper form.

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

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.

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

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

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 $\left(0,-2\right).$

## Practice set b

Solve each of the following systems using the addition method.

$\left\{\begin{array}{l}3x+y=1\\ 5x+y=3\end{array}$

$\left(1,-2\right)$

$\left\{\begin{array}{l}x+4y=1\\ x-2y=-5\end{array}$

$\left(-3,1\right)$

$\left\{\begin{array}{l}2x+3y=-10\\ -x+2y=-2\end{array}$

$\left(-2,-2\right)$

$\left\{\begin{array}{l}5x-3y=1\\ 8x-6y=4\end{array}$

$\left(-1,-2\right)$

$\left\{\begin{array}{l}3x-5y=9\\ 4x+8y=12\end{array}$

$\left(3,0\right)$

## Addition and parallel or coincident lines

When the lines of a system are parallel or coincident, the method of elimination produces results identical to that of the method of elimination by substitution.

If computations eliminate all variables and produce a contradiction, the two lines of the system are parallel and the system is called inconsistent.

If computations eliminate all variables and produce an identity, the two lines of the system are coincident and the system is called dependent.

## Sample set c

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

Step 1: The equations are in the proper form.

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

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

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

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

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

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.

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

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

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

This system is dependent.

## Practice set c

Solve each of the following systems using the addition method.

$\left\{\begin{array}{c}-x+2y=6\\ -6x+12y=1\end{array}$

inconsistent

$\left\{\begin{array}{c}4x-28y=-4\\ x-7y=-1\end{array}$

dependent

## Exercises

For the following problems, solve the systems using elimination by addition.

$\left\{\begin{array}{c}x+y=11\\ x-y=-1\end{array}$

$\left(5,6\right)$

$\left\{\begin{array}{c}x+3y=13\\ x-3y=-11\end{array}$

$\left\{\begin{array}{c}3x-5y=-4\\ -4x+5y=2\end{array}$

$\left(2,2\right)$

$\left\{\begin{array}{c}2x-7y=1\\ 5x+7y=-22\end{array}$

$\left\{\begin{array}{c}-3x+4y=-24\\ 3x-7y=42\end{array}$

$\left(0,-6\right)$

$\left\{\begin{array}{c}8x+5y=3\\ 9x-5y=-71\end{array}$

$\left\{\begin{array}{c}-x+2y=-6\\ x+3y=-4\end{array}$

$\left(2,-2\right)$

$\left\{\begin{array}{c}4x+y=0\\ 3x+y=0\end{array}$

$\left\{\begin{array}{c}x+y=-4\\ -x-y=4\end{array}$

dependent

$\left\{\begin{array}{c}-2x-3y=-6\\ 2x+3y=6\end{array}$

$\left\{\begin{array}{c}3x+4y=7\\ x+5y=6\end{array}$

$\left(1,1\right)$

$\left\{\begin{array}{c}4x-2y=2\\ 7x+4y=26\end{array}$

$\left\{\begin{array}{c}3x+y=-4\\ 5x-2y=-14\end{array}$

$\left(-2,2\right)$

$\left\{\begin{array}{c}5x-3y=20\\ -x+6y=-4\end{array}$

$\left\{\begin{array}{c}6x+2y=-18\\ -x+5y=19\end{array}$

$\left(-4,3\right)$

$\left\{\begin{array}{c}x-11y=17\\ 2x-22y=4\end{array}$

$\left\{\begin{array}{c}-2x+3y=20\\ -3x+2y=15\end{array}$

$\left(-1,6\right)$

$\left\{\begin{array}{c}-5x+2y=-4\\ -3x-5y=10\end{array}$

$\left\{\begin{array}{c}-3x-4y=2\\ -9x-12y=6\end{array}$

dependent

$\left\{\begin{array}{c}3x-5y=28\\ -4x-2y=-20\end{array}$

$\left\{\begin{array}{c}6x-3y=3\\ 10x-7y=3\end{array}$

$\left(1,\text{\hspace{0.17em}}1\right)$

$\left\{\begin{array}{c}-4x+12y=0\\ -8x+16y=0\end{array}$

$\left\{\begin{array}{c}3x+y=-1\\ 12x+4y=6\end{array}$

inconsistent

$\left\{\begin{array}{c}8x+5y=-23\\ -3x-3y=12\end{array}$

$\left\{\begin{array}{c}2x+8y=10\\ 3x+12y=15\end{array}$

dependent

$\left\{\begin{array}{c}4x+6y=8\\ 6x+8y=12\end{array}$

$\left\{\begin{array}{c}10x+2y=2\\ -15x-3y=3\end{array}$

inconsistent

$\left\{\begin{array}{c}x+\frac{3}{4}y=-\frac{1}{2}\\ \frac{3}{5}x+y=-\frac{7}{5}\end{array}$

$\left\{\begin{array}{c}x+\frac{1}{3}y=\frac{4}{3}\\ -x+\frac{1}{6}y=\frac{2}{3}\end{array}$

$\left(0,4\right)$

$\left\{\begin{array}{c}8x-3y=25\\ 4x-5y=-5\end{array}$

$\left\{\begin{array}{c}-10x-4y=72\\ 9x+5y=39\end{array}$

$\left(-\frac{258}{7},\frac{519}{7}\right)$

$\left\{\begin{array}{c}12x+16y=-36\\ -10x+12y=30\end{array}$

$\left\{\begin{array}{c}25x-32y=14\\ -50x+64y=-28\end{array}$

dependent

## Exercises for review

( [link] ) Simplify and write ${\left(2{x}^{-3}{y}^{4}\right)}^{5}{\left(2x{y}^{-6}\right)}^{-5}$ so that only positive exponents appear.

( [link] ) Simplify $\sqrt{8}+3\sqrt{50}.$

$17\sqrt{2}$

( [link] ) Solve the radical equation $\sqrt{2x+3}+5=8.$

( [link] ) Solve by graphing $\left\{\begin{array}{c}x+y=4\\ 3x-y=0\end{array}$ $\left(1,3\right)$ ( [link] ) Solve using the substitution method: $\left\{\begin{array}{c}3x-4y=-11\\ 5x+y=-3\end{array}$

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