# 11.3 Elimination by addition  (Page 2/2)

 Page 2 / 2

## Practice set a

Solve each system by addition.

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

$\left(2,4\right)$

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

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

## 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 $a$ .

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

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.

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

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.

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

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

## 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.

## Addition and parallel lines

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

## Addition and coincident lines

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.

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

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

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.

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

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.

## 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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