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Solve each system by addition.
$\{\begin{array}{l}x+y=6\\ 2x-y=0\end{array}$
$\left(2,4\right)$
$\{\begin{array}{l}x+6y=8\\ -x-2y=0\end{array}$
$\left(-4,2\right)$
Solve the following systems using the addition method.
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 each of the following systems using the addition method.
$\{\begin{array}{l}3x+y=1\\ 5x+y=3\end{array}$
$\left(1,-2\right)$
$\{\begin{array}{l}x+4y=1\\ x-2y=-5\end{array}$
$\left(-3,1\right)$
$\{\begin{array}{l}2x+3y=-10\\ -x+2y=-2\end{array}$
$\left(-2,-2\right)$
$\{\begin{array}{l}5x-3y=1\\ 8x-6y=4\end{array}$
$\left(-1,-2\right)$
$\{\begin{array}{l}3x-5y=9\\ 4x+8y=12\end{array}$
$\left(3,0\right)$
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.
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.
Solve each of the following systems using the addition method.
$\{\begin{array}{c}-x+2y=6\\ -6x+12y=1\end{array}$
inconsistent
$\{\begin{array}{c}4x-28y=-4\\ x-7y=-1\end{array}$
dependent
For the following problems, solve the systems using elimination by addition.
$\{\begin{array}{c}x+y=11\\ x-y=-1\end{array}$
$\left(5,6\right)$
$\{\begin{array}{c}x+3y=13\\ x-3y=-11\end{array}$
$\{\begin{array}{c}3x-5y=-4\\ -4x+5y=2\end{array}$
$\left(2,2\right)$
$\{\begin{array}{c}2x-7y=1\\ 5x+7y=-22\end{array}$
$\{\begin{array}{c}-3x+4y=-24\\ 3x-7y=42\end{array}$
$\left(0,-6\right)$
$\{\begin{array}{c}8x+5y=3\\ 9x-5y=-71\end{array}$
$\{\begin{array}{c}-x+2y=-6\\ x+3y=-4\end{array}$
$\left(2,-2\right)$
$\{\begin{array}{c}4x+y=0\\ 3x+y=0\end{array}$
$\{\begin{array}{c}-2x-3y=-6\\ 2x+3y=6\end{array}$
$\{\begin{array}{c}3x+4y=7\\ x+5y=6\end{array}$
$\left(1,1\right)$
$\{\begin{array}{c}4x-2y=2\\ 7x+4y=26\end{array}$
$\{\begin{array}{c}3x+y=-4\\ 5x-2y=-14\end{array}$
$\left(-2,2\right)$
$\{\begin{array}{c}5x-3y=20\\ -x+6y=-4\end{array}$
$\{\begin{array}{c}6x+2y=-18\\ -x+5y=19\end{array}$
$\left(-4,3\right)$
$\{\begin{array}{c}x-11y=17\\ 2x-22y=4\end{array}$
$\{\begin{array}{c}-2x+3y=20\\ -3x+2y=15\end{array}$
$\left(-1,6\right)$
$\{\begin{array}{c}-5x+2y=-4\\ -3x-5y=10\end{array}$
$\{\begin{array}{c}-3x-4y=2\\ -9x-12y=6\end{array}$
dependent
$\{\begin{array}{c}3x-5y=28\\ -4x-2y=-20\end{array}$
$\{\begin{array}{c}6x-3y=3\\ 10x-7y=3\end{array}$
$\left(1,\text{\hspace{0.17em}}1\right)$
$\{\begin{array}{c}-4x+12y=0\\ -8x+16y=0\end{array}$
$\{\begin{array}{c}3x+y=-1\\ 12x+4y=6\end{array}$
inconsistent
$\{\begin{array}{c}8x+5y=-23\\ -3x-3y=12\end{array}$
$\{\begin{array}{c}2x+8y=10\\ 3x+12y=15\end{array}$
dependent
$\{\begin{array}{c}4x+6y=8\\ 6x+8y=12\end{array}$
$\{\begin{array}{c}10x+2y=2\\ -15x-3y=3\end{array}$
inconsistent
$\{\begin{array}{c}x+\frac{3}{4}y=-\frac{1}{2}\\ \frac{3}{5}x+y=-\frac{7}{5}\end{array}$
$\{\begin{array}{c}x+\frac{1}{3}y=\frac{4}{3}\\ -x+\frac{1}{6}y=\frac{2}{3}\end{array}$
$\left(0,4\right)$
$\{\begin{array}{c}8x-3y=25\\ 4x-5y=-5\end{array}$
$\{\begin{array}{c}-10x-4y=72\\ 9x+5y=39\end{array}$
$\left(-\frac{258}{7},\frac{519}{7}\right)$
$\{\begin{array}{c}12x+16y=-36\\ -10x+12y=30\end{array}$
$\{\begin{array}{c}25x-32y=14\\ -50x+64y=-28\end{array}$
dependent
( [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] ) Solve the radical equation $\sqrt{2x+3}+5=8.$
(
[link] ) Solve by graphing
$\{\begin{array}{c}x+y=4\\ 3x-y=0\end{array}$
$\left(1,3\right)$
( [link] ) Solve using the substitution method: $\{\begin{array}{c}3x-4y=-11\\ 5x+y=-3\end{array}$
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