Multiplication with Addition Method Practice

1. Solve the system of equations.

   $\begin{aligned}&10y+7x = 29\\\\&-5y-9x=2\end{aligned}$

   $x=$____

   $y=$____
   

2. Solve the system of equations.

   $\begin{aligned}&-3y+4x = 13\\\\&-5y-6x=-67\end{aligned}$

   $x=$____

   $y=$____
   

3. Solve the system of equations.

   $\begin{aligned}&14x+5y = 31\\\\&2x-3y=-29\end{aligned}$

   $x=$____

   $y=$____
   

4. Solve the system of equations.

   $\begin{aligned}&-4y+11x - 67=0\\\\&2y+5x-19=0\end{aligned}$

   $x=$____

   $y=$____
   

Source: Khan Academy, https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-equations/x2f8bb11595b61c86:solving-systems-elimination/e/systems_of_equations_with_elimination
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1. Let's solve the system by using the elimination method. In order to eliminate one of the variables, we need to manipulate the equations so that variable has coefficients of the same size but opposite signs in each equation.

   Let's take a look at the system:

   $\begin{aligned}&10y+7x = 29\\\\&-5y-9x=2\end{aligned}$

   The coefficient of ‍$y$ in the first equation, 10, is exactly ‍-2 times the coefficient of ‍$y$ in the second equation, ‍-5. Therefore, we can multiply the second equation by ‍2 in order to eliminate $y$.

   $\begin{aligned} {2}\cdot(-5y)- {2}\cdot9x&= {2}\cdot2\\\\-10y-18x&=4\end{aligned}$
   

   Now we can eliminate $y$ as follows:

   $\begin{aligned} {10y}+7x &= 29 \\\\ {+}\ {-10y}-18x&=4\\\hline\\0-11x &=33\end{aligned}$
   

   When we solve the resulting equation we obtain that $x = -3$. Then, we can substitute this into one of the original equations and solve for ‍$y$ to obtain $y=5$‍.
   

   This is the solution of the system:

   $\begin{aligned}&x=-3\\\\&y=5\end{aligned}$

   ---

2. Let's solve the system by using the elimination method. In order to eliminate one of the variables, we need to manipulate the equations so that variable has coefficients of the same size but opposite signs in each equation.

   Let's take a look at the system:

   $\begin{aligned}&-3y+4x = 13\\\\&-5y-6x=-67\end{aligned}$

   Since the coefficients of either variable aren't multiples of each other, we must multiply both equations in order to eliminate a variable of our choice. Let's eliminate $y$.

   Since the least common multiple of the coefficients of $y$ in the two equations is -15, we can multiply the first equation by ‍5 and the second equation by ‍-3 in order to eliminate $y$.
   

   $\begin{aligned} {5}\!\cdot\!(-3y)+ {5}\!\cdot\!4x&= {5}\cdot13\\\\-15y+20x&=65\end{aligned}$

   $\begin{aligned} {-3}\!\cdot\!(-5y)-( {-3})\!\cdot\!6x&= {-3}\cdot(-67)\\\\15y+18x&=201\end{aligned}$

   Now we can eliminate $y$ as follows:

   $\begin{aligned} {-15y}+20x &= 65 \\\\ {+}\ {15y}+18x&=201\\\hline\\0+38x &=266\end{aligned}$
   

   When we solve the resulting equation we obtain that $x = 7$. Then, we can substitute this into one of the original equations and solve for ‍$y$ to obtain $y=5$‍ .

   This is the solution of the system:

   $\begin{aligned}&x=7\\\\&y=5\end{aligned}$

   ---

3. Let's solve the system by using the elimination method. In order to eliminate one of the variables, we need to manipulate the equations so that variable has coefficients of the same size but opposite signs in each equation.

   Let's take a look at the system:

   $\begin{aligned}&14x+5y = 31\\\\&2x-3y=-29\end{aligned}$

   The coefficient of ‍$x$ in the first equation, 14, is exactly ‍7 times the coefficient of ‍$x$ in the second equation, ‍2. Therefore, we can multiply the second equation by ‍-7 in order to eliminate $x$.

   $\begin{aligned} {-7}\cdot2x-( {-7})\cdot3y&= {-7}\cdot(-29)\\\\-14x+21y&=203\end{aligned}$
   

   Now we can eliminate $x$ as follows:

   $\begin{aligned} {14x}+5y &= 31 \\\\ {+}\ {-14x}+21y&=203\\\hline\\0+26y &=234\end{aligned}$

   When we solve the resulting equation we obtain that $y = 9$. Then, we can substitute this into one of the original equations and solve for ‍$x$ to obtain $x=-1$‍ .

   This is the solution of the system:

   $\begin{aligned}&x=-1\\\\&y=9\end{aligned}$

   ---

4. Let's solve the system by using the elimination method. In order to eliminate one of the variables, we need to manipulate the equations so that variable has coefficients of the same size but opposite signs in each equation.

   Let's take a look at the system:

   $\begin{aligned}&-4y+11x - 67=0\\\\&2y+5x-19=0\end{aligned}$

   The coefficient of ‍$y$ in the first equation, -4, is exactly ‍-2 times the coefficient of ‍$y$ in the second equation, ‍2. Therefore, we can multiply the second equation by ‍2 in order to eliminate $y$.

   $\begin{aligned} {2}\cdot2y+ {2}\cdot5x- {2}\cdot19&=0\\\\4y+10x-38&=0\end{aligned}$
   

   Now we can eliminate $y$ as follows:

   $\begin{aligned} {-4y}+11x - 67&=0 \\\\ {+}\ {4y}+10x-38&=0\\\hline\\0+21x -105&=0\end{aligned}$
   

   When we solve the resulting equation we obtain that $x = 5$. Then, we can substitute this into one of the original equations and solve for ‍$y$ to obtain $y=-3$‍.
   

   This is the solution of the system:

   $\begin{aligned}&x=5\\\\&y=-3\end{aligned}$