Addition Method Practice

We know that we can replace one of the equations with the sum of the two equations and obtain an equivalent system that has the same solution.

If, by doing that, we obtain an equation with a single variable, we will be able to find the solution for that variable. This is called the elimination method.

We can see that the coefficients of the variable $y$
have opposite signs in the two equations. By adding the equations, we can eliminate ‍$y$ as follows:

$\begin{aligned} 13x+{2y} &= 1 \\{+}5x{-2y}&=-19\\\hline\\18x+0 &=-18 \end{aligned}$

Now we can solve the obtained equation for $x$:

$\begin{aligned} 18x&=-18\\\\x&=-1\end{aligned}$

Since we know that ‍$x=-1$, we can substitute this into one of the original equations and solve for ‍‍$y$. Let's use the equation ‍‍$13x+2y=1$:

$\begin{aligned} 13{x}+2y&=1\\\\13\cdot({-1})+2y&=1\\\\-13+2y& = 1\\\\2y&=14\\\\y&=7\end{aligned}$

You can try using the other equation yourself and see that you obtain the same solution!

This is the solution of the system:

$\begin{aligned}&x = -1\\\\&y = 7\end{aligned}$

We know that we can replace one of the equations with the sum of the two equations and obtain an equivalent system that has the same solution.

If, by doing that, we obtain an equation with a single variable, we will be able to find the solution for that variable. This is called the elimination method.

We can see that the coefficients of the variable $x$
have opposite signs in the two equations. By adding the equations, we can eliminate ‍$x$ as follows:

$\begin{aligned} {-2x}+{15y} &= -24 \\{+}{2x}+{9y}&=24\\\hline\\0+24y &=0 \end{aligned}$

Now we can solve the obtained equation for $y$:

$\begin{aligned} 24y&=0\\\\y&=0\end{aligned}$

Since we know that ‍$y=0$, we can substitute this into one of the original equations and solve for ‍‍$x$. Let's use the equation ‍‍$2x+9y=24$:

$\begin{aligned} 2x+9y&=24\\\\2x+9\cdot{0}&=24\\\\2x+0& = 24\\\\2x&=24\\\\x&=12\end{aligned}$

You can try using the other equation yourself and see that you obtain the same solution!

This is the solution of the system:

$\begin{aligned}&x = 12\\\\&y = 0\end{aligned}$

We know that we can replace one of the equations with the sum of the two equations and obtain an equivalent system that has the same solution.

If, by doing that, we obtain an equation with a single variable, we will be able to find the solution for that variable. This is called the elimination method.

We can see that the coefficients of the variable $y$
have opposite signs in the two equations. By adding the equations, we can eliminate ‍$y$ as follows:

$\begin{aligned} {-5y}+8x &= -18 \\{+}{5y}+{2x}&=58\\\hline\\0+10x &=40 \end{aligned}$

Now we can solve the obtained equation for $x$:

$\begin{aligned} 10x&=40\\\\x&=4\end{aligned}$

Since we know that ‍$x=4$, we can substitute this into one of the original equations and solve for ‍‍$y$. Let's use the equation ‍‍$-5y+8x=-18$:

$\begin{aligned} -5y+8{x}&=-18\\\\-5y+8\cdot{4}&=-18\\\\-5y+32& = -18\\\\-5y&=-50\\\\y&=10\end{aligned}$

You can try using the other equation yourself and see that you obtain the same solution!

This is the solution of the system:

$\begin{aligned}&x = 4\\\\&y = 10\end{aligned}$

We know that we can replace one of the equations with the sum of the two equations and obtain an equivalent system that has the same solution.

If, by doing that, we obtain an equation with a single variable, we will be able to find the solution for that variable. This is called the elimination method.

We can see that the coefficients of the variable $x$
have opposite signs in the two equations. By adding the equations, we can eliminate ‍$x$ as follows:

$\begin{aligned}5y{-4x} &=-7\\{+}2y+{4x} &=14\\\hline\\7y+0&=7\end{aligned}$

Now we can solve the obtained equation for $y$:

$\begin{aligned} 7y&=7\\\\y&=1\end{aligned}$

Since we know that ‍$y=1$, we can substitute this into one of the original equations and solve for ‍‍$x$. Let's use the equation ‍‍$5y-4x=-7$:

$\begin{aligned} 5{y}-4x&=-7\\\\5\cdot{1}-4x&=-7\\\\5-4x& = -7\\\\-4x&=-12\\\\x&=3\end{aligned}$

You can try using the other equation yourself and see that you obtain the same solution!

This is the solution of the system:

$\begin{aligned}&x = 3\\\\&y = 1\end{aligned}$