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Step 1: Write both equations in standard form. If any coefficients are fractions, clear them.	Both equations are in standard form, $A x + B y = C$ . There are no fractions.	$\begin{cases} 2 x + y = 7 \\ x - 2 y = 6 \end{cases}$
Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites.	We can eliminate the $y$ ’s by multiplying the first equation by 2. Multiply both sides of $2 x + y = 7$ by 2.	$\begin{cases} 2 x + \mathbf{\mathit{y}} = 7 \\ x - 2 \mathbf{\mathit{y}} = 6 \end{cases}$ $\begin{cases} 2 \left(2 x + y\right) & = & 2 \left(7\right) \\ x - 2 y & = & 6 \end{cases}$
Step 3: Add the equations resulting from Step 2 to eliminate one variable.	We add the $x$ ’s, $y$ ’s, and constants.	$\begin{matrix}\begin{cases} 4 x + 2 y = 14 \\ \underset{\underline}{x - 2 y = 6 } \end{cases} \\ 5 x = 20 \end{matrix}$
Step 4: Solve for the remaining variable.	Solve for $x$ .	$x = 4$
Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.	Substitute $x = 4$ into the second equation, $x - 2 y = 6$ . Then solve for $y$ .	$\begin{aligned}x - 2 y & = & 6 \\ 4 - 2 y & = & 6 \\ - 2 y & = & 2 \\ y & = & - 1\end{aligned}$
Step 6: Write the solution as an ordered pair.	Write it as $\left(x , y\right)$ .	$\left(4 , - 1\right)$
Step 7: Check that the ordered pair is a solution to both original equations.	Substitute $x = 4$ , $y = - 1$ into $2 x + y = 7$ and $x - 2 y = 6$ . Do they make both equations true? Yes!	$\begin{matrix}\begin{matrix}2 x + y & = & 7 \\ 2 \left(4\right) + \left(- 1\right) & \overset{?}{=} & 7 \\ 7 & = & 7 \checkmark\end{matrix} & \begin{matrix}x - 2 y & = & 6 \\ 4 - 2 \left(- 1\right) & \overset{?}{=} & 6 \\ 6 & = & 6 \checkmark\end{matrix}\end{matrix}$ The solution is $\left(4 , - 1\right)$ .

58:T9bd,

Step 1: Write both equations in standard form. If any coefficients are fractions, clear them.	Both equations are in standard form, $A x + B y = C$. There are no fractions.	$\begin{cases} 2 x + y = 7 \\ x - 2 y = 6 \end{cases}$
Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites.	We can eliminate the $y$’s by multiplying the first equation by 2. Multiply both sides of $2 x + y = 7$ by 2.	$\begin{cases} 2 x + \mathbf{\mathit{y}} = 7 \\ x - 2 \mathbf{\mathit{y}} = 6 \end{cases}$ $\begin{cases} 2 \left(2 x + y\right) & = & 2 \left(7\right) \\ x - 2 y & = & 6 \end{cases}$
Step 3: Add the equations resulting from Step 2 to eliminate one variable.	We add the $x$’s, $y$’s, and constants.	$\begin{matrix}\begin{cases} 4 x + 2 y = 14 \\ \underset{\underline}{x - 2 y = 6 } \end{cases} \\ 5 x = 20 \end{matrix}$
Step 4: Solve for the remaining variable.	Solve for $x$.	$x = 4$
Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.	Substitute $x = 4$ into the second equation, $x - 2 y = 6$. Then solve for $y$.	$\begin{aligned}x - 2 y & = & 6 \\ 4 - 2 y & = & 6 \\ - 2 y & = & 2 \\ y & = & - 1\end{aligned}$
Step 6: Write the solution as an ordered pair.	Write it as $\left(x , y\right)$.	$\left(4 , - 1\right)$
Step 7: Check that the ordered pair is a solution to both original equations.	Substitute $x = 4$, $y = - 1$ into $2 x + y = 7$ and $x - 2 y = 6$. Do they make both equations true? Yes!	$\begin{matrix}\begin{matrix}2 x + y & = & 7 \\ 2 \left(4\right) + \left(- 1\right) & \overset{?}{=} & 7 \\ 7 & = & 7 \checkmark\end{matrix} & \begin{matrix}x - 2 y & = & 6 \\ 4 - 2 \left(- 1\right) & \overset{?}{=} & 6 \\ 6 & = & 6 \checkmark\end{matrix}\end{matrix}$ The solution is $\left(4 , - 1\right)$ .

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Solve this system of linear equations by elimination:

\\begin{cases} 2 x + y = 7 \\\\ x - 2 y = 6 \\end{cases}

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Step 1: Write both equations in standard form. If any coefficients are fractions, clear them.	Both equations are in standard form, \(A x + B y = C\). There are no fractions.	\(\begin{cases} 2 x + y = 7 \\ x - 2 y = 6 \end{cases}\)
Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites.	We can eliminate the \(y\)’s by multiplying the first equation by 2. Multiply both sides of \(2 x + y = 7\) by 2.	\(\begin{cases} 2 x + \mathbf{\mathit{y}} = 7 \\ x - 2 \mathbf{\mathit{y}} = 6 \end{cases}\) \(\begin{cases} 2 \left(2 x + y\right) & = & 2 \left(7\right) \\ x - 2 y & = & 6 \end{cases}\)
Step 3: Add the equations resulting from Step 2 to eliminate one variable.	We add the \(x\)’s, \(y\)’s, and constants.	\(\begin{matrix}\begin{cases} 4 x + 2 y = 14 \\ \underset{\underline}{x - 2 y = 6 } \end{cases} \\ 5 x = 20 \end{matrix}\)
Step 4: Solve for the remaining variable.	Solve for \(x\).	\(x = 4\)
Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.	Substitute \(x = 4\) into the second equation, \(x - 2 y = 6\). Then solve for \(y\).	\(\begin{aligned}x - 2 y & = & 6 \\ 4 - 2 y & = & 6 \\ - 2 y & = & 2 \\ y & = & - 1\end{aligned}\)
Step 6: Write the solution as an ordered pair.	Write it as \(\left(x , y\right)\).	\(\left(4 , - 1\right)\)
Step 7: Check that the ordered pair is a solution to both original equations.	Substitute \(x = 4\), \(y = - 1\) into \(2 x + y = 7\) and \(x - 2 y = 6\). Do they make both equations true? Yes!	\(\begin{matrix}\begin{matrix}2 x + y & = & 7 \\ 2 \left(4\right) + \left(- 1\right) & \overset{?}{=} & 7 \\ 7 & = & 7 \checkmark\end{matrix} & \begin{matrix}x - 2 y & = & 6 \\ 4 - 2 \left(- 1\right) & \overset{?}{=} & 6 \\ 6 & = & 6 \checkmark\end{matrix}\end{matrix}\) The solution is \(\left(4 , - 1\right)\) .