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The substitution method is a technique for solving systems of linear equations. Start by solving one of the equations for one variable. For example, if you have the system: \(\begin{aligned} 4x + 5y &= 5 \ 2x + 3y &= 1 \end{aligned} \) \( \ \ \)First, solve the second equation for \(x\): \(2x + 3y = 1 \ 2x = 1 - 3y \ x = \frac{1 - 3y}{2}\) \( \ \ \)Next, substitute the expression for \(x\) into the first equation: \( 4\left(\frac{1 - 3y}{2}\right) + 5y = 5\) \( \ \ \)Simplify and solve for \(y\): \(2(1 - 3y) + 5y = 5 \ 2 - 6y + 5y = 5 \ 2 - y = 5 \ -y = 3 \ y = -3\) \( \ \ \)Now that we've found \(y = -3\), substitute it back into the equation for \(x\): \( x = \frac{1 - 3(-3)}{2} \ x = \frac{1 + 9}{2} \ x = \frac{10}{2} \ x = 5\) \( \ \ \)Our solution is \((x, y) = (5, -3)\). This method helps to simplify the equations one variable at a time, making the solving process more straightforward.

Verification is a crucial step to ensure the solution is correct. Once you've found your values for \(x\) and \(y\), substitute them back into the original equations. For our system: \( \ \ \)The original equations are: \( 4x + 5y = 5 \ 2x + 3y = 1\) \( \ \ \)Substitute \(x = 5\) and \(y = -3\) into both equations: \( 4(5) + 5(-3) = 5\ 20 - 15 = 5\ 5 = 5\) \( \ \ \)And \( 2(5) + 3(-3) = 1\ 10 - 9 = 1\ 1 = 1\) \( \ \ \)Both equations hold true, confirming our solution \((5, -3)\). Verification ensures that your solution satisfies all equations in the system. This is a vital step to prevent errors and confirm the work.

Graphing linear equations is a useful way to visualize solutions. Plot each equation on a graph, and the point where they intersect is the solution to the system. For our example: \( \begin{aligned} 4x + 5y &= 5 \ 2x + 3y &= 1\end{aligned} \) \( \ \ \)Here's how to do it: \(\begin{aligned} y_1 &= \frac{5 - 4x}{5} \ y_2 &= \frac{1 - 2x}{3} \end{aligned} \) \( \ \ \)Plot \(y_1\) and \(y_2\) on a graphing calculator or a graphing tool. The intersection point of these lines will be \((5, -3)\), which is our solution. \( \ \ \)Using the graphing calculator: \

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• Enter both equations.\
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• Use the intersection feature.\
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• The calculator should show the point \((5, -3)\).\
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\Visually confirming the intersection supports our algebraic solution and provides double assurance of correctness. Graphing is beneficial for understanding concepts of linear systems and their solutions.