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The substitution method is a technique used to solve systems of equations, especially when you have a system of linear equations. This involves solving one of the equations for one variable and then substituting that expression into the other equation. It helps to reduce the system to one equation with one unknown, simplifying the process.
For example, in our exercise, we first solve one equation for 'x'. After isolating 'x' in the first equation as \( x = 3y - 17 \), we substitute this expression into the second equation. With this step, we were able to eliminate 'x' from the second equation, allowing us to solve for 'y' directly.
By substituting back, we ensure that each variable satisfies both equations in the original system, leading us to the correct solution for both 'x' and 'y'. This method is effective for systems where one equation is easy to manipulate, making it ideal for linear equations.

Linear equations are polynomial equations with the highest degree of one. They produce a straight line graphically and have the general form \( ax + by = c \), where 'a', 'b', and 'c' are constants.
In the context of solving systems, linear equations represent lines. When solving two linear equations together, such as:

• \(-x + 3y = 17\)
• \(4x + 3y = 7\)

We are looking for the point where these two lines intersect. This point represents the values of 'x' and 'y' that satisfy both equations simultaneously.
Understanding the nature of linear equations is crucial because it allows us to leverage geometric insights into finding solutions. Depending on how the lines intersect (either one point, parallel with no intersection, or the same line), we determine if there is a single solution, no solution, or infinitely many solutions.

Solving equations is a fundamental skill in algebra, involving finding the values of variables that make the equation true. For linear equations, this often involves rearranging terms to isolate the variable of interest.
Consider the steps taken in solving for 'y' in our exercise. The substitution led to an equation where 'y' could be isolated: \( 15y - 68 = 7 \). We handled it by balancing operations - adding 68 to both sides and then dividing by 15.
This process is systematic and requires careful operations on both sides of the equation. It ensures that the equality is maintained while progressively simplifying until the variable is isolated. Once one variable is found, substitute it back into the earlier derived equations to solve for the other variables, as we did by finding 'x' after determining 'y'.
Always double-check your solutions by plugging the values back into the original equations to confirm they satisfy all given conditions. This ensures accuracy and provides confidence in the derived solution.