91Ó°ÊÓ

The Substitution Method is a handy tool when solving systems of linear equations. It involves solving one equation for a single variable, and then substituting this expression into the other equation. The idea is to simplify the system into one equation with a single variable. In our exercise, we solved the first equation for \( y \), which gave us \( y = 3x - 7 \). This expression for \( y \) was then substituted into the second equation \( x + y = 5 \). By doing this, we replace \( y \) in the second equation, which leaves us with one equation involving only \( x \). This reduces the complexity of the original system, making it easier to solve for \( x \) and subsequently \( y \). The method is particularly beneficial when one equation is already solved for a single variable, like \( y = 3x - 7 \). This allows us to quickly substitute and proceed to finding our values.

Algebraic equations in a system like the one we have consist of relationships between variables expressed with equal signs. These equations involve various operations such as addition, subtraction, and multiplication which help in forming the relationships between the unknowns. In our problem, we have two linear equations: \( y = 3x - 7 \) and \( x + y = 5 \).

• Linear equations represent straight lines when graphed on a coordinate plane.
• The coefficients (like the 3 in \( 3x \) and the \(-7\)) define the slope and position of these lines.
• Understanding the structure of algebraic equations helps in rearranging and solving them.

In our system, the substitution and algebraic manipulation helped step-by-step to isolate and solve for the unknowns \( x \) and \( y \). Once an equation is simplified, it can be solved using basic arithmetic operations.

A solution of a system of linear equations is an ordered pair \((x, y)\) that satisfies both equations at the same time. The purpose of solving such systems is to find this unique point of intersection of the lines represented by the equations.

• Our original system was: \( y = 3x - 7 \) and \( x + y = 5 \).
• Using substitution, we simplified the system to find \( x = 3 \).
• We then substituted \( x = 3 \) back into the first equation to find \( y = 2 \).

Thus, the solution to our system is \((3, 2)\), meaning that this coordinate works for both equations and represents the intersection point of the lines \( y = 3x - 7 \) and \( x + y = 5 \) on a graph. Solving systems like these is a critical skill in algebra, as it provides insights into both graphical and algebraic problem-solving techniques.