System of Equations Notes

Objective: Solve and graph a system of equations.
Definition: Solving a system of equations means finding the x and y values that satisfy both equations.
Method Overview: Use one equation to solve for either x or y and substitute that value into the other equation.

Isolate x:
Subtract 15 from both sides:
[ 12x = 0 ]
Solve for x:
Divide by 12:
[ x = 0 ]
Finding y:
Substitute ( x = 0 ) back into either original equation:
- Using ( 9(0) + 3y = 15 ):
  [ 3y = 15 ]
  [ y = 5 ]
Results:
Solution: ( x = 0, y = 5 ).
Verifies both equations: ( y - x = 5 ) holds true.

Intersecting Point:
Solution occurs at the intersection of the two graphed lines.
Confirmed through algebraic substitution showing ( x = 0, y = 5 ).
Interpretation:
Intersection corresponds to the values that satisfy both original equations, represented graphically.

Objective: To effectively solve and graph a system of equations using algebraic methods.
Definition: Solving a system of equations involves finding the unique values for x and y that satisfy both equations simultaneously. Systems can consist of linear equations, which typically form straight lines when graphed.
Method Overview: Employ the substitution method by manipulating one equation to express either x or y, and then substitute that expression into the other equation for resolution.

Starting Point: The given equation is ( y - x = 5 ). This represents a linear relationship between x and y.
Solving for y:
To express y in terms of x, add x to both sides:
[ y = 5 + x ]
Applying Substitution:
Use the expression for y in the second equation: ( 9x + 3y = 15 ). Substitute for y:
[ 9x + 3(5 + x) = 15 ]
Expansion and Simplification:
Distribute to eliminate the parentheses:
[ 9x + 15 + 3x = 15 ]
Combine like terms:
[ 12x + 15 = 15 ]

Isolate x:
To solve for x, first reduce the equation by subtracting 15 from both sides:
[ 12x = 0 ]
Solve for x:
Next, divide both sides by 12 to find:
[ x = 0 ]
Finding y:
Substitute the value obtained for x back into either of the original equations.
Using the first equation ( 9(0) + 3y = 15 ), resolve for y:
[ 3y = 15 ]
[ y = 5 ]

Results:
The system of equations has been solved, yielding the solution point: ( x = 0, y = 5 ).
To verify the correctness of this solution, substitute the values back into both original equations to ensure they hold true.

Graphing:
To analyze the second equation further, rewrite it in slope-intercept form ( y = mx + b), which facilitates the graphing process and interpretation of the slope and intercept values.
Transform the original equation from ( y - x = 5 ) to:
[ y = x + 5 ].
Identify slope & y-intercept:
From the equation ( y = x + 5 ), extract the slope (m) which is equal to 1, indicating a positive diagonal line.
The y-intercept occurs at the coordinate (0, 5), signifying the point where the line crosses the y-axis.

Drawing the Graph:
Begin the graphing process by plotting the y-intercept at (0, 5).
From this point, utilize the slope to move: one unit right on the x-axis and one unit up on the y-axis to locate another point on the graph.
Draw a straight line through these defined points to represent the equation ( y = x + 5 ).

Graphing the first equation:
The original equation to be graphed is ( 9x + 3y = 15 ).
Simplifying this equation is essential: divide all terms by 3 to achieve:
[ 3x + y = 5 ].
Rewrite in slope-intercept form:
Then, rewrite the simplified equation into slope-intercept form:
[ y = -3x + 5 ].
Identify slope & y-intercept:
Analyze the new slope (m) which is -3, indicating a downward slope.
The y-intercept is at (0, 5), similar to the first equation.
Graph the second line:
Starting from the y-intercept (0, 5), move left one unit and down three units to plot another point, depicting how the line decreases as x increases, reflecting all combinations satisfying the equation ( y = -3x + 5 ).

Intersecting Point:
The solution to the system of equations can be visualized as the intersection point of the two graphed lines, which occurs at the coordinates ( x = 0, y = 5 ).
This intersection has been confirmed through algebraic substitution, demonstrating that it satisfies both equations.
Interpretation:
The intersection point is significant as it represents the unique values for x and y that satisfy both original equations, providing valuable insight into the relationship between the two linear equations in the graph.