Comprehensive Study Notes: Algebra 1 Systems of Equations and Inequalities

Student Name: Sylvia Jasso (referenced as Sylvia Jasso and Sylvia Jos on different pages).
Total Course Assessments Documented: 4 Assessments.
Dates Recorded: * 11/22/2025 (Quiz 6.1-6.4) * 11/15/2005 (Quiz 6.1-6.2) * 1/29/2025 (Quiz 6.1-6.6) * 2/4/2026 (Test: Ch. 6)
Block Identification: Block D.
Scoring Summary: * Quiz: 6.1-6.4 Score: $19$ out of $24$ * Quiz: 6.1-6.2 Score: $18.5$ out of $20$ * Quiz: 6.1-6.6 Score: $23.5$ out of $24$ * Test: Ch. 6 Score: $33$ out of $43$

One Solution (Independent Solution): A system where the lines intersect at exactly one coordinate point $(x, y)$ . This occurs when the lines have different slopes.
No Solution (Inconsistent System): A system where the lines are parallel. They have the same slope but different $y$ -intercepts, so they never intersect.
Infinitely Many Solutions (Dependent System): A system where both equations represent the same line. They have the same slope and the same $y$ -intercept.

Objective: Graph both linear equations on a single coordinate plane and identify the point of intersection.
Case Study: System with One Solution * Equations: $y = x + 2$ and $y = 2x$ * Graphing results in an intersection at the point $(2, 4)$ .
Case Study: Infinitely Many Solutions * Equation 1: $3x - 3y = 9$ * Equation 2: $-y = 3 - x$ * By rewriting Equation 1: $3y = 3x - 9 \rightarrow y = x - 3$ * By rewriting Equation 2: $y = x - 3$ * Since the equations are identical, the result is "infinitely many solutions."
Case Study: Different Slopes * Equations: $y = -2x - 1$ and $y = x + 4$ * Graphing results in an intersection at $(-2, 2)$ .

Objective: Solve one equation for one variable, then replace that variable's expression in the second equation to solve for the remaining variable.
Procedural Example 1: * Equation 1: $x = 5y - 7$ * Equation 2: $-2x - 3y = -12$ * Step 1: Substitute Equation 1 into Equation 2: $-2(5y - 7) - 3y = -12$ * Step 2: Distribute: $-10y + 14 - 3y = -12$ * Step 3: Combine like terms: $-13y + 14 = -12$ * Step 4: Solve for $y$ : $-13y = -26 \rightarrow y = 2$ * Step 5: Solve for $x$ : $x = 5(2) - 7 \rightarrow x = 3$ * Final Solution: $(3, 2)$ .
Procedural Example 2: * Equation 1: $y = x - 1$ * Equation 2: $3x + 2y = 8$ * Substitute: $3x + 2(x - 1) = 8 \rightarrow 3x + 2x - 2 = 8 \rightarrow 5x = 10 \rightarrow x = 2$ * Find $y$ : $y = 2 - 1 = 1$ * Final Solution: $(2, 1)$ .

Objective: Add or subtract the equations to "eliminate" one variable, often requiring multiplication of one or both equations by a constant first.
Procedural Example 1: * Equation 1: $x + 4y = 1$ * Equation 2: $3x - 5y = -23$ * Multiply Equation 1 by $-3$ to align $x$ : $-3x - 12y = -3$ * Add to Equation 2: $-17y = -34 \rightarrow y = 2$ * Find $x$ : $x + 4(2) = 1 \rightarrow x + 8 = 1 \rightarrow x = -7$ * Final Solution: $(-7, 2)$ .
Procedural Example 2: * Equation 1: $4x + 3y = 8$ * Equation 2: $x - 2y = -9$ * Multiply Equation 2 by $-4$ : $-4x + 8y = 36$ * Add to Equation 1: $11y = 44 \rightarrow y = 4$ * Find $x$ : $x - 2(4) = -9 \rightarrow x - 8 = -9 \rightarrow x = -1$ * Final Solution: $(-1, 4)$ .

Graphing Requirements: * Use a Dashed Line for inequalities with $<$ or $>$ . * Use a Solid Line for inequalities with $\leq$ or $\geq$ .
Shading Rules: * Shade Above the line for $>$ or $\geq$ . * Shade Below the line for $<$ or $\leq$ .
Example 1: * $y \leq x + 5$ (Solid line, shade below) * $y > 3x + 2$ (Dashed line, shade above)
Example 2: * $y \leq -x - 2$ * $2x - 3y < 9 \rightarrow -3y < -2x + 9 \rightarrow y > \frac{2}{3}x - 3$
No Solution System: * $y < -x - 5$ * $y > -x + 5$ * Because the lines are parallel and the shaded regions move away from each other, there is no overlapping region of solutions.

Scenario: A number minus twice another number is $4$ . Three times the first number plus two times the second number is $12$ .
System Formulation: * $x - 2y = 4$ * $3x + 2y = 12$
Solution Process: Using elimination, * $(x - 2y) + (3x + 2y) = 4 + 12 \rightarrow 4x = 16 \rightarrow x = 4$ * Substituting back: $4 - 2y = 4 \rightarrow -2y = 0 \rightarrow y = 0$ * Numbers: $x = 4$ , $y = 0$ .

Scenario: One-fourth a number plus twice another number is negative one. The difference between the numbers is $5$ .
System Formulation: * $\frac{1}{4}x + 2y = -1$ * $x - y = 5$
Solution Process: Solve Equation 2 for $x$ : $x = y + 5$ . Substitute into Equation 1: * $\frac{1}{4}(y + 5) + 2y = -1 \rightarrow y + 5 + 8y = -4 \rightarrow 9y = -9 \rightarrow y = -1$ * Find $x$ : $x = -1 + 5 = 4$ * Numbers: $x = 4$ , $y = -1$ .

Scenario: The sum of two numbers is $70$ . One number is five less than one-half the other.
System Formulation: * $x + y = 70$ * $x = \frac{1}{2}y - 5$
Solution Process: Substitute into Equation 1: * $\frac{1}{2}y - 5 + y = 70 \rightarrow 1.5y = 75 \rightarrow y = 50$ * Find $x$ : $x + 50 = 70 \rightarrow x = 20$ * Numbers: $x = 20$ , $y = 50$ .

Variable Substitution: If $y = 4 - x$ , the expression $4 - x$ is substituted into the second equation for the variable $y$ .
Elimination Constants: To eliminate $y$ in the system ( $6x + 4y = 22$ and $2x - y = 1$ ), the second equation should be multiplied by $4$ .
Testing Solutions: To determine if a point $(x, y)$ is a solution to a system of inequalities, substitute the coordinates into both inequalities and check for validity. (e.g., $(1, 1)$ satisfies the test system).

Task: Find the area of the polygon formed by the system of equations: * $y = 0$ * $y = 2x + 4$ * $3y - 4x = -12$
Instructional Context: No partial credit awarded for this problem.