Gridlock Clash Worksheet
Algebra II & Honors Unit 2: Linear Systems
Solving Linear Systems by Graphing
Name:
Date: Period:
SYSTEM CLASSIFICATION:
Intersecting: 1 Sol. (Consistent & Indep.) Parallel: No Sol. (Inconsistent) Coincident: \(\infty\) Sol. (Consistent & Dep.)
Problem 1 Slope-Intercept Form
\(\begin{cases} y = 2x - 3 \\ y = -\frac{1}{2}x + 2 \end{cases}\)
Line 1: \(m=\underline{\quad}, b=\underline{\quad}\)
Line 2: \(m=\underline{\quad}, b=\underline{\quad}\)
x y -4 -2 2 4 -4 -2 2 4
Solution: ( , )
Type: Indep. Incons. Dep.
Problem 2 Standard & Slope-Intercept
\(\begin{cases} 2x + y = 4 \\ x - y = -1 \end{cases}\)
Eq 1: \(y=\underline{\qquad\quad}\)
Eq 2: \(y=\underline{\qquad\quad}\)
x y -4 -2 2 4 -4 -2 2 4
Solution: ( , )
Type: Indep. Incons. Dep.
Problem 3 Parallel Analysis
\(\begin{cases} 2x - 3y = 6 \\ y = \frac{2}{3}x + 2 \end{cases}\)
Eq 1: \(y=\underline{\qquad\quad}\)
Eq 2: \(m=\underline{\quad}, b=\underline{\quad}\)
x y -4 -2 2 4 -4 -2 2 4
Solution: _______________
Type: Indep. Incons. Dep.
Problem 4 Coincident Lines
\(\begin{cases} 3x - y = 3 \\ -6x + 2y = -6 \end{cases}\)
Eq 1: \(y=\underline{\qquad\quad}\)
Eq 2: \(y=\underline{\qquad\quad}\)
x y -4 -2 2 4 -4 -2 2 4
Solution: _______________
Type: Indep. Incons. Dep.
Check work by substituting your \((x, y)\) coordinates back into both original equations! Page 1 of 2
PART II
Advanced & Honors Applications
Algebra II / Honors Curriculum
Problem 5 Standard Form Conversion
\(\begin{cases} 3x + 4y = 8 \\ x - 2y = -4 \end{cases}\)
Transform each to slope-intercept form:
\(L_1:\)
\(L_2:\)
x y -4 -2 2 4 -4 -2 2 4
Intersection: ( , )
Problem 6 Flight Route Modeling
Surveillance radar tracks two automated survey drones:
• Drone Alpha: Starts at \((-4, -5)\), climbs at slope \(m = \frac{3}{2}\).
• Drone Bravo: Follows flight path \(x + y = 6\).
Alpha Eq: \(y=\)
Bravo Eq: \(y=\)
x y -4 -2 2 4 -4 -2 2 4
Waypoint / Crossing: ( , )
Honors Analytical Extension
Problem 7: Parametric System Investigation
Algebraic Rigor & Classification
Consider the linear system containing unknown parameter \(k\): \[\begin{cases} 4x - ky = 12 \\ 6x + 3y = 9 \end{cases}\]
Part A: Inconsistent System
Find the exact value of \(k\) such that the system has no solution (parallel lines):
\(k = \underline{\qquad\qquad\qquad}\)
Part B: Dependent System Analysis
Can \(k\) be chosen so the system has infinitely many solutions? Explain why or why not:
Conclusion: [ ] Possible [ ] Impossible
Honors Tip: Express lines in slope-intercept form \(y = mx + b\) to compare slopes \(m_1, m_2\) and intercepts \(b_1, b_2\). Page 2 of 2
Gridlock Clash Answer Key
Teacher Resource Answer Key & Solutions
Solving Linear Systems by Graphing
Master Solution Key
Algebra II / Honors Ready
GRADING MATRIX:
P1: \((2, 1)\) • Indep. P2: \((1, 2)\) • Indep. P3: \(\emptyset\) (No Sol.) • Incons. P4: \(\infty\) Sol. • Dep.
Problem 1 1 Solution
\(\begin{cases} y = 2x - 3 & \color{#2563eb}{(L_1)} \\ y = -\frac{1}{2}x + 2 & \color{#7c3aed}{(L_2)} \end{cases}\)
\(L_1: m=2, b=-3\)
\(L_2: m=-\frac{1}{2}, b=2\)
(2, 1)
Solution: ( 2 , 1 )
Type: ☑ Consistent & Independent
Problem 2 Standard Form
\(\begin{cases} 2x + y = 4 \\ x - y = -1 \end{cases}\)
\(y = -2x + 4\)
\(y = x + 1\)
(1, 2)
Solution: ( 1 , 2 )
Type: ☑ Consistent & Independent
Problem 3 Parallel Lines
\(\begin{cases} 2x - 3y = 6 \\ y = \frac{2}{3}x + 2 \end{cases}\)
\(y = \frac{2}{3}x - 2\)
\(m = \frac{2}{3}, b = 2\)
Parallel (m = 2/3)
Solution: No Solution (\(\emptyset\))
Type: ☑ Inconsistent (Same slope, \(b_1 \neq b_2\))
Problem 4 Coincident Lines
\(\begin{cases} 3x - y = 3 \\ -6x + 2y = -6 \end{cases}\)
\(y = 3x - 3\)
\(y = 3x - 3\)
Same Line!
Solution: \(\infty\) Solutions
Type: ☑ Consistent & Dependent (Identical)
Teacher Verification Note: Both equations in Problem 4 are scalar multiples of each other (\(\times -2\)). Answer Key • Page 1 of 2
KEY: PART II
Advanced & Honors Solutions
Step-by-Step Teacher Notes
Problem 5 Shared Intercept: (0, 2)
\(\begin{cases} 3x + 4y = 8 \\ x - 2y = -4 \end{cases}\)
\(L_1: 4y = -3x + 8 \implies \mathbf{y = -\frac{3}{4}x + 2}\)
\(L_2: -2y = -x - 4 \implies \mathbf{y = \frac{1}{2}x + 2}\)
(0, 2)
Intersection: ( 0 , 2 )
Problem 6 Waypoint: (2, 4)
• Alpha: point \((-4, -5)\), \(m = \frac{3}{2}\)
• Bravo: path \(x + y = 6\)
Alpha: \(y - (-5) = \frac{3}{2}(x - (-4)) \implies \mathbf{y = \frac{3}{2}x + 1}\)
Bravo: \(x + y = 6 \implies \mathbf{y = -x + 6}\)
(2, 4)
Waypoint / Crossing: ( 2 , 4 )
Honors Complete Solution
Problem 7: Parametric System Analysis
Algebraic Proof
System: \(\begin{cases} 4x - ky = 12 \implies y = \frac{4}{k}x - \frac{12}{k} \\ 6x + 3y = 9 \implies y = -2x + 3 \end{cases}\) Slopes: \(m_1 = \frac{4}{k}\), \(m_2 = -2\)
