Add directly (\(y\) eliminated): \(10x = 90 \implies \mathbf{x = 9}\)
Back-sub: \(2(9) + 3y = 6 \implies 18 + 3y = 6 \implies \mathbf{y = -4}\)
Check: \(\tfrac{2}{3}(9) - \tfrac{1}{4}(-4) = 6 - (-1) = 7\) ✓
Note: Clearing fractions produces opposite \(y\) terms (\(-3y\) and \(+3y\)) without secondary scaling!
Problem 5: Decimals Solution: \((8, 2)\)
Scale by powers of 10:
\[\begin{aligned} 10 \cdot (0.4x - 0.3y = 2.6) &\implies 4x - 3y = 26 \\ 100 \cdot (0.15x + 0.25y = 1.7) &\implies 15x + 25y = 170 \end{aligned}\]
Divide Eq 2 by 5: \(3x + 5y = 34\). Multiply Eq 1 by 5, Eq 2 by 3:
\(20x - 15y = 130\) & \(9x + 15y = 102 \implies 29x = 232 \implies \mathbf{x = 8}\).
Back-sub: \(4(8) - 3y = 26 \implies 32 - 3y = 26 \implies \mathbf{y = 2}\)
Pitfall: Students often multiply \(1.7\) by 10 instead of 100, getting \(17\) instead of \(170\).
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Teacher Solutions Key | Linear Systems: Degenerate Analysis & Modeling Key
Page 2 of 2
Problems 6–7
Problem 6: Inconsistent No Solution (\(\emptyset\))
\[\begin{cases} 6x - 4y = 10 & (\times 3) \implies 18x - 12y = 30 \\ -9x + 6y = 15 & (\times 2) \implies -18x + 12y = 30 \end{cases}\] \[\text{Add: } 0x + 0y = 60 \implies \mathbf{0 = 60} \quad (\text{False})\]
• Resulting Statement: \(0 = 60\) (Contradiction)
• Number of Solutions: Zero / No Solution (\(\emptyset\))
• Classification: Inconsistent
• Geometric Graph: Parallel lines (equal slope \(m = \frac{3}{2}\), distinct intercepts \(-\frac{5}{2} \neq \frac{5}{2}\))
Award full credit only if algebraic contradiction is explicitly stated.
Problem 7: Dependent Infinitely Many Solutions
\[\begin{cases} 4x - 10y = 14 & (\times 3) \implies 12x - 30y = 42 \\ -6x + 15y = -21 & (\times 2) \implies -12x + 30y = -42 \end{cases}\] \[\text{Add: } 0x + 0y = 0 \implies \mathbf{0 = 0} \quad (\text{Identity})\]
• Resulting Statement: \(0 = 0\) (Identity)
• Number of Solutions: Infinitely Many Solutions
• Classification: Consistent Dependent
• Set-Builder Form: \(\{(x, y) \mid y = \frac{2}{5}x - \frac{7}{5}\}\) or \((t, \frac{2}{5}t - \frac{7}{5})\)
Honors check: Deduct 1 pt if student writes "all real numbers" instead of points on the line.
12 pts total
(a) Variable Definitions & System Formulation 3 pts (1.5 pts defs, 1.5 pts system)
Definitions:
Let \(T\) = energy consumption of one Titan-4 drone per circuit (in kWh).
Let \(S\) = energy consumption of one Swift-X drone per circuit (in kWh).
\[\begin{cases} 5T + 8S = 142 & \text{(Mission Alpha)} \\ 7T + 4S = 134 & \text{(Mission Beta)} \end{cases}\]
(b) Elimination Solution & Unit Values 6 pts (2 pts multiplier, 2 pts 1st var, 2 pts back-sub)
\[\begin{aligned} 5T + 8S &= 142 \\ -2 \cdot (7T + 4S = 134) &\implies -14T - 8S = -268 \\ \hline -9T &= -126 \\ \mathbf{T} &= \mathbf{14\text{ kWh}} \end{aligned}\]
Back-substitute \(T = 14\) into Mission Alpha:
\(5(14) + 8S = 142 \implies 70 + 8S = 142\)
\(8S = 72 \implies \mathbf{S = 9\text{ kWh}}\)
Verification: \(7(14) + 4(9) = 98 + 36 = 134\) kWh ✓
Titan-4 Unit Consumption: 14 kWh per circuit Swift-X Unit Consumption: 9 kWh per circuit
(c) Honors Constraint Analysis (Charging Capacity Limit: 220 kWh) 3 pts
Dispatch Energy Calculation: \(E = 10 \times T + 8 \times S = 10(14) + 8(9) = 140 + 72 = \mathbf{212\text{ kWh}}\).
Comparison: \(212\text{ kWh} \le 220\text{ kWh}\) (Condition satisfied). FEASIBLE: YES Surplus Reserve: \(220 - 212 = \mathbf{8\text{ kWh}}\)
Honors Algebra 1 — Elimination Method Teacher Guide & Master Solutions Key
