System Showdown Station Cards
Station Activity Gallery Walk Series
Room Station 01
Method: Substitution
STATION 01
Two-Variable Linear System
Station 01
Solve the System: Substitution Friendly
Eq 1 \(y = 2x - 3\)
Eq 2 \(3x + 4y = 32\)
Station Clue Notice that \(y\) is already isolated in Equation 1. Substitute the expression \((2x - 3)\) in place of \(y\) in Equation 2, solve for \(x\), then find \(y\).
Show all substitution and back-substitution steps on your recording sheet.
Card 1 of 12 (Front)
Station 01 Key Flip-Side Verification
Card 1 of 12 (Back)
Final Coordinate Solution
\((x, y) = (4, 5)\)
System Type Consistent & Independent
1 Substitute \(y\) into Equation 2 Replace \(y\) with \((2x - 3)\)
\(3x + 4(2x - 3) = 32\)
2 Distribute, Combine & Solve for \(x\) Isolate variable \(x\)
\(3x + 8x - 12 = 32\)
\(11x - 12 = 32 \implies 11x = 44 \implies \mathbf{x = 4}\)
3 Back-Substitute to Solve for \(y\) Use Eq 1: \(y = 2x - 3\)
\(y = 2(4) - 3 = 8 - 3 \implies \mathbf{y = 5}\)
Self-Check Verification:
Eq 1: \(5 = 2(4) - 3 = 5\) ✓
Eq 2: \(3(4) + 4(5) = 12 + 20 = 32\) ✓
Geometry: The two linear equations intersect at exactly one coordinate point: \((4, 5)\). Verified
Station Activity Gallery Walk Series
Room Station 02
Method: Substitution
STATION 02
Two-Variable Linear System
Station 02
Solve the System: \(x\) Variable Isolated
Eq 1 \(x = -3y + 1\)
Eq 2 \(2x + 5y = -1\)
Station Clue Here \(x\) is isolated in Equation 1. Substitute \((-3y + 1)\) into Equation 2 in place of \(x\). Be mindful when distributing the factor of 2!
Remember to state your final answer as an ordered pair \((x, y)\).
Card 2 of 12 (Front)
Station 02 Key Flip-Side Verification
Card 2 of 12 (Back)
Final Coordinate Solution
\((x, y) = (-8, 3)\)
System Type Consistent & Independent
1 Substitute \(x\) into Equation 2 Replace \(x\) with \((-3y + 1)\)
\(2(-3y + 1) + 5y = -1\)
2 Distribute, Combine & Solve for \(y\) Isolate variable \(y\)
\(-6y + 2 + 5y = -1\)
\(-y + 2 = -1 \implies -y = -3 \implies \mathbf{y = 3}\)
3 Back-Substitute to Solve for \(x\) Use Eq 1: \(x = -3y + 1\)
\(x = -3(3) + 1 = -9 + 1 \implies \mathbf{x = -8}\)
Self-Check Verification:
Eq 1: \(-8 = -3(3) + 1 = -8\) ✓
Eq 2: \(2(-8) + 5(3) = -16 + 15 = -1\) ✓
Geometry: Single intersection in Quadrant II at \((-8, 3)\). Verified
Station Activity Gallery Walk Series
Room Station 03
Method: Substitution
STATION 03
Two-Variable Linear System
Station 03
Solve the System: Special Case Alert
Eq 1 \(y = -2x + 5\)
Eq 2 \(4x + 2y = 7\)
Station Clue Substitute \((-2x + 5)\) into Equation 2. What happens to the variable terms? Pay close attention to whether the resulting equation is true or false!
If variables cancel out, interpret what the mathematical statement means.
Card 3 of 12 (Front)
Station 03 Key Flip-Side Verification
Card 3 of 12 (Back)
System Result
NO SOLUTION (\(\emptyset\))
System Type Inconsistent
1 Substitute \(y\) into Equation 2 Replace \(y\) with \((-2x + 5)\)
\(4x + 2(-2x + 5) = 7\)
2 Distribute & Simplify Observe variable elimination
\(4x - 4x + 10 = 7\)
\(0 + 10 = 7 \implies \mathbf{10 = 7 \quad (\text{FALSE!})}\)
3 Mathematical Conclusion Contradiction
Because the statement \(10 = 7\) is fundamentally impossible (a false mathematical statement) and contains no variables, there is no coordinate pair \((x, y)\) that satisfies both equations simultaneously.
Geometric Slope Comparison:
Line 1: \(y = -2x + 5 \implies m = -2, b = 5\)
Line 2: \(y = -2x + 3.5 \implies m = -2, b = 3.5\)
Geometry: The lines have identical slopes but different y-intercepts: they are parallel and never intersect. Inconsistent
Station Activity Gallery Walk Series
Room Station 04
Method: Substitution
STATION 04
Two-Variable Linear System
Station 04
Solve the System: Binomial Expression
Eq 1 \(x = 4y - 6\)
Eq 2 \(3x - 2y = 12\)
Station Clue Substitute \((4y - 6)\) for \(x\) in Equation 2: \(3(4y - 6) - 2y = 12\). Distribute the 3 across both terms inside parentheses before combining like terms.
Double-check your negative signs during multi-step distribution.
Card 4 of 12 (Front)
Station 04 Key Flip-Side Verification
Card 4 of 12 (Back)
Final Coordinate Solution
\((x, y) = (6, 3)\)
System Type Consistent & Independent
1 Substitute \(x\) into Equation 2 Replace \(x\) with \((4y - 6)\)
\(3(4y - 6) - 2y = 12\)
2 Distribute, Combine & Solve for \(y\) Combine \(12y - 2y\)
\(12y - 18 - 2y = 12\)
\(10y - 18 = 12 \implies 10y = 30 \implies \mathbf{y = 3}\)
3 Back-Substitute to Solve for \(x\) Use Eq 1: \(x = 4y - 6\)
\(x = 4(3) - 6 = 12 - 6 \implies \mathbf{x = 6}\)
Self-Check Verification:
Eq 1: \(6 = 4(3) - 6 = 6\) ✓
Eq 2: \(3(6) - 2(3) = 18 - 6 = 12\) ✓
Geometry: Single intersection in Quadrant I at \((6, 3)\). Verified
Station Activity Gallery Walk Series
Room Station 05
Method: Elimination (Direct Cancel)
STATION 05
Two-Variable Linear System
Station 05
Solve the System: Opposite Coefficients Ready
Eq 1 \(4x + 3y = 18\)
Eq 2 \(2x - 3y = 0\)
Station Clue Examine the \(y\) coefficients: \(+3y\) and \(-3y\) are exact opposites! Add the two equations vertically to immediately eliminate \(y\).
No initial multiplication required. Add vertically and solve!
Card 5 of 12 (Front)
Station 05 Key Flip-Side Verification
Card 5 of 12 (Back)
Final Coordinate Solution
\((x, y) = (3, 2)\)
Elimination Step Direct Addition
1 Add the Equations Vertically \(+3y\) and \(-3y\) cancel out
\(\phantom{+} (4x + 3y = 18)\)
\(+ (2x - 3y = 0)\)
\(6x + 0y = 18 \implies 6x = 18\)
2 Solve for \(x\) Divide by 6
\(x = \frac{18}{6} \implies \mathbf{x = 3}\)
3 Back-Substitute to Solve for \(y\) Use Eq 2: \(2x - 3y = 0\)
\(2(3) - 3y = 0 \implies 6 - 3y = 0 \implies 3y = 6 \implies \mathbf{y = 2}\)
Self-Check Verification:
Eq 1: \(4(3) + 3(2) = 12 + 6 = 18\) ✓
Eq 2: \(2(3) - 3(2) = 6 - 6 = 0\) ✓
Geometry: Single intersection in Quadrant I at \((3, 2)\). Verified
Station Activity Gallery Walk Series
Room Station 06
Method: Elimination (Direct Cancel)
STATION 06
Two-Variable Linear System
Station 06
Solve the System: Opposite Coefficients Ready
Eq 1 \(5x + 2y = 0\)
Eq 2 \(-5x + 6y = 40\)
Station Clue Look at the \(x\) variable coefficients: \(+5x\) and \(-5x\). Add the two equations together to eliminate \(x\) instantly and solve for \(y\).
Watch zero on the right side of Equation 1 when adding.
Card 6 of 12 (Front)
Station 06 Key Flip-Side Verification
Card 6 of 12 (Back)
Final Coordinate Solution
\((x, y) = (-2, 5)\)
Elimination Step Direct Addition (\(x\))
1 Add the Equations Vertically \(5x\) and \(-5x\) cancel out
\(\phantom{+} (5x + 2y = 0)\)
\(+ (-5x + 6y = 40)\)
\(0x + 8y = 40 \implies 8y = 40\)
2 Solve for \(y\) Divide by 8
\(y = \frac{40}{8} \implies \mathbf{y = 5}\)
3 Back-Substitute to Solve for \(x\) Use Eq 1: \(5x + 2y = 0\)
\(5x + 2(5) = 0 \implies 5x + 10 = 0 \implies 5x = -10 \implies \mathbf{x = -2}\)
Self-Check Verification:
Eq 1: \(5(-2) + 2(5) = -10 + 10 = 0\) ✓
Eq 2: \(-5(-2) + 6(5) = 10 + 30 = 40\) ✓
Geometry: Single intersection in Quadrant II at \((-2, 5)\). Verified
Station Activity Gallery Walk Series
Room Station 07
Method: Elimination (Multiply One Equation)
STATION 07
Two-Variable Linear System
Station 07
Solve the System: Scale 1 Equation
Eq 1 \(x - 4y = 9\)
Eq 2 \(3x + 2y = 13\)
Station Clue Notice that multiplying Equation 2 by \(2\) produces \(+4y\), which will cancel directly with \(-4y\) in Equation 1! Alternatively, multiply Equation 1 by \(-3\).
Remember to multiply EVERY term on both sides of the equation.
Card 7 of 12 (Front)
Station 07 Key Flip-Side Verification
Card 7 of 12 (Back)
Final Coordinate Solution
\((x, y) = (5, -1)\)
Elimination Step Multiply Eq 2 by 2
1 Multiply Equation 2 by 2 Create \(+4y\)
\(2 \times (3x + 2y = 13) \implies \mathbf{6x + 4y = 26}\)
2 Add to Equation 1 & Solve for \(x\) \(-4y\) and \(+4y\) cancel
\(\phantom{+} (x - 4y = 9)\)
\(+ (6x + 4y = 26)\)
\(7x = 35 \implies \mathbf{x = 5}\)
3 Back-Substitute to Solve for \(y\) Use Eq 1: \(x - 4y = 9\)
\(5 - 4y = 9 \implies -4y = 4 \implies \mathbf{y = -1}\)
Self-Check Verification:
Eq 1: \(5 - 4(-1) = 5 + 4 = 9\) ✓
Eq 2: \(3(5) + 2(-1) = 15 - 2 = 13\) ✓
Geometry: Single intersection in Quadrant IV at \((5, -1)\). Verified
Station Activity Gallery Walk Series
Room Station 08
Method: Elimination (Multiply One Equation)
STATION 08
Two-Variable Linear System
Station 08
Solve the System: Negative Multiplier
Eq 1 \(3x + 8y = 38\)
Eq 2 \(2x + 4y = 20\)
Station Clue Equation 1 has \(+8y\), while Equation 2 has \(+4y\). Multiply Equation 2 by \(-2\) to produce \(-8y\), then add the equations together to eliminate \(y\).
Be sure to multiply \(-2\) by the constant 20 as well!
Card 8 of 12 (Front)
Station 08 Key Flip-Side Verification
Card 8 of 12 (Back)
Final Coordinate Solution
\((x, y) = (2, 4)\)
Elimination Step Multiply Eq 2 by \(-2\)
1 Multiply Equation 2 by \(-2\) Create \(-8y\)
\(-2 \times (2x + 4y = 20) \implies \mathbf{-4x - 8y = -40}\)
2 Add to Equation 1 & Solve for \(x\) \(+8y\) and \(-8y\) cancel
\(\phantom{+} (3x + 8y = 38)\)
\(+ (-4x - 8y = -40)\)
\(-x = -2 \implies \mathbf{x = 2}\)
3 Back-Substitute to Solve for \(y\) Use Eq 2: \(2x + 4y = 20\)
\(2(2) + 4y = 20 \implies 4 + 4y = 20 \implies 4y = 16 \implies \mathbf{y = 4}\)
Self-Check Verification:
Eq 1: \(3(2) + 8(4) = 6 + 32 = 38\) ✓
Eq 2: \(2(2) + 4(4) = 4 + 16 = 20\) ✓
Geometry: Single intersection in Quadrant I at \((2, 4)\). Verified
Station Activity Gallery Walk Series
Room Station 09
Method: Elimination (Multiply Both Equations)
STATION 09
Two-Variable Linear System
Station 09
Solve the System: Scale Both Equations
Eq 1 \(3x + 5y = -3\)
Eq 2 \(4x + 3y = 7\)
Station Clue Find a common multiple! To eliminate \(y\), find the LCM of 5 and 3 (which is 15). Multiply Equation 1 by \(3\) and Equation 2 by \(-5\) to make the \(y\) coefficients \(+15\) and \(-15\).
Carefully check signs when multiplying by negative five!
Card 9 of 12 (Front)
Station 09 Key Flip-Side Verification
Card 9 of 12 (Back)
Final Coordinate Solution
\((x, y) = (4, -3)\)
Elimination Step Scale by 3 and \(-5\)
1 Multiply Both Equations Target \(y\): LCM \(= 15\)
\(3 \times (3x + 5y = -3) \implies \mathbf{9x + 15y = -9}\)
\(-5 \times (4x + 3y = 7) \implies \mathbf{-20x - 15y = -35}\)
2 Add the Transformed Equations \(+15y\) and \(-15y\) cancel
\(\phantom{+} (9x + 15y = -9)\)
\(+ (-20x - 15y = -35)\)
\(-11x = -44 \implies \mathbf{x = 4}\)
3 Back-Substitute to Solve for \(y\) Use Eq 2: \(4x + 3y = 7\)
\(4(4) + 3y = 7 \implies 16 + 3y = 7 \implies 3y = -9 \implies \mathbf{y = -3}\)
Self-Check Verification:
Eq 1: \(3(4) + 5(-3) = 12 - 15 = -3\) ✓
Eq 2: \(4(4) + 3(-3) = 16 - 9 = 7\) ✓
Geometry: Single intersection in Quadrant IV at \((4, -3)\). Verified
Station Activity Gallery Walk Series
Room Station 10
Method: Elimination (Multiply Both Equations)
STATION 10
Two-Variable Linear System
Station 10
Solve the System: Scale Both Equations
Eq 1 \(5x - 4y = -7\)
Eq 2 \(2x - 3y = 0\)
Station Clue To eliminate \(y\), target LCM 12. Multiply Equation 1 by \(3\) (giving \(-12y\)) and Equation 2 by \(-4\) (giving \(+12y\)). Then add!
Both coordinate values in your solution will be negative integers.
Card 10 of 12 (Front)
Station 10 Key Flip-Side Verification
Card 10 of 12 (Back)
Final Coordinate Solution
\((x, y) = (-3, -2)\)
Elimination Step Scale by 3 and \(-4\)
1 Multiply Both Equations Target \(y\): LCM \(= 12\)
\(3 \times (5x - 4y = -7) \implies \mathbf{15x - 12y = -21}\)
\(-4 \times (2x - 3y = 0) \implies \mathbf{-8x + 12y = 0}\)
2 Add the Transformed Equations \(-12y\) and \(+12y\) cancel
\(\phantom{+} (15x - 12y = -21)\)
\(+ (-8x + 12y = 0)\)
\(7x = -21 \implies \mathbf{x = -3}\)
3 Back-Substitute to Solve for \(y\) Use Eq 2: \(2x - 3y = 0\)
\(2(-3) - 3y = 0 \implies -6 - 3y = 0 \implies -3y = 6 \implies \mathbf{y = -2}\)
Self-Check Verification:
Eq 1: \(5(-3) - 4(-2) = -15 + 8 = -7\) ✓
Eq 2: \(2(-3) - 3(-2) = -6 + 6 = 0\) ✓
Geometry: Single intersection in Quadrant III at \((-3, -2)\). Verified
Station Activity Gallery Walk Series
Room Station 11
Method: Elimination (Special Case)
STATION 11
Two-Variable Linear System
Station 11
Solve the System: Special Case Alert
Eq 1 \(2x - 5y = 8\)
Eq 2 \(-6x + 15y = -24\)
Station Clue Try multiplying Equation 1 by \(3\) to eliminate \(x\). What occurs to both variables AND the constant term when you add the equations?
Analyze whether the final resulting statement is always true or false.
Card 11 of 12 (Front)
Station 11 Key Flip-Side Verification
Card 11 of 12 (Back)
System Result
INFINITELY MANY SOLUTIONS
System Type Consistent & Dependent
1 Multiply Equation 1 by 3 Target \(x\) elimination
\(3 \times (2x - 5y = 8) \implies \mathbf{6x - 15y = 24}\)
2 Add the Equations Vertically All terms eliminate
\(\phantom{+} (6x - 15y = 24)\)
\(+ (-6x + 15y = -24)\)
\(0x + 0y = 0 \implies \mathbf{0 = 0 \quad (\text{ALWAYS TRUE!})}\)
3 Mathematical Conclusion Identity statement
Because \(0 = 0\) is an identity (always true for any real values), both equations represent the exact same line. Every coordinate point lying on \(2x - 5y = 8\) is a valid solution to the system.
Coincident Line Check:
Eq 2 Divided by \(-3\): \(2x - 5y = 8\)
Slope-Intercept: \(y = \frac{2}{5}x - \frac{8}{5}\) (Identical)
Geometry: The two lines are coincident (lie directly on top of each other with infinite shared points). Dependent
Station Activity Gallery Walk Series
Room Station 12
Method: Elimination (Grand Finale)
STATION 12
Two-Variable Linear System
Station 12
Solve the System: Mastery Challenge
Eq 1 \(7x - 2y = 40\)
Eq 2 \(3x + 5y = 23\)
Station Clue Eliminate \(y\) with opposite signs already present! Multiply Equation 1 by \(5\) (to get \(-10y\)) and Equation 2 by \(2\) (to get \(+10y\)).
Final station! Complete your answer sheet and check all solutions.
Card 12 of 12 (Front)
Station 12 Key Flip-Side Verification
Card 12 of 12 (Back)
Final Coordinate Solution
\((x, y) = (6, 1)\)
Elimination Step Scale by 5 and 2
1 Multiply Both Equations Target \(y\): LCM \(= 10\)
\(5 \times (7x - 2y = 40) \implies \mathbf{35x - 10y = 200}\)
\(2 \times (3x + 5y = 23) \implies \mathbf{6x + 10y = 46}\)
2 Add the Transformed Equations \(-10y\) and \(+10y\) cancel
\(\phantom{+} (35x - 10y = 200)\)
\(+ (6x + 10y = 46)\)
\(41x = 246 \implies \mathbf{x = 6}\)
3 Back-Substitute to Solve for \(y\) Use Eq 2: \(3x + 5y = 23\)
\(3(6) + 5y = 23 \implies 18 + 5y = 23 \implies 5y = 5 \implies \mathbf{y = 1}\)
Self-Check Verification:
Eq 1: \(7(6) - 2(1) = 42 - 2 = 40\) ✓
Eq 2: \(3(6) + 5(1) = 18 + 5 = 23\) ✓
Geometry: Single intersection in Quadrant I at \((6, 1)\). Verified
