Grid Lock Worksheet System Solvers • Blueprint 1
Grid Lock
Goal: Solve systems of linear equations by graphing and find where paths collide.
Name:
Date:
Concept Sandbox: Graphing Systems
Step 1: Identify the slope \(m\) and \(y\)-intercept \(b\) for each equation.
Step 2: Plot the \(y\)-intercept \((0, b)\) on the coordinate grid.
Step 3: Use the slope \(\frac{\text{rise}}{\text{run}}\) to plot at least 2 more points and draw a line.
Step 4: Find the point of intersection \((x, y)\). This is the solution! Verify it in both equations.
Visual Key:
Line A
Line B
Intersection = Solution
Level 1: Scaffolded Base
Problem 1
Solve the system: \(\begin{cases} y = x + 2 \\ y = -x + 4 \end{cases}\)
Line A: \(y = x + 2\)
Slope (\(m\)): 1 / 1
\(y\)-int (\(b\)): (0, 2)
Line B: \(y = -x + 4\)
Slope (\(m\)): -1 / 1
\(y\)-int (\(b\)): (0, 4)
Intersection: (____ , ____)
Algebraic Verification Check:
Line A: \((\quad) = (\quad) + 2\)
Line B: \((\quad) = -(\quad) + 4\)
x y 0 5 -5 5 -5 Scale: 1 square = 1 unit
Level 1: Scaffolded Base
Problem 2
Solve the system: \(\begin{cases} y = 2x - 3 \\ y = -x + 3 \end{cases}\)
Line A: \(y = 2x - 3\)
Slope (\(m\)): /
\(y\)-int (\(b\)): (0, )
Line B: \(y = -x + 3\)
Slope (\(m\)): /
\(y\)-int (\(b\)): (0, )
Intersection: (____ , ____)
Algebraic Verification Check:
Line A: \((\quad) = 2(\quad) - 3\)
Line B: \((\quad) = -(\quad) + 3\)
x y 0 Scale: 1 square = 1 unit
GL
Grid Lock • Practice Workspace
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Level 2: Step Up
Problem 3
Solve the system: \(\begin{cases} y = \frac{1}{2}x + 1 \\ y = -x - 2 \end{cases}\)
Line A: \(y = \frac{1}{2}x + 1\)
\(m = \rule{20px}{0.5px}\)
\(b = \rule{20px}{0.5px}\)
Line B: \(y = -x - 2\)
\(m = \rule{20px}{0.5px}\)
\(b = \rule{20px}{0.5px}\)
Intersection: (____ , ____)
Check: Plug values to verify!
Level 2: Step Up
Problem 4
Solve the system: \(\begin{cases} y = -2x \\ y = x - 3 \end{cases}\)
Line A: \(y = -2x\)
\(m = \rule{20px}{0.5px}\)
\(b = \rule{20px}{0.5px}\)
Line B: \(y = x - 3\)
\(m = \rule{20px}{0.5px}\)
\(b = \rule{20px}{0.5px}\)
Intersection: (____ , ____)
Check: Plug values to verify!
Level 3: Critical Thinking (Special Case)
Problem 5
Solve the system: \(\begin{cases} y = x - 1 \\ y = x + 3 \end{cases}\)
Line A: \(y = x - 1\)
\(m = \rule{20px}{0.5px}\)
\(b = \rule{20px}{0.5px}\)
Line B: \(y = x + 3\)
\(m = \rule{20px}{0.5px}\)
\(b = \rule{20px}{0.5px}\)
How do the lines relate geometrically?
Parallel Intersecting
Solution: ____________________________
Drafting Rubric self-evaluation:
Sharp lines drawn Both equations checked Solution written as (x,y)
Teacher Approval
Swap Out Worksheet System Solvers • Blueprint 2
The Swap Out
Goal: Master basic substitution where one equation acts as a direct plug-in key.
Name:
Date:
Concept Sandbox: Plugs & Sockets
Substitution is like wiring an electrical circuit. When you see an equation like \(y = 3x - 1\) , that variable \(y\) is a plug . Its wire is the expression \(3x - 1\) .
Locate the socket \(y\) in the other equation (e.g., \(2x + y = 9\)), and plug the expression directly into it: \(2x + (3x - 1) = 9\). Now you have a single equation to solve!
The Swap Rule:
Eq A: \(y = [\text{Key}]\)
Eq B: \(x + \mathbf{y} = 10\)
\(x + \mathbf{[\text{Key}]} = 10\)
Level 1: Scaffolded Wireframe
Problem 1
Solve the system: \(\begin{cases} y = 2x + 1 \quad \text{(Eq A)} \\ 3x + y = 11 \quad \text{(Eq B)} \end{cases}\)
Plug: \(y\) → socket in Eq B
Step 1: Write equation with socket plugged
\(3x + \mathbf{(2x + 1)} = 11\)
Step 2: Solve for \(x\) (Show your work)
Step 3: Solve for \(y\) (Plug \(x\) into Eq A)
\(y = 2(\quad) + 1\)
Step 4: Solve for \(y\) & Check
Final Coordinate Solution: (x, y) = (____ , ____)
Level 1: Scaffolded Wireframe
Problem 2
Solve the system: \(\begin{cases} x = 3y - 2 \quad \text{(Eq A)} \\ x + 2y = 13 \quad \text{(Eq B)} \end{cases}\)
Plug: \(x\) → socket in Eq B
Step 1: Write equation with socket plugged
\(\mathbf{(3y - 2)} + 2y = 13\)
Step 2: Solve for \(y\) (Show your work)
Step 3: Solve for \(x\) (Plug \(y\) into Eq A)
\(x = 3(\quad) - 2\)
Step 4: Solve for \(x\) & Check
Final Coordinate Solution: (x, y) = (____ , ____)
SO
The Swap Out • Mastery Space
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Level 2: Step Up
Problem 3
Solve the system: \(\begin{cases} y = -4x + 6 \\ 2x - y = 6 \end{cases}\)
Write & Solve Plugged Equation:
Calculated Value: \(x = \) _____
Calculate Second Variable:
Calculated Value: \(y = \) _____
Solution: (____ , ____)
Verify check: \(2(\quad) - (\quad) \stackrel{?}{=} 6\)
Level 2: Step Up
Problem 4
Solve the system: \(\begin{cases} x = 2y + 5 \\ 3x + y = 1 \end{cases}\)
Target Acquired Worksheet System Solvers • Blueprint 3
Target Acquired
Goal: Isolate a variable with coefficient 1 or -1, then execute the swap sequence.
Name:
Date:
Concept Sandbox: Finding the Best Target
If neither equation has a variable isolated (like \(x = \dots\) or \(y = \dots\)), you must isolate a target first.
Look for any variable with a coefficient of 1 or -1 . It is the easiest to isolate because you won't have to divide by anything, avoiding fractions!
For example, in \(2x + \mathbf{y} = 8\), the \(y\) is an excellent target. Isolate it by subtracting \(2x\): \(\mathbf{y} = -2x + 8\).
Target Finder:
\(x + 3y = 7 \quad \mathbf{\color{emerald}(x)}\)
\(4x - y = 5 \quad \mathbf{\color{emerald}(-y)}\)
\(2x + 3y = 12 \quad \text{(Fractions!)}\)
Level 2: Target Acquired
Problem 1
Solve: \(\begin{cases} x + 2y = 8 \quad \text{(Eq A)} \\ 3x - 2y = 8 \quad \text{(Eq B)} \end{cases}\)
Ideal Target: \(x\) in Eq A
Step 1: Isolate \(x\) in Equation A
\(x = -2y + 8\)
Step 2: Plug into Eq B & Solve for \(y\)
\(3\mathbf{(-2y+8)} - 2y = 8\)
Step 3: Solve for \(x\) using isolated Eq A
\(x = -2(\quad) + 8\)
Step 4: Solve for \(x\) & Check
Coordinate Solution: (x, y) = (____ , ____)
Level 2: Target Acquired
Problem 2
Solve: \(\begin{cases} 2x + y = 1 \quad \text{(Eq A)} \\ 5x - 3y = 19 \quad \text{(Eq B)} \end{cases}\)
Ideal Target: \(y\) in Eq A
Step 1: Isolate \(y\) in Equation A
\(y = -2x + 1\)
Step 2: Plug into Eq B & Solve for \(x\)
\(5x - 3\mathbf{(-2x+1)} = 19\)
Step 3: Solve for \(y\) using isolated Eq A
\(y = -2(\quad) + 1\)
Step 4: Solve for \(y\) & Check
Coordinate Solution: (x, y) = (____ , ____)
TA
Target Acquired • Intermediate Lab
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Level 2: Step Up
Problem 3
Solve: \(\begin{cases} 2x - y = -1 \\ 3x + 2y = 23 \end{cases}\)
Find target: \(-y\) in Eq A
1. Isolate target & write plugged equation:
Value 1: \(x = \) _____
2. Calculate second variable:
Value 2: \(y = \) _____
Solution: (____ , ____)
Check: \(3(\quad) + 2(\quad) \stackrel{?}{=} 23\)
Line Em Up Worksheet System Solvers • Blueprint 4
Line Em Up
Goal: Solve systems of equations by stacking standard-form equations to eliminate opposites.
Name:
Date:
Concept Sandbox: Stack, Add, Strike
Elimination works beautifully when equations are in Standard Form : \(Ax + By = C\). Stack the equations so like-terms are perfectly aligned in columns.
If you see opposite coefficients (like \(+3y\) and \(-3y\)), you can add the entire system vertically. The opposites will cancel out (\(0\)), leaving you with just one variable to solve!
Column Alignment:
\(2x + \mathbf{3y} = 11\)
\(+\;[5x - \mathbf{3y} = 10]\)
\(7x + \;0\; = 21\)
Level 1: Opposite Strike
Problem 1
Solve: \(\begin{cases} 3x + y = 13 \quad \text{(Eq A)} \\ 2x - y = 2 \quad \text{(Eq B)} \end{cases}\)
Opposites: \(+y\) and \(-y\) (Add)
Step 1: Add stacked equations
\(3x + y = 13\)
\(+\;(2x - y = 2)\)
\(5x + 0 = 15\)
Step 2: Solve for \(x\)
x = _____
Step 3: Plug \(x\) value into Eq A to solve for \(y\)
\(3(\quad) + y = 13\)
Step 4: Solve for \(y\) & Check
Final Coordinate Solution: (x, y) = (____ , ____)
Level 1: Opposite Strike
Problem 2
Solve: \(\begin{cases} -x + 4y = 6 \quad \text{(Eq A)} \\ x + 2y = 9 \quad \text{(Eq B)} \end{cases}\)
Opposites: \(-x\) and \(+x\) (Add)
Step 1: Add stacked equations
\(-x + 4y = 6\)
\(+\;(x + 2y = 9)\)
\(0 + 6y = 15\)
Step 2: Solve for \(y\)
y = _____
Step 3: Plug \(y\) value into Eq B to solve for \(x\)
\(x + 2(\quad) = 9\)
Step 4: Solve for \(x\) & Check
Final Coordinate Solution: (x, y) = (____ , ____)
LE
Line Em Up • Eliminator Desk
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Level 1: Opposite Strike
Problem 3
Solve: \(\begin{cases} 4x + 3y = 5 \\ -4x + 2y = 10 \end{cases}\)
1. Stack equations & add vertically:
Striate Value: \(y = \) _____
2. Back-solve for second variable:
Striate Value: \(x = \) _____
Solution: (____ , ____)
Verify check: \(4(\quad) + 3(\quad) \stackrel{?}{=} 5\)
Level 2: subtraction twist (Match Signs)
Problem 4
Scale and Strike Worksheet System Solvers • Blueprint 5
Scale and Strike
Goal: Scale equations with multiplier factors to force coefficients into matching opposites.
Name:
Date:
Concept Sandbox: Scaling the Blueprint
If stacking equations does not instantly eliminate a variable (coefficients do not match), you must scale one or both equations first!
Multiply the entire equation (every single term on both sides of the equals sign) by a scale factor. This creates an equivalent equation with matching coefficients.
For example, in \(x - 3y = 4\), multiply the entire equation by \(2\) to get \(\mathbf{2x - 6y = 8}\). This now matches a \(2x\) in another equation!
Distribute the Scale:
Multiply Eq A by \(3\):
\(\mathbf{3} \times (x + 2y = 5)\)
Level 2: Scale One Equation
Problem 1
Solve: \(\begin{cases} x + 2y = 8 \quad \text{(Eq A)} \\ 3x - 4y = 4 \quad \text{(Eq B)} \end{cases}\)
Strategy: Multiply Eq A by \(-3\) to cancel \(3x\)
Step 1: Scale Eq A & Stack
\(-3(x + 2y = 8)\)
\(-3x - 6y = -24\)
\(+\;(3x - 4y = 4)\)
\(0 - 10y = -20\)
Step 2: Solve for \(y\)
y = _____
Step 3: Plug \(y\) value into original Eq A to solve for \(x\)
\(x + 2(\quad) = 8\)
Step 4: Solve for \(x\) & Check
Coordinate Solution: (x, y) = (____ , ____)
Level 2: Scale One Equation
Problem 2
Solve: \(\begin{cases} 4x + 3y = 1 \quad \text{(Eq A)} \\ 2x - y = -7 \quad \text{(Eq B)} \end{cases}\)
Strategy: Multiply Eq B by \(3\) to cancel \(3y\)
Step 1: Scale Eq B & Stack
\(3(2x - y = -7)\)
\(6x - 3y = -21\)
\(+\;(4x + 3y = 1)\)
\(10x + 0 = -20\)
Step 2: Solve for \(x\)
x = _____
Step 3: Plug \(x\) value into original Eq B to solve for \(y\)
\(2(\quad) - y = -7\)
Step 4: Solve for \(y\) & Check
Coordinate Solution: (x, y) = (____ , ____)
SS
Scale and Strike • Heavy Machinery
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Level 2: Scale One Equation
Problem 3
Solve: \(\begin{cases} 3x - y = 9 \\ 2x + 3y = -5 \end{cases}\)
1. Scale Eq A by \(3\) & stack-add to Eq B:
Calculated Value: \(x = \) _____
