1. Equation: \( 2.50m + 5 = 22.50 \)
2. Subtract 5: \( 2.50m = 17.50 \)
3. Divide by 2.50: \( m = \frac{17.50}{2.50} \)
4. Solve: \( \mathbf{m = 7} \) miles
SPED Checkpoint
Have students use the color-coded visual template provided on their worksheet page to align numbers correctly.
UNIT: ALGEBRA 1 REINFORCEMENT KEY PAGE 1 OF 6 ENG-TEACHER-REV.01
Guided Teacher Solution Key
6-Week Pacing
SPED Strategy for Week 2: Emphasize that the solution represents one shared point \( (x, y) \). Use visual highlighters to map substitute components into equations.
Question 6
Substitution
\( y = 2x - 3 \)
\( x + y = 9 \)
Ans: \( (4, 5) \)
Step-by-Step Solution
1. Substitute: \( x + (2x - 3) = 9 \)
2. Combine: \( 3x - 3 = 9 \Rightarrow 3x = 12 \Rightarrow x = 4 \)
3. Find y: \( y = 2(4) - 3 = 5 \)
4. Coordinate: \( \mathbf{(4, 5)} \)
SPED Checkpoint
Remind students they are not done when they find \( x = 4 \). They must plug it back in to get their \( y \) coordinate!
Question 7
Elimination
\( 2x + 3y = 11 \)
\( 2x - y = 3 \)
Ans: \( (2.5, 2) \)
Step-by-Step Solution
1. Subtract: \( (2x-2x) + (3y-(-y)) = 11-3 \)
2. Simplify: \( 4y = 8 \Rightarrow y = 2 \)
3. Find x: \( 2x - (2) = 3 \Rightarrow 2x = 5 \Rightarrow x = 2.5 \)
4. Coordinate: \( \mathbf{(2.5, 2)} \)
SPED Checkpoint
Double negative error: students subtract \( 3y - y \) instead of \( 3y - (-y) \). Remind them: subtraction flips the sign!
Question 8
Solve by Graphing
\( y = x + 2 \)
\( y = -x + 4 \)
Ans: \( (1, 3) \)
Step-by-Step Solution
1. Graph line 1: y-int = 2, slope = 1.
2. Graph line 2: y-int = 4, slope = -1.
3. Find their intersection. It is point \( \mathbf{(1, 3)} \).
4. Verify coordinates using expanded axis numbers (-5 to 5).
SPED Checkpoint
Students struggle to locate points on empty grids. Have them point out the printed vertical axis label \( 3 \) and horizontal \( 1 \) explicitly on the expanded graph.
Question 9
System Word Prob
Sum = 15, Diff = 3
Ans: \( 9 \) and \( 6 \)
Step-by-Step Solution
1. System: \( x+y=15 \), \( x-y=3 \)
2. Add equations: \( 2x = 18 \Rightarrow x = 9 \)
3. Substitute: \( 9 + y = 15 \Rightarrow y = 6 \)
4. Solution: The numbers are \( \mathbf{9} \) and \( \mathbf{6} \).
SPED Checkpoint
For SPED support, the pre-written system formulas on their sheets prevent writing frustration and speed up lesson pacing.
Question 10
Special Systems
\( 2x - y = 4 \)
\( -4x + 2y = -8 \)
Ans: Infinite Sol.
Step-by-Step Solution
1. Multiply Eq 1 by -2: \( -4x + 2y = -8 \)
2. Notice it is identical to Eq 2.
3. Subtracting gives \( 0 = 0 \) (True).
4. Conclusion: Infinite Solutions.
SPED Checkpoint
If equations are completely identical (same slope and same intercept), it's the exact same line, meaning Infinite Solutions.
UNIT: ALGEBRA 1 REINFORCEMENT KEY PAGE 2 OF 6 ENG-TEACHER-REV.01
Guided Teacher Solution Key
6-Week Pacing
SPED Strategy for Week 3: Provide the quadratic formula on the whiteboard as well. Have students use different colors for coefficient inputs \( a, b, c \).
Question 11
Trinomial Factoring
\( x^2 - 5x - 6 \)
Ans: \( (x-6)(x+1) \)
Step-by-Step Solution
1. Multiply to \( c = -6 \).
2. Add to \( b = -5 \).
3. Test factors: \( -6 \times 1 = -6 \) and \( -6 + 1 = -5 \).
4. Solution: \( \mathbf{(x - 6)(x + 1)} \)
SPED Checkpoint
Show students why \( (x-2)(x-3) \) fails: their product is \( +6 \) instead of \( -6 \). Emphasize sign verification.
Question 12
Quadratic Formula
\( x^2 - 4x + 3 = 0 \)
Ans: \( x = 3, 1 \)
Step-by-Step Solution
1. Plug in values: \( a=1, b=-4, c=3 \).
2. Discriminant: \( \sqrt{(-4)^2 - 4(1)(3)} = \sqrt{16-12} = 2 \).
3. Split: \( x = \frac{4 \pm 2}{2} \).
4. Answer: \( \mathbf{x = 3} \) or \( \mathbf{x = 1} \).
SPED Checkpoint
For SPED accommodation, the printed formula template in their packet reduces cognitive load and visual search blocks.
Question 13
Parabola Features
\( y = x^2 - 2x - 3 \)
Ans: Vertex \( (1, -4) \)
Step-by-Step Solution
1. Axis of symmetry: \( x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1 \).
2. Vertex y: \( y = (1)^2 - 2(1) - 3 = -4 \).
3. Vertex coordinates: \( \mathbf{(1, -4)} \).
4. y-intercept is \( (0, -3) \).
SPED Checkpoint
Check their calculation of axis of symmetry: \( -b/2a \). Double negatives often turn positive here: \( -(-2) = +2 \).
Question 14
Solve by Factoring
\( 2x^2 + 5x - 3 = 0 \)
Ans: \( x = \frac{1}{2}, -3 \)
Step-by-Step Solution
1. Factor: \( (2x - 1)(x + 3) = 0 \).
2. Set 1: \( 2x - 1 = 0 \Rightarrow x = 0.5 \).
3. Set 2: \( x + 3 = 0 \Rightarrow x = -3 \).
4. Roots: \( \mathbf{x = 0.5} \) and \( \mathbf{x = -3} \).
SPED Checkpoint
For SPED support, we provide the factored template clue so they can practice the critical zero-product property without getting stuck factoring.
Question 15
Vertex Form
\( y = 2(x + 3)^2 - 4 \)
Ans: Vertex \( (-3, -4) \)
Step-by-Step Solution
1. Compare to \( y = a(x-h)^2 + k \).
2. Find \( h \): \( x+3 \rightarrow h = -3 \) (sign flips!).
3. Find \( k \): \( k = -4 \).
4. Axis of symmetry: \( \mathbf{x = -3} \), Opens: UP.
SPED Checkpoint
Remind students: "Inside parentheses, the sign is the opposite of what you see. Outside parentheses, the sign stays same."
UNIT: ALGEBRA 1 REINFORCEMENT KEY PAGE 3 OF 6 ENG-TEACHER-REV.01
Guided Teacher Solution Key
6-Week Pacing
SPED Strategy for Week 4: Reference the perfect squares strip provided on student sheets. Encourage students to check the root rules inside Question 20.
Question 16
Exponents Rules
\( (2x^3 y^{-2})^4 \)
Ans: \( \frac{16x^{12}}{y^8} \)
Step-by-Step Solution
1. Apply power: \( 2^4 \cdot (x^3)^4 \cdot (y^{-2})^4 \).
2. Coefficient: \( 2^4 = 16 \).
3. Powers: \( 16 \cdot x^{12} \cdot y^{-8} \).
4. Move negative: \( \mathbf{\frac{16x^{12}}{y^8}} \).
SPED Checkpoint
Check that students apply the power of 4 to the number coefficient \( 2 \), not just the variables. Students often write \( 2x^{12} \).
Question 17
Simplifying Radicals
\( \sqrt{72} \) and \( \sqrt{48x^5} \)
Ans: \( 6\sqrt{2} \), \( 4x^2\sqrt{3x} \)
Step-by-Step Solution
Part A: \( \sqrt{36 \cdot 2} = \mathbf{6\sqrt{2}} \).
Part B: \( \sqrt{16x^4 \cdot 3x} = \sqrt{16} \cdot \sqrt{x^4} \cdot \sqrt{3x} \)
\( = \mathbf{4x^2\sqrt{3x}} \).
SPED Checkpoint
Use the printed perfect squares reference on their sheets to identify factors: \( 36 \) for \( 72 \) and \( 16 \) for \( 48 \).
Question 18
Radical Operations
\( \sqrt{3} \cdot \sqrt{12} \) & \( \frac{\sqrt{50}}{\sqrt{2}} \)
Ans: \( 6 \) and \( 5 \)
Step-by-Step Solution
Part A: \( \sqrt{3 \cdot 12} = \sqrt{36} = \mathbf{6} \).
Part B: \( \sqrt{\frac{50}{2}} = \sqrt{25} = \mathbf{5} \).
SPED Checkpoint
Ensure students fully simplify. They often write \( \sqrt{36} \) and stop, forgetting that \( 36 \) has a clean square root of \( 6 \).
Question 19
Scientific Notation
\( (3 \times 10^4) \times (4 \times 10^5) \)
Ans: \( 1.2 \times 10^{10} \)
Step-by-Step Solution
1. Multiply values: \( 3 \times 4 = 12 \).
2. Add powers: \( 10^{4+5} = 10^9 \).
3. Combine: \( 12 \times 10^9 \).
4. Shift decimal: \( \mathbf{1.2 \times 10^{10}} \).
SPED Checkpoint
Check the final conversion step. Students will often leave the result as \( 12 \times 10^9 \), violating standard scientific notation rules.
Question 20
Fractional Exp
\( 27^{2/3} \) and \( 16^{-1/2} \)
Ans: \( 9 \), \( \frac{1}{4} \)
Step-by-Step Solution
Part A: \( (\sqrt[3]{27})^2 = 3^2 = \mathbf{9} \).
Part B: \( \frac{1}{16^{1/2}} = \frac{1}{\sqrt{16}} = \mathbf{\frac{1}{4}} \).
SPED Checkpoint
Model the fraction conversion rules clearly. Check that they do not multiply the base by the fraction exponent (e.g. \( 27 \times 2/3 = 18 \)).
UNIT: ALGEBRA 1 REINFORCEMENT KEY PAGE 4 OF 6 ENG-TEACHER-REV.01
Guided Teacher Solution Key
6-Week Pacing
SPED Strategy for Week 5: Check GCF variables carefully in Question 21. For FOIL (Question 22), encourage students to draw connecting arcs to prevent missing multiplications.
Question 21
Factor out a GCF
\( 6x^3 - 15x^2 + 9x \)
Ans: \( 3x(2x^2 - 5x + 3) \)
Step-by-Step Solution
1. Coefficients: \( 6, -15, 9 \rightarrow \) GCF is \( 3 \).
2. Variable parts: \( x^3, x^2, x \rightarrow \) GCF is \( x \).
3. Combine: GCF is \( \mathbf{3x} \).
4. Divide terms: \( \mathbf{3x(2x^2 - 5x + 3)} \).
SPED Checkpoint
Students commonly factor out only the number \( 3 \) and forget to pull out the variable \( x \), leaving higher powers inside.
Question 22
Multiply Binomials
\( (2x + 3)(x - 5) \)
Ans: \( 2x^2 - 7x - 15 \)
Step-by-Step Solution
1. First: \( 2x \cdot x = 2x^2 \).
2. Outer/Inner: \( -10x + 3x = -7x \).
3. Last: \( 3 \cdot (-5) = -15 \).
4. Combine: \( \mathbf{2x^2 - 7x - 15} \).
SPED Checkpoint
Check the Last term: students may add \( 3 + (-5) = -2 \) instead of multiplying to get \( -15 \). Emphasize multiplication.
Question 23
Difference of Squares
\( 4x^2 - 25 \)
Ans: \( (2x-5)(2x+5) \)
Step-by-Step Solution
1. Form: \( a^2 - b^2 \).
2. Find \( a \): \( \sqrt{4x^2} = 2x \).
3. Find \( b \): \( \sqrt{25} = 5 \).
4. Plug in: \( \mathbf{(2x - 5)(2x + 5)} \).
SPED Checkpoint
Students often forget to write the two opposite signs, incorrectly factoring to \( (2x - 5)^2 \) which has a middle term.
Question 24
Subtract Polynomials
\( (3x^2 + 5x - 6) \)
\( - (x^2 - 2x + 4) \)
Ans: \( 2x^2 + 7x - 10 \)
Step-by-Step Solution
1. Distribute: \( -x^2 + 2x - 4 \).
2. Combine \( x^2 \): \( 3x^2 - x^2 = 2x^2 \).
3. Combine \( x \): \( 5x + 2x = 7x \).
4. Combine constants: \( -6 - 4 = -10 \).
5. Result: \( \mathbf{2x^2 + 7x - 10} \).
SPED Checkpoint
The warning clue box on the student sheet is crucial. Ensure they distribute the negative to ALL three terms inside the second set.
Question 25
Perfect Squares
\( x^2 + 10x + 25 \)
Ans: \( (x + 5)^2 \)
Step-by-Step Solution
1. Multiply to \( c = 25 \), Add to \( b = 10 \).
2. Factors: \( 5 \) and \( 5 \).
3. Write factors: \( (x + 5)(x + 5) \).
4. Simplify: \( \mathbf{(x + 5)^2} \).
SPED Checkpoint
Some students write \( (x + 25)(x + 1) \) because they see \( 25 \). Direct them to verify if those terms add up to \( 10 \).
UNIT: ALGEBRA 1 REINFORCEMENT KEY PAGE 5 OF 6 ENG-TEACHER-REV.01
Guided Teacher Solution Key
6-Week Pacing
SPED Strategy for Week 6: Check that students list domain elements without repeats (Question 27). Practice finding differences in y-values vertically for tables (Question 28).
Question 26
Function Notation
\( f(x) = -3x + 8 \)
Ans: \( f(4) = -4 \)
Step-by-Step Solution
1. Plug in \( x = 4 \).
2. Compute: \( -3(4) + 8 \).
3. Multiply: \( -12 + 8 \).
4. Solve: \( \mathbf{f(4) = -4} \).
SPED Checkpoint
Check for integer addition errors: \( -12 + 8 \). Students might mistakenly calculate positive \( 4 \) or \( -20 \).
Question 27
Domain and Range
\( \{ (2,3), (4,-1), (6,5), (8,3) \} \)
D:\(\{2,4,6,8\}\) | R:\(\{-1,3,5\}\)
Step-by-Step Solution
1. Domain = list of inputs: \( \mathbf{\{2, 4, 6, 8\}} \).
2. Range = list of outputs: \( \mathbf{\{-1, 3, 5\}} \).
3. Note: Repeats in range (the \( 3 \)) are only listed once.
SPED Checkpoint
Make sure students write the range values from least to greatest and do not write duplicate \( 3 \)'s.
Question 28
Slope from Table
y increases by 6
x increases by 2
Ans: \( m = 3 \)
Step-by-Step Solution
1. Change in \( y = 11 - 5 = 6 \).
2. Change in \( x = 3 - 1 = 2 \).
3. Slope \( m = \frac{\Delta y}{\Delta x} = \frac{6}{2} \).
4. Solve: \( \mathbf{m = 3} \).
SPED Checkpoint
A common mistake is putting change of x over change of y, resulting in a slope of \( 1/3 \). Emphasize "rise over run".
Question 29
Linear vs. Exponential
A: Doubles
B: constant rate
A: Exp | B: Linear
Step-by-Step Solution
A: "Doubles" indicates geometric growth, which represents an Exponential model.
B: Constant adding rate of "+12 gallons" represents an arithmetic Linear model.
SPED Checkpoint
Use real-life visual analogies: Linear increases like standard stairs; Exponential multiplies like bacteria spreading rapidly.
Question 30
Write Linear Eq
Point: (0, 4) | m = 3
Ans: \( y = 3x + 4 \)
Step-by-Step Solution
1. Form: \( y = mx + b \).
2. Identify y-intercept: \( (0, 4) \rightarrow b = 4 \).
3. Slope \( m = 3 \).
4. Equation: \( \mathbf{y = 3x + 4} \).
SPED Checkpoint
Make sure students place coordinates correctly. They sometimes write \( y = 4x + 3 \) instead, swapping variables \( m \) and \( b \).
UNIT: ALGEBRA 1 REINFORCEMENT KEY PAGE 6 OF 6 ENG-TEACHER-REV.01