Algebra Scaffolded Assessment Level 1 • Supported Algebra I • Quarter 1 Review
Linear Equations & Foundations
CALCULATOR: OFF
Total Points: 25
Name:
Date:
Period:
Solving Roadmap:
1. Distribute \(a(b+c)\) 2. Combine Like Terms 3. Move Variables to One Side 4. Isolate \(x\)
1 Solve for \(x\): \(2x - 3 = 4(x + 1)\)
3 Points
Step Checklist:
Distribute the \(4\) on the right Subtract \(2x\) from both sides Divide by the coefficient
Show Work:
Final Answer: \(x =\)
2 Find the value of \(h\): \(24 = \frac{h}{10} - 6\)
3 Points
Hint & Clues:
First, eliminate \(- 6\) by adding 6 to both sides.
Then, eliminate division by 10 by multiplying by 10 .
Show Work:
Final Answer: \(h =\)
3 Solve the equation: \(\frac{2}{3}(x - 5) = 6\)
3 Points
Helpful Strategy:
Multiply both sides by reciprocal \(\frac{3}{2}\) , or multiply by \(3\) then divide by \(2\).
\(6 \times \frac{3}{2} = \frac{18}{2} = 9\)
Show Work:
Final Answer: \(x =\)
Algebra I • Modified Interim Assessment Page 1 of 3
Section 2
Expressions & Real-World Models
Name:
4 Simplify by combining like terms: \(3x + 5 - 2x + 7\)
3 Points
Step 1: Group Like Terms
Variable Terms (\(x\)) \(3x\) and \(-2x\)
Constant Numbers \(+5\) and \(+7\)
Step 2: Combine Each Group
Variables: \(3x - 2x =\)
Constants: \(5 + 7 =\)
Choose the correct equivalent expression:
(A) \(5x + 12\) (B) \(x + 12\) (C) \(x + 2\)
5 Modeling a Word Problem:
3 Points
A repair technician charges a one-time visit fee of $50 plus an hourly rate of $30 per hour . Let \(h\) represent the number of hours worked.
Breakdown:
Repeating cost (per hour): \(30 \cdot h\)
Starting fee (added once): \(+ 50\)
Which function models the total cost \(C(h)\)?
(A) \(C(h) = 50h + 30\) (B) \(C(h) = 30h + 50\) (C) \(C(h) = 80h\) (D) \(C(h) = 80 + h\)
6 Equivalent Equations:
3 Points
Given the base equation: \(4x + 8 = 0\) (Its solution is \(x = -2\) ).
Which of the following equations have the exact same solution (\(x = -2\)) ? (Check all that apply)
\(2x + 4 = 0\) Notice: divided every term by 2
\(8x + 16 = 0\) Notice: multiplied every term by 2
\(-(x + 2) = 0\) Test: \(-(-2 + 2) = -(0) = 0\)
\(-4x + 8 = 0\) Test: \(-4(-2) + 8 = 8 + 8 = 16 \neq 0\)
Algebra I • Modified Interim Assessment Page 2 of 3
Section 3
Functions & Applications
Name:
7 Evaluating Functions: Let \(f(x) = 2x^2 + 4x - 3\). Find the value of \(f(2)\) .
3 Points
Guided Substitution Template: Replace every \(x\) with 2
\(f(2) = 2(\underline{\phantom{x}})^2 + 4(\underline{\phantom{x}}) - 3\)
Step 1: Exponent \(2^2 = 4\)
Step 2: Multiply \(2(4) = 8\) | \(4(2) = 8\)
Step 3: Add / Subtract \(8 + 8 - 3 =\) ___
8 Reading a Function Graph: Use the graph of \(y = f(x)\) below:
4 Points
0 2 4 6 8 10 y 2 4 5 6 8 x
Part A: What is the value of \(f(2)\)? Look at \(x = 2\) on the horizontal axis and read up to the graph.
Answer: \(f(2) =\)
Part B: Find an \(x\)-value where \(f(x) = 6\). Look across line \(y = 6\). Name one \(x\) coordinate.
Answer: \(x =\)
9 Open Response Application: Savings Goal
6 Points
You currently have $50 already saved in your bank account. You plan to save $5 each week .
Part A: Write an equation Total amount \(T\) after \(w\) weeks:
\(T =\) ______________________
Part B: How many weeks to save $85? Set \(T = 85\) and solve for \(w\):
Solution: \(w =\) weeks
You reached the end of this assessment! Double check your work.
Page 3 of 3
Algebra Challenge Assessment Level 2 • Challenge Algebra I • Quarter 1 Review
Linear Systems & Structures
CALCULATOR: OFF
Total Points: 25
Name:
Date:
Period:
1 Solve for \(x\): \(\frac{4}{x + 1} = \frac{6}{2x - 3}\)
3 Points
Show algebraic work (cross-multiply or common denominator):
Answer: \(x =\)
2 Simplify completely: \(2(x^2 - 3x) + 5x - x^2\)
3 Points
Show work (distribute then combine like terms):
Simplified Form: ________________
Checklist Reminder: • Distribute \(2\) to both terms • Group \(x^2\) terms • Group linear \(x\) terms
3 Multi-Step Rational Equation: \(\frac{1}{2}(18 - 5x) = 2 - \frac{4}{3}x\)
5 Points
Find the value of \(x\) that satisfies the equation. Show all intermediate steps clearly and express your final answer as an integer or reduced improper fraction.
Full Algebraic Solution Area:
Final Solution: \(x =\)
Algebra I • Challenge Assessment Page 1 of 2
Level 2 • Challenge
Parameters & Function Comparison
Name:
4 Parameter Analysis: \(21x + 14 = 7(3x + a)\)
4 Points
In the equation above, \(a\) is a real constant. For what value of \(a\) does the equation have infinitely many solutions ? Justify your reasoning mathematically.
Justification & Solution:
Value of \(a\): \(a =\)
5 Real-World Savings Model & Extension:
4 Points
Jordan starts with $50 in savings and adds $5 every week.
Part A: Model & Target Write an equation for total \(T\) after \(w\) weeks, and calculate weeks needed to reach $85.
Eq: \(T = 5w + 50\) | \(w =\) _____ wks
Part B: Comparative Rate If Jordan increases weekly savings to $10, how many fewer weeks will it take to reach the $85 target?
Takes _____ fewer weeks
6 Dual-Function Graph Comparison:
6 Points
0 2 4 6 8 y 2 4 6 8 x g(x) f(x)
Part A: Evaluate Function Values Find \(f(3)\) and \(g(3)\) using the graph:
\(f(3) =\) ____ \(g(3) =\) ____
Part B: System Intersection State the \(x\)-value where \(f(x) = g(x)\):
\(x \approx\) _______
Part C: Inequality Comparison Identify an \(x\)-value where \(f(x) > g(x)\). Explain why:
Because \(f(x)\) is graphed (above / below) \(g(x)\) at \(x =\) _____.
Assessment Complete. Verify all calculations.
Algebra Assessment Scoring Guide Teacher Master Key Algebra I • Quarter 1 Interim
Scoring Guide: Level 1 Supported
Total Points: 25
Target: 80% Mastery
1 Distributive Equation: \(2x - 3 = 4(x + 1)\)
Ans: \(x = -\frac{7}{2}\) or \(-3.5\) \(2x - 3 = 4x + 4 \implies -2x = 7 \implies x = -\frac{7}{2}\)
3 Pts: 1 pt distribute, 1 pt isolate, 1 pt ans.
2 Two-Step Fraction: \(24 = \frac{h}{10} - 6\)
Ans: \(h = 300\) Add 6 \((30 = \frac{h}{10})\), multiply by 10 \((h = 300)\).
3 Pts: 1 pt add 6, 1 pt multiply 10, 1 pt ans.
3 Equation with Factor: \(\frac{2}{3}(x - 5) = 6\)
Ans: \(x = 14\) \(x - 5 = 6 \cdot \frac{3}{2} = 9 \implies x = 9 + 5 = 14\).
3 Pts: 1 pt reciprocal, 1 pt add 5, 1 pt ans.
4 Combine Like Terms: \(3x + 5 - 2x + 7\)
Ans: (B) \(x + 12\) \((3x - 2x) + (5 + 7) = x + 12\).
3 Pts: 1 pt group, 1 pt combine, 1 pt choice.
5 Linear Model: $50 fee + $30/hr
Ans: (B) \(C(h) = 30h + 50\) Rate 30 is attached to \(h\); starting fee 50 is constant.
3 Pts: Full credit for correct equation model.
6 Equivalent to \(4x + 8 = 0\)
Ans: 1st, 2nd, and 3rd checked (\(2x+4=0\), \(8x+16=0\), \(-(x+2)=0\)). 4th is false.
3 Pts: 1 pt each correct selection.
7 Evaluate: \(f(x) = 2x^2 + 4x - 3\) at \(x = 2\)
Ans: \(f(2) = 13\) \(2(4) + 4(2) - 3 = 8 + 8 - 3 = 13\).
3 Pts: 1 pt substitution, 1 pt PEMDAS, 1 pt ans.
8 Read Graph: Coordinates
Part A: \(f(2) = 5\) | Part B: \(x = 4\) or \(x = 6\)
4 Pts: 2 pts Part A; 2 pts Part B.
9 Open Response: Savings Plan
Part A: \(T = 5w + 50\) | Part B: \(w = 7\) weeks (\(85 = 5w + 50 \implies 35 = 5w \implies w = 7\))
6 Pts: 3 pts Part A equation; 3 pts Part B solution.
Algebra I Scoring Guide • Level 1 Answer Key Page 1 of 3
Teacher Master Key Algebra I • Quarter 1 Interim
Scoring Guide: Level 2 Challenge
Total Points: 25
Extension Standard
1 Rational Proportion: \(\frac{4}{x + 1} = \frac{6}{2x - 3}\) 3 Points
Cross multiply: \(4(2x - 3) = 6(x + 1) \implies 8x - 12 = 6x + 6 \implies 2x = 18 \implies\) \(x = 9\) .
Rubric: 1 pt cross product setup; 1 pt distributing; 1 pt final value \(x = 9\).
2 Polynomial Simplification: \(2(x^2 - 3x) + 5x - x^2\) 3 Points
Distribute: \(2x^2 - 6x + 5x - x^2\). Group like terms: \((2x^2 - x^2) + (-6x + 5x) \implies\)
