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Monday — Linear-Quadratic Systems
Tuesday — Systems of Inequalities (Algebra + Graph)
Wednesday — Word-to-System Translation + Q35 Redo
Thursday — Graphing Functions from Scratch
Friday — Timed Part III + Part IV
Saturday — Full-Length Regents Simulation (3 hours)
After Week 3

Week 3 — Systems, Graphing, and Full Simulation

Goal this week: convert the biggest point-loss area on the June 25 exam (Part III multi-step systems and Part IV graphing) into reliable points. End with a full-length simulation.

Time: ~60 min weekdays, ~3 hours Saturday.

Heuristics in heavy use: H1 (read prompt twice, circle every noun), H4 (simplest form), plus the graphing rubric checklist.

How to use the practice problems: every problem has a Hint and a Show answer drop-down. Attempt the problem on paper first; open Hint only if stuck; reveal Show answer only after committing to one of your own.

Monday — Linear-Quadratic Systems

Objective

Solve linear-quadratic systems all the way through to ordered pairs $(x, y)$ . The June 25 Q34 lost a point because $x$ -values were found but $y$ -values were missed or had wrong signs.

The completion checklist (use on EVERY problem)

Did I find every $x$ value the quadratic produces?
Did I back-substitute each $x$ into the linear equation (it’s easier)?
Did I keep the negative signs through back-substitution?
Did I write my final answer as ordered pairs $(x, y)$ , not just $x$ values?

Practice problems

Solve each system algebraically. Report all solutions as ordered pairs.

1. $\begin{cases} y = x^2 - 5x + 6 \\ y = x + 1 \end{cases}$

Hint

Set equal: $x^2 - 5x + 6 = x + 1 \implies x^2 - 6x + 5 = 0$ . Factors nicely.

Show answer

$(1, 2)$ and $(5, 6)$ .

$(x - 1)(x - 5) = 0$ , then $y = x + 1$ gives the pairs.

2. $\begin{cases} y = x^2 + 9x + 4 \\ y - 2x = -6 \end{cases}$ (Q34)

Hint

Rewrite linear as $y = 2x - 6$ . Sub in and collect.

Show answer

$(-5, -16)$ and $(-2, -10)$ .

$2x - 6 = x^2 + 9x + 4 \implies 0 = x^2 + 7x + 10 \implies (x + 5)(x + 2) = 0$ . Back-substitute into $y = 2x - 6$ (H3 — keep signs).

3. $\begin{cases} y = x^2 - 4 \\ y = 3x \end{cases}$

Hint

$x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0$ .

Show answer

$(4, 12)$ and $(-1, -3)$ .

4. $\begin{cases} y = 2x^2 + 3x - 1 \\ y = x + 2 \end{cases}$

Hint

Set equal and collect: $2x^2 + 2x - 3 = 0$ . Doesn’t factor over integers — use the formula.

Show answer

$x = \dfrac{-1 \pm \sqrt{7}}{2}$ , giving $\left(\dfrac{-1 + \sqrt{7}}{2},\ \dfrac{3 + \sqrt{7}}{2}\right)$ and $\left(\dfrac{-1 - \sqrt{7}}{2},\ \dfrac{3 - \sqrt{7}}{2}\right)$ .

$a = 2, b = 2, c = -3$ . Disc $= 4 + 24 = 28$ . $x = \dfrac{-2 \pm 2\sqrt{7}}{4} = \dfrac{-1 \pm \sqrt{7}}{2}$ . Back-substitute into $y = x + 2$ .

5. $\begin{cases} y = -x^2 + 4x \\ y = -x + 4 \end{cases}$

Hint

$-x^2 + 4x = -x + 4 \implies 0 = x^2 - 5x + 4 \implies (x - 1)(x - 4) = 0$ .

Show answer

$(1, 3)$ and $(4, 0)$ .

Back-substitute into $y = -x + 4$ :

$x = 1 \implies y = 3$
$x = 4 \implies y = 0$

6. $\begin{cases} y = x^2 + 2x - 8 \\ y = 2x + 1 \end{cases}$

Hint

$x^2 + 2x - 8 = 2x + 1 \implies x^2 = 9$ .

Show answer

$(3, 7)$ and $(-3, -5)$ .

Self-check question

After solving the quadratic, which equation should you back-substitute into and why? (The linear equation — arithmetic is simpler and less error-prone than re-squaring.)

Tuesday — Systems of Inequalities (Algebra + Graph)

Objective

Graph a system of inequalities with correct line types, correct shading, and labels — the three things graders specifically check.

Graph rubric (every problem must satisfy this)

Element	Required
Axes labeled	" $x$ " and " $y$ " or context labels
Scale	tick marks at consistent intervals
Line	solid for $\leq$ or $\geq$ ; dashed for $<$ or $>$
Plotted points	at least 2 per line, both intercepts when possible
Arrows	both ends of every infinite line
Shading	correct side of each boundary; overlap distinctly visible
Labels	each inequality named on its line (e.g., " $y \leq 2x + 1$ ")

Workflow

Convert each inequality to slope-intercept form.
Decide solid vs dashed from the inequality sign.
Plot the line.
Test the point $(0, 0)$ — if it satisfies, shade that side; if not, shade the other side.
Label the line.
Identify the overlap and outline it.

Practice problems

Graph each system. Show your work on each step.

1. $y \leq 2x + 1$ and $y > -x + 3$

Hint

Line A solid ( $\leq$ ), line B dashed ( $>$ ). Test $(0, 0)$ in each to determine shading.

Show answer

Line A: $y \leq 2x + 1$ — solid, through $(0, 1)$ and $(1, 3)$ . Test $(0,0)$ : $0 \leq 1$ ✓ → shade below (toward origin).

Line B: $y > -x + 3$ — dashed, through $(0, 3)$ and $(3, 0)$ . Test $(0,0)$ : $0 > 3$ ✗ → shade above (away from origin).

Feasible region: the overlap of “below A” and “above B.”

2. $y \geq x - 2$ and $y \leq -2x + 8$

Hint

Both solid. Both lines fail to pass through the origin, so $(0,0)$ is a clean test point for each.

Show answer

Line A: $y \geq x - 2$ — solid, through $(0, -2)$ and $(2, 0)$ . Test $(0,0)$ : $0 \geq -2$ ✓ → shade above.

Line B: $y \leq -2x + 8$ — solid, through $(0, 8)$ and $(4, 0)$ . Test $(0,0)$ : $0 \leq 8$ ✓ → shade below.

Feasible region: the overlap — above A and below B.

3. $x + y < 10$ and $y > 2x - 1$

Hint

Rewrite the first: $y < -x + 10$ . Both dashed (strict).

Show answer

Line A: $y < -x + 10$ — dashed, through $(0, 10)$ and $(10, 0)$ . Test $(0,0)$ : $0 < 10$ ✓ → shade below.

Line B: $y > 2x - 1$ — dashed, through $(0, -1)$ and $(1, 1)$ . Test $(0,0)$ : $0 > -1$ ✓ → shade above.

Feasible region: below A and above B.

4. $2x + y \leq 12$ and $x \geq 0$ and $y \geq 0$

Hint

The last two restrict to the first quadrant. The first is a solid line.

Show answer

Line: $y \leq -2x + 12$ — solid, through $(0, 12)$ and $(6, 0)$ . Test $(0,0)$ : $0 \leq 12$ ✓ → shade below.

Combined with $x \geq 0$ and $y \geq 0$ , the feasible region is the triangle with vertices $(0, 0)$ , $(6, 0)$ , $(0, 12)$ .

5. $3x - y > 6$ and $y \geq -x + 2$

Hint

Rewrite the first: $y < 3x - 6$ (dashed). The second stays $y \geq -x + 2$ (solid).

Show answer

Line A: $y < 3x - 6$ — dashed, through $(0, -6)$ and $(2, 0)$ . Test $(0,0)$ : $0 < -6$ ✗ → shade below the line (away from origin).

Line B: $y \geq -x + 2$ — solid, through $(0, 2)$ and $(2, 0)$ . Test $(0,0)$ : $0 \geq 2$ ✗ → shade above (away from origin).

Feasible region: below A and above B.

Self-check question

What test point should you use to determine which side to shade? ( $(0, 0)$ is almost always easiest — unless the line passes through the origin, in which case use $(1, 0)$ or $(0, 1)$ .)

Wednesday — Word-to-System Translation + Q35 Redo

Objective

Translate two-quantity word problems into a full system: inequalities + graph + valid combination + justification. Then re-do Q35 end-to-end.

Translation template

Every two-quantity word problem decomposes into:

Money / value inequality: $\text{rate}_x \cdot x + \text{rate}_y \cdot y\ [\geq,\ \leq,\ =]\ \text{total}$
Quantity / capacity inequality: $x + y\ [\leq,\ =]\ \text{cap}$
Domain limits if $x, y$ are real-world counts: $x \geq 0$ , $y \geq 0$

Solution template (4 deliverables)

System of inequalities written clearly.
Graph following the Tuesday rubric.
Combination — a point in the feasible region. Verify both inequalities numerically.
Justification — a sentence stating which inequalities are satisfied and why the combination is valid.

Practice problems

1. Marcus sells lemonade at $2/cup and cookies at $3/cookie. He wants to earn at least $60 at the fair. He can carry max 40 items total. Let $x$ = cups, $y$ = cookies. Write the system, graph it, state a valid combination, justify.

Hint

Money inequality: $2x + 3y \geq 60$ . Quantity inequality: $x + y \leq 40$ . Domain: $x, y \geq 0$ .

Show answer

System: $2x + 3y \geq 60$ ; $x + y \leq 40$ ; $x \geq 0$ ; $y \geq 0$ .

Combination: $x = 10$ cups, $y = 20$ cookies.

Money: $2(10) + 3(20) = 20 + 60 = 80 \geq 60$ ✓
Quantity: $10 + 20 = 30 \leq 40$ ✓

Justification: “Selling 10 cups and 20 cookies earns $80 (meets the $60 minimum) and uses 30 of the 40-item capacity.”

2. A bookstore sells novels for $15 and cookbooks for $25. To break even today they must earn at least $300. They have shelf space for at most 30 books. Let $x$ = novels, $y$ = cookbooks. System, graph, combination, justification.

Hint

$15x + 25y \geq 300$ ; $x + y \leq 30$ .

Show answer

System: $15x + 25y \geq 300$ ; $x + y \leq 30$ ; $x \geq 0$ ; $y \geq 0$ .

Combination: $x = 10$ novels, $y = 10$ cookbooks.

Money: $15(10) + 25(10) = 150 + 250 = 400 \geq 300$ ✓
Quantity: $10 + 10 = 20 \leq 30$ ✓

Justification: “10 novels and 10 cookbooks earn $400 (above the $300 break-even) using 20 of the 30 shelf slots.”

3. Sarah’s babysitting + tutoring problem (Q35 redo): $6/hr babysitting ( $x$ ), $12/hr tutoring ( $y$ ), at least $120/week, max 14 hours total. System, graph, combination, justification.

Hint

$6x + 12y \geq 120$ (simplifies to $y \geq -\tfrac{1}{2}x + 10$ ); $x + y \leq 14$ .

Show answer

System:

\begin{cases} 6x + 12y \geq 120 & \text{(money: at least \$120)} \\ x + y \leq 14 & \text{(time: at most 14 hours)} \\ x \geq 0,\ y \geq 0 & \text{(non-negative hours)} \end{cases}

Convert for graphing:

$6x + 12y \geq 120 \implies y \geq -\tfrac{1}{2}x + 10$ . Solid line. Test $(0, 0)$ : $0 \geq 10$ ? No — shade above (away from origin).
$x + y \leq 14 \implies y \leq -x + 14$ . Solid line. Test $(0, 0)$ : $0 \leq 14$ ? Yes — shade below (toward origin).

Combination: $x = 4$ babysitting hours, $y = 10$ tutoring hours.

Money: $6(4) + 12(10) = 24 + 120 = 144 \geq 120$ ✓
Hours: $4 + 10 = 14 \leq 14$ ✓

Justification: “Working 4 hours babysitting and 10 hours tutoring satisfies both constraints. Sarah earns $144 (which meets the $120 goal) and works exactly 14 hours (within the 14-hour limit).”

4. A coffee shop has two drinks: lattes ($4) and teas ($3). The owner needs to sell at least $200 worth in an hour. The shop can make at most 60 drinks per hour. Let $x$ = lattes, $y$ = teas. System, graph, combination, justification.

Hint

$4x + 3y \geq 200$ ; $x + y \leq 60$ .

Show answer

System: $4x + 3y \geq 200$ ; $x + y \leq 60$ ; $x \geq 0$ ; $y \geq 0$ .

Combination: $x = 40$ lattes, $y = 20$ teas.

Money: $4(40) + 3(20) = 160 + 60 = 220 \geq 200$ ✓
Quantity: $40 + 20 = 60 \leq 60$ ✓

Justification: “Selling 40 lattes and 20 teas earns $220 (above the $200 minimum) using all 60 of the hourly drink capacity.”

Self-check question

How do you decide whether to shade above or below a boundary line? (Pick a test point not on the line — usually $(0, 0)$ . If it satisfies the inequality, shade that side; otherwise, the other side.)

Thursday — Graphing Functions from Scratch

Objective

Drill graphing fidelity. Every graph must satisfy the rubric: labeled axes, scale, plotted points, arrows, and (for inequalities) shading.

Instructions

Graph all 8 functions on separate coordinate grids.
For each, plot at least 3 points and the key feature (intercept for lines, vertex for parabolas, etc.).
Apply arrows on both ends of every infinite curve.
Self-check against the rubric checklist after each one.

Practice — graph these

1. $y = -3x + 1$

Hint

Slope-intercept form: $y$ -intercept is $1$ , slope is $-3$ .

Show answer

$y$ -intercept $(0, 1)$ ; slope $-3$ → from $(0,1)$ go right 1, down 3 to $(1, -2)$ ; also $(-1, 4)$ . Straight line, arrows both ends, label " $y = -3x + 1$ ".

2. $f(x) = 2x - 4$

Hint

Slope $2$ , $y$ -intercept $-4$ . Find the $x$ -intercept by setting $y = 0$ .

Show answer

$y$ -int $(0, -4)$ ; $x$ -int $(2, 0)$ ; also $(1, -2)$ and $(3, 2)$ . Straight line, arrows, label.

3. $y = x^2 - 4$

Hint

Parabola opening up; vertex shifted down 4 from $(0, 0)$ .

Show answer

Vertex $(0, -4)$ ; $x$ -intercepts $(\pm 2, 0)$ ; symmetric points $(\pm 1, -3)$ , $(\pm 3, 5)$ . Opens up, smooth curve, arrows on both branches, label.

4. $g(x) = -(x - 1)^2 + 4$

Hint

Vertex form: vertex at $(1, 4)$ . Negative coefficient → opens down.

Show answer

Vertex $(1, 4)$ ; opens down; $x$ -intercepts $(-1, 0)$ and $(3, 0)$ ; symmetric points $(0, 3)$ and $(2, 3)$ .

5. $y = |x - 2|$

Hint

V-shape, vertex shifted right 2.

Show answer

V-shape, vertex $(2, 0)$ , opens up; $(0, 2), (1, 1), (3, 1), (4, 2)$ .

6. $y = |x| + 3$

Hint

V-shape shifted up 3 (still opens up).

Show answer

V-shape, vertex $(0, 3)$ , opens up; $(1, 4), (-1, 4), (2, 5), (-2, 5)$ .

7. $f(x) = -3x$ and $g(x) = x^2 + 2$ on same axes (Q31!) — find intersection.

Hint

Set equal: $-3x = x^2 + 2 \implies x^2 + 3x + 2 = 0 \implies (x + 1)(x + 2) = 0$ .

Show answer

Line $f(x) = -3x$ : through $(0, 0)$ , $(1, -3)$ , $(-1, 3)$ , $(-2, 6)$ .

Parabola $g(x) = x^2 + 2$ : vertex $(0, 2)$ ; $(\pm 1, 3)$ , $(\pm 2, 6)$ , $(\pm 3, 11)$ .

Intersections: $(-1, 3)$ and $(-2, 6)$ .

8. $y = (x + 2)(x - 4)$

Hint

Roots at $x = -2$ and $x = 4$ ; vertex sits midway at $x = 1$ .

Show answer

$x$ -intercepts $(-2, 0)$ and $(4, 0)$ ; vertex at $x = 1$ , $y = (3)(-3) = -9$ , so $(1, -9)$ ; opens up.

Self-check rubric

After each graph, score yourself:

☐ Both axes labeled
☐ Scale visible (tick marks)
☐ Key feature plotted (vertex / intercept)
☐ At least 2 other points plotted
☐ Arrows on infinite ends
☐ Line/curve labeled with its equation

A graph that misses any of these would lose points on the Regents.

Friday — Timed Part III + Part IV

Objective

Final dress rehearsal before Saturday’s full simulation. Apply every checklist under timed pressure.

Instructions

Pull Part III (Q31–Q34) and Part IV (Q35) from a prior Regents administration.
Timer: 60 minutes.
Work in pen. Use the linear-quadratic completion checklist on every system problem.
Use the graphing rubric on every graph.
Score strictly against the Model Response Set.

Self-score targets

Part III: 12+ / 16
Part IV: 5+ / 6

What to do with misses

For every lost point, identify which checklist item failed:

Did you skip the back-substitution step? (Completion checklist #2)
Did you forget a negative sign? (Checklist #3)
Did you give $x$ -values only? (Checklist #4)
Did you forget to shade or label a graph? (Graphing rubric)
Did you forget the justification on Q35-type problems? (Solution template #4)

Each of these is a process failure, not a knowledge failure. Re-do the problem on Saturday morning before the full simulation.

Saturday — Full-Length Regents Simulation (3 hours)

Objective

Realistic, end-to-end measurement of progress.

Instructions

Setup: print a complete prior Regents Algebra I (any released year). Sit at a clear table. No phone, no distractions.
Timer: 3 hours, no break. Bring a graphing calculator, ruler, pen, and pencil (graphs only).
Work the test in order. Apply every heuristic and checklist from the past three weeks.
Score immediately afterward against the Model Response Set.

Score breakdown to record

Part	Possible	Earned	vs. June 25 baseline
I (MC)	48	__	44
II (2-pt)	12	__	10
III (4-pt)	16	__	~5
IV (6-pt)	6	__	~3
Raw total	82	__	62

Success bar for Week 3

Raw score $\geq$ 70 (up from 62) — that’s a scale score of ~85+, the Mastery threshold.
Zero “didn’t finish” errors on multi-step problems.
Zero graphing rubric misses.

Post-test reflection

Write 1–2 paragraphs in the Mistake Log:

What worked? Which heuristics fired automatically?
What still needs work? Which patterns showed up multiple times?
What’s the next focus? Based on remaining gaps.

If raw score is below 70, repeat the day(s) of the corresponding weak area before retaking the test.

After Week 3

You’ve completed 21 sessions of focused work. The next 2–3 weeks before the actual Regents should be:

One full-length simulation per week under timed conditions.
Daily 20-minute mistake-log review — flip through the entries and re-do any problem you no longer instantly see the solution to.
Targeted drills on whatever pattern shows up most in your latest simulation.

The pattern is durable: every Regents mistake fits into one of the H1–H5 buckets. If you keep tagging and reviewing, the failure modes get harder and harder to repeat.

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