MAT8 q2W6D3: Exterior Angle Theorem and Exterior Angle Inequality Theorem

Learning Goals

By the end of the lesson, you will be able to:

1. Solve for an unknown variable in triangle side expressions by forming and solving the inequality set $a+b>c,a+c>b,b+c>a$ and compressing it to the range form $\left|p\right(x)-q(x\left)\right|<r\left(x\right)<p\left(x\right)+q\left(x\right)$, producing a correct solution set for $x$.
2. Translate real constraints (catalog steps, unit conversions, perimeter or measurement bounds) into mathematics, intersect with the triangle-inequality range, and state final feasible values using interval notation and integers when required.
3. Rank angles by comparing algebraic side expressions (for example, $2x+3$ vs. $3x-4$) and justify the ordering using the side–angle relationship.

Key Ideas & Terms

• Triangle inequality (algebraic form): For expressions $s_1\left(x\right)$, $s_2\left(x\right)$, and $s_3\left(x\right)$, all must be positive and satisfy $s_1+s_2>s_3,s_1+s_3>s_2,s_2+s_3>s_1$.
• Range form: Compress to $\left|a\right(x)-b(x\left)\right|<c\left(x\right)<a\left(x\right)+b\left(x\right)$ when a third side is designated.
• Strict inequality: Use $<$ and $>$ only for nondegenerate triangles.
• Feasible set or Intersection: Combine triangle range with external constraints such as $[20,40]$ or integers in steps.
• Compound inequality solving: Solve $L<f\left(x\right)<U$ and give the answer in interval notation.
• Angle–side relationship (algebraic): If $s_1\left(x\right)>s_2\left(x\right)>s_3\left(x\right)$, then the angle opposite $s_1$ is largest and the angle opposite $s_3$ is smallest.

Quick Recall / Prior Knowledge

A. Inequality refresher

1. Solve: $3y-4>5$
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   $3y>9$ so $y>3$.
2. Solve the compound inequality: $-2<5t-7<8$
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   Add 7: $5<5t<15$; divide 5: $1<t<3$. Interval: $(1,3)$.

B. Triangle inequality check

3. Can 6 cm, 9 cm, 14 cm form a triangle?
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   $6+9=15>14$, $6+14=20>9$, $9+14=23>6$. Yes.

4. State the open interval for the third side if two sides are 8 cm and 15 cm.
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   $|8-15|<x<8+15$ $\to$ $7<x<23$, interval $(7,23)$.

C. Angle–side relationship

For sides 5 cm, 12 cm, 13 cm, which angle is largest?

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Largest side is 13 cm, so the angle opposite 13 cm is largest.

D. Units and feasibility

A supplier sells bars in steps of 0.5 m, from 1.0 m to 9.0 m inclusive. If two triangle sides are 3.8 m and 4.1 m, list the smallest and largest supplier lengths for the third side that could work.

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Triangle interval: $(0.3,7.9)$ m. Supplier set: 1.0, 1.5, …, 7.5, 8.0, 8.5, 9.0. Feasible intersection gives smallest 1.0 m and largest 7.5 m.

Explore the Lesson

Solving Triangle Inequalities with Algebraic Sides

Purpose: Learn how to solve for a variable when triangle side lengths are given as algebraic expressions. Turn three inequalities into a single, usable solution set for the variable, check positivity of sides, meet real-life constraints (catalog steps, units, bounds), and rank angles using side expressions. All equations use MathML. All answers are hidden in collapsible details.

1) Why this skill matters

Imagine configuring a foldable metal frame whose bars depend on a setting $x$: one bar is $2x+3$, another is $x+7$, and the third is $18$ cm. You need the full set of $x$-values that produce a valid nondegenerate triangle, not just a single guess.

Guiding question: If sides are functions of $x$, should we test single values or solve for all $x$ that work?

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We solve for all $x$ that work and report a solution set for $x$.

Mini-summary: We will derive the full set of $x$ that makes the triangle valid.

2) The triangle inequalities with algebraic sides

For sides $s_1\left(x\right)$, $s_2\left(x\right)$, and $s_3\left(x\right)$ to form a nondegenerate triangle, we need positivity and the three strict inequalities. When one side is designated as the “third side,” the three inequalities can be compressed into a range:

$\left|s_1\right(x)-s_2(x\left)\right|<s_3\left(x\right)<s_1\left(x\right)+s_2\left(x\right)$

Checkpoint: If $s_1=2x+3$, $s_2=x+7$, $s_3=18$, which method is faster?

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The compressed range method, because $18$ is constant: $\left|\right(2x+3)-(x+7\left)\right|<18<(2x+3)+(x+7)$.

Mini-summary: We will use both methods and confirm that they match.

3) Method A: Three-inequality intersection

Let $s_1=2x+3$, $s_2=x+7$, $s_3=18$.

1. Positivity: $2x+3>0$ $\Rightarrow$ $x>-\frac{3}{2}$; $x+7>0$ $\Rightarrow$ $x>-7$.
2. Triangle inequalities:
   $(2x+3)+(x+7)>18$, $(2x+3)+18>(x+7)$, $(x+7)+18>(2x+3)$
3. Solve each:
   • $3x+10>18$ $\Rightarrow$ $3x>8$ $\Rightarrow$ $x>\frac{8}{3}$
   • $2x+21>x+7$ $\Rightarrow$ $x>-14$
   • $x+25>2x+3$ $\Rightarrow$ $x<22$
4. Intersect: Final $\frac{8}{3}<x<22$.

Checkpoint: Does the final solution imply positivity?

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Yes. $x>\frac{8}{3}$ is stronger than the positivity bounds.

Mini-summary: Intersecting three linear inequalities plus positivity gives a tight open interval for $x$.

4) Method B: Compressed range for a designated third side

Designate $s_3=18$. Then

$\left|\right(2x+3)-(x+7\left)\right|<18<(2x+3)+(x+7)$

Left part: $|x-4|<18$ $\Rightarrow$ $-18<x-4<18$ $\Rightarrow$ $-14<x<22$. Right part: $18<3x+10$ $\Rightarrow$ $\frac{8}{3}<x$. Intersection again gives $\frac{8}{3}<x<22$.

Guiding question: Why still check positivity?

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To exclude any values where a side expression might be nonpositive.

Mini-summary: The range form is quick; still intersect with positivity.

5) A general procedure

1. Identify side expressions and any constants.
2. Write positivity conditions $s_i\left(x\right)>0$.
3. Choose a method: three inequalities or a compressed range for a chosen third side.
4. Intersect results with positivity.
5. Apply real-world constraints (steps, units, bounds).
6. For angle ordering, compare side expressions after choosing a valid $x$.

Checkpoint: If all three sides depend on $x$, is the range method still possible?

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Yes. Choose any side as the third side and solve the double inequality plus positivity.

Mini-summary: The two methods are equivalent; pick the cleaner algebra.

6) Full guided example (all sides variable)

Let $s_1=3x-4$, $s_2=2x+5$, $s_3=x+2$.

Positivity: $3x-4>0$ $\Rightarrow$ $x>\frac{4}{3}$; $2x+5>0$ $\Rightarrow$ $x>-\frac{5}{2}$; $x+2>0$ $\Rightarrow$ $x>-2$. Strongest is $x>\frac{4}{3}$.

Inequalities: $(3x-4)+(2x+5)>(x+2)$, $(3x-4)+(x+2)>(2x+5)$, $(2x+5)+(x+2)>(3x-4)$.

Solving: From the second inequality, $4x-2>2x+5$ $\Rightarrow$ $x>\frac{7}{2}$. Intersection yields $x>\frac{7}{2}$.

Checkpoint: Test $x=4$.

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At $x=4$: $3x-4=8$, $2x+5=13$, $x+2=6$. All inequalities hold.

Mini-summary: With all sides variable, intersection is often cleaner than forcing a range.

7) Compressed range when all sides are variable

Pick $s_3=x+2$ and solve $\left|\right(3x-4)-(2x+5\left)\right|<x+2<(3x-4)+(2x+5)$ with positivity. This again yields $x>\frac{7}{2}$.

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Both methods give the same solution: $x>\frac{7}{2}$.

Mini-summary: The compressed range reproduces the same solution as the intersection method.

8) Practical constraints: catalogs, unit conversions, bounds

Overlay real constraints after obtaining the mathematical interval. Example with sides 2.5 m, 180 cm, and $1.2x+10$ cm results in $50<x<350$, then apply steps such as multiples of 5.

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Smallest multiple strictly greater than $50$ is $55$; largest strictly less than $350$ is $345$.

Mini-summary: Convert units first, then apply the inequality and catalog steps.

9) Edge cases: equality and degeneracy

Equality in any inequality flattens the triangle. Keep endpoints out of the interval.

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Endpoints where equality holds are excluded from the solution set.

Mini-summary: Strict inequalities ensure a nondegenerate triangle.

10) Angle ordering with algebraic sides

Largest angle is opposite the longest side; smallest angle is opposite the shortest side.

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Compare side expressions within the valid $x$-range to determine which angle is largest.

Mini-summary: Order angles by ordering side expressions.

11) Second guided example with bounds

Let $4x+1$, $2x+9$, $3x+4$ be sides (cm), all between $12$ and $60$ inclusive, with integer $x$. Intersection yields integers 3 through 14.

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Testing $x=14$ gives sides $57$, $37$, $46$ which satisfy all inequalities.

Mini-summary: Intersect triangle-inequality, bounds, and step constraints.

12) Strategy choice

Use the range method when a constant or simple side appears; use intersection when expressions are all variable or algebra gets cleaner that way.

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The three-inequality intersection often reduces mistakes with large coefficients.

Mini-summary: Choose the method that keeps algebra clean.

13) Common pitfalls

• Forgetting positivity.
• Dropping strictness.
• Mishandling absolute value.
• Mixing units without converting.
• Forgetting real-world filters after solving.

Show Answer

Correct $|x-4|<18$ as $-18<x-4<18$ $\Rightarrow$ $-14<x<22$.

Mini-summary: A checklist of positivity, strictness, absolute value, units, and filters prevents most errors.

14) Reverse problems

Given $9<2x+1<31$, solve $x$: $4<x<15$. With integers, $5,6,7,8,9,10,11,12,13,14$.

15) Synthesis challenge

With sides $5x-2$, $3x+4$, $2x+9$ and bounds $[20,80]$ cm with step $0.5$, the continuous feasible interval is $5.5\le x\le 16.4$, so allowable settings are 5.5 to 16.0 in 0.5 steps. The longest side is $5x-2$ on the entire feasible set.

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Sample checks at $x=6$ and $x=16$ confirm $5x-2$ is longest.

16) Real-world modeling: tolerances

Use conservative bounds by maximizing differences and minimizing sums when applying tolerances. Example reduces to approximately $x>3.28$.

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Replace informal subscripts by MathML: $|A_{\text{high}}-B_{\text{low}}|<C<A_{\text{low}}+B_{\text{low}}$. Solving yields $x>3.28$.

17) Choice problems: target a largest angle

For sides $4x+3$, $3x+8$, $2x+15$, requiring the largest angle opposite $4x+3$ leads to $x>6$.

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At $x=6$ the two sides tie; strict inequality requires $x>6$.

18) Quick reference

• Positivity + triangle inequalities (strict) + real constraints.
• $\left|A\right|<B$ with $B>0$ becomes $-B<A<B$.
• Order angles by ordering sides.

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Compute $\text{Positivity}\cap \text{Triangle inequalities}\cap \text{Real constraints}$.

19) Your turn, scaffolded

Three tasks with hidden guidance.

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Task 1: With sides $2x+7$, $x+12$, $25$, final: $2<x<30$.

Task 2: With sides $3x+2$, $2x+11$, $x+8$, final: $x>\frac{1}{2}$.

Task 3: With $1.5x+20$ cm, $120$ cm, $0.6x+30$ cm, final: $x>33.333$, so integers $34,35,\dots ,\text{etc.}$

20) References

• Textbook discussions of triangle inequality and variable expressions.
• Algebra references for compound and absolute value inequalities.
• Curriculum notes on translating measurement constraints to intervals.

Example in Action

Five worked examples followed by five "Now You Try" tasks. Reveal each solution after attempting.

Worked Example 1 – Compressed range with one constant side

Sides: $2x+5$, $x+9$, $24$.

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Left: $|x-4|<24$ $\Rightarrow$ $-20<x<28$. Right: $24<3x+14$ $\Rightarrow$ $\frac{10}{3}<x$. Final: $\frac{10}{3}<x<28$.

Worked Example 2 – Three-inequality intersection

Sides: $3x-1$, $2x+6$, $x+8$. Final: $x>\frac{3}{4}$.

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Work through positivity and three inequalities; tightest lower bound is $\frac{3}{4}$.

Worked Example 3 – Units + compressed range + integer catalog

Sides: $120$ cm, $1.2x+10$ cm, $80$ cm.

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Intersection yields integer $26,27,\dots ,158$.

Worked Example 4 – Piecewise angle ordering

Sides: $2x+3$, $4x-5$, $x+10$. See panel for piecewise result.

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On $(\frac{12}{5},5)$, longest is $x+10$. On $(5,18)$, longest is $4x-5$. Tie at $x=5$.

Worked Example 5 – Reverse from a given third-side range

Fixed sides $15$ cm and $28$ cm; third side $3x+1$ cm.

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Range gives $4<x<14$.

Now You Try – 5 Tasks

1. Intersection method for $4x+2$, $x+11$, $3x-1$.
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   Final: $x>\frac{5}{3}$; one integer is $2$.
2. Units + range for $90$ cm, $x+40$ cm, $75$ cm.
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   Interval $(-25,125)$.
3. Mixed sides $18$ cm, $x+7$ cm, $2x+1$ cm.
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   Continuous: $\frac{10}{3}<x<24$, integers $4,5,\dots ,23$.
4. Angle ordering for $5x-6$, $3x+2$, $2x+15$.
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   Domain $\frac{6}{5}<x$; longest $2x+15$ for $(\frac{6}{5},7)$, longest $5x-6$ for $(7,\infty )$.
5. Reverse from fixed pair $22$ cm and $9$ cm; third side $2x+9$ cm.
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   Integer $x$ in $\{3,4,5,6,7,8,9,10\}$.

Try It Out

1. Compressed range with $25$ as third side for $2x+3$, $x+11$.
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   $\frac{10}{3}<x<33$.
2. Three-inequality for $3x-1$, $2x+6$, $x+8$.
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   $x>\frac{3}{4}$.
3. Mixed units: $0.9$ m, $1.5x+20$ cm, third side $140$ cm.
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   $20<x<140$.
4. Integer catalog 0–50 for $2x+8$, $35$, $x+20$.
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   Integers $3,4,\dots ,46$.
5. Reverse from fixed pair $12$ and $25$; third side $x+9$.
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   $4<x<28$.
6. Angle ordering domain for $4x-3$, $x+12$, $2x+7$.
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   Domain $\frac{8}{5}<x<22$; longest $x+12$ for $(\frac{8}{5},5)$, longest $4x-3$ for $(5,22)$; tie at $x=5$.
7. Fractions with constant third side $20$ for $\frac{x}{2}+3$, $\frac{x}{3}+5$.
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   $\frac{72}{5}<x<132$.
8. Step size $0.5$ for $50$ cm, $2x+5$ cm, $x+32$ cm.
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   Smallest $4.5$, largest $40.0$.
9. Largest angle at A with $a=x+2$, $b=18$, $c=25$.
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   $23<x<41$.
10. Perimeter bound $120$ for $3x+1$, $2x+9$, $x+8$.
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    $2<x\le 21.5$.

Check Yourself

1. Why strict inequalities? Open or closed endpoints?
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   Open; equality degenerates the triangle.
2. Range method with $30$, $2x+9$, $x+6$.
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   $5<x<27$.
3. Three-inequality for $4x-1$, $3x+10$, $x+7$.
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   $x>2$.
4. Mixed units: $1.2$ m, $70$ cm, third side $0.5x+10$ cm.
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   $80<x<360$.
5. Angle ordering piecewise for $3x+2$, $5x-7$, $2x+20$.
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   Longest $2x+20$ for $(\frac{25}{6},9)$, longest $5x-7$ for $(9,\infty )$.
6. Reverse from fixed pair $15$ and $25$; third side $x+4$.
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   $6<x<36$.
7. Integer catalog for $24$, $x+3$, $2x+5$, integers 0–50.
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   { 6, 7, …, 21 }.
8. Perimeter $\le 100$ for $2x+1$, $x+9$, $x+4$.
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   $2<x\le 21.5$.
9. Degeneracy check for $12$, $x+3$, $2x+9$.
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   Degenerate at $x=0$ and $x=6$.
10. Method choice with $5x-1$, $3x+2$, $40$.
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    Compressed range is simpler.
11. Fractions with $40$: $\frac{x}{2}+6$, $\frac{x}{3}+10$.
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    $28.8<x<264$.
12. Tolerance concept.
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Use worst-case difference and minimum sum when forming the inequality.
13. Angle location domain for $7x-8$, $4x+11$, $2x+25$.
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    Domain $\frac{22}{9}<x<44$; $7x-8$ longest for $(6.6,44)$.
14. Step size $0.25$ with bounds for $40$, $x+6$, $3x+2$.
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    Smallest $8.25$, largest $21.75$.
15. Equivalence of methods.
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The compressed range pair is equivalent to the three inequalities under positivity.

Go Further

1. Manufacturing with tolerances and step sizes.
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   Safe continuous interval $5.48<x<30.0$; feasible steps 5.5 to 30.0 in 0.5 increments.
2. Optimize a perimeter under triangle and catalog constraints.
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   Max at $x=20$ with perimeter $156$ cm.
3. Show equivalence of methods (general linear case).
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   Derive $|a-b|<c$ and $c<a+b$ from the three inequalities and vice versa.
4. Mixed units, supplier steps, and a target angle.
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   Integers $\{37,38,\dots ,90\}$.
5. Force a middle-length side (ordering windows).
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   $(\frac{18}{7},3)\cup (13,\infty )$.

My Reflection

Instructions (Notebook): Open your notebook, write today’s date and the title Triangle Inequalities with Algebraic Sides — Reflection (Day 3). Complete the 3–2–1 prompts in full sentences.

3–2–1 Reflection

3 things I can now do confidently

1. __________________________________________________________
2. __________________________________________________________
3. __________________________________________________________

2 common mistakes I will avoid next time

1. __________________________________________________________
2. __________________________________________________________

1 question I still have (or want to explore next)

____________________________________________________________

Show Sample Responses

3 things I can now do confidently

1. Form the triangle-inequality system for algebraic sides and keep each side positive.
2. Use a compressed range such as $|a-b|<c<a+b$ and intersect with positivity.
3. Convert mixed units and apply step constraints after the continuous interval.

2 common mistakes I will avoid

1. Forgetting strict inequalities; I will keep endpoints open.
2. Ignoring unit conversions before forming inequalities.

1 question I still have — when all three sides depend on $x$, how do I quickly choose the method that minimizes algebra?

Evidence Snapshot (Notebook)

Choose one problem you solved today and justify your interval for $x$ in 3–5 steps: positivity, inequalities or compressed range, solving, intersection with constraints, final set.

Show a Model Snapshot

Example: With $2x+5$, $x+9$, $24$, use $\left|\right(2x+5)-(x+9\left)\right|<24<(2x+5)+(x+9)$ $\Rightarrow$ $|x-4|<24$ and $24<3x+14$ $\Rightarrow$ $\frac{10}{3}<x<28$.

Commitment Plan (Notebook)

• Action 1 (Tomorrow): _____________________________________________
• Action 2 (This week): ____________________________________________

Show Example Commitments

• Review 3 absolute-value inequalities and rewrite each as a compound inequality before solving.
• Create one mixed-units problem and solve it twice to compare incorrect vs. correct unit handling.