Solving Linear Equations in One Variable

[EXEC: MICRO_CORE]

✖️ 1. Solving multi-step equations by systematically isolating the variable

🎯 Isolating the Variable Step-by-Step

Work backwards through the order of operations.
Undo addition and subtraction first.
Then undo multiplication and division.
Always perform the same operation on both sides.
Simplify after each step to avoid mistakes.

Example: Solve $3x + 7 = 22$ . Subtract 7 from both sides: $3x = 15$ . Divide both sides by 3: $x = 5$ .

💡 Think of it like unpacking a suitcase: remove outer layers first, then inner layers.

[EXEC: DEEP_COMPUTE]

1. Solving multi-step equations by systematically isolating the variable

Solving Multi-Step Equations by Systematically Isolating the Variable

A multi-step linear equation is an equation of the form $ax + b = c$ (or more complex variants) requiring multiple inverse operations to solve. The goal is to isolate the variable $x$ on one side of the equation.

Intuition: Each operation performed on one side must be performed on the other to maintain equality. We reverse the order of operations: undo addition/subtraction first, then multiplication/division.

Core Rules:

Simplify both sides by combining like terms and distributing if necessary.
Eliminate constants from the variable's side using inverse operations (add/subtract).
Eliminate coefficients by multiplying or dividing both sides by the coefficient of $x$ .
Check the solution by substituting back into the original equation.

Consequence: Every linear equation with one variable has exactly one solution unless it is a special case (identity or contradiction).

Example: Solve $3x + 7 = 22$ . Subtract 7: $3x = 15$ . Divide by 3: $x = 5$ .

EXECUTION_BLOCK

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LVL_2

EXEC: ALGORITHM

Solve the inequality: $x - 7 > 12$ . Write the final isolated inequality for $x$ .

DEEP_COMPUTE

ULTRA

[EXEC: MICRO_CORE]

✖️ 2. Handling equations with variables on both sides of the equals sign

⚖️ Variables on Both Sides

Move all variable terms to one side by adding or subtracting.
Move all constant terms to the opposite side.
Choose the side that keeps coefficients positive when possible.
Combine like terms after moving.
Then isolate the variable as usual.

Example: Solve $5x - 3 = 2x + 9$ . Subtract $2x$ from both sides: $3x - 3 = 9$ . Add 3: $3x = 12$ . Divide by 3: $x = 4$ .

💡 Gather all x's on one side like herding sheep into one pen.

[EXEC: DEEP_COMPUTE]

2. Handling equations with variables on both sides of the equals sign

Handling Equations with Variables on Both Sides

An equation with variables on both sides has the form $ax + b = cx + d$ , where variable terms appear on each side of the equals sign. The strategy is to collect all variable terms on one side and all constants on the other.

Intuition: Move terms strategically to simplify the equation into the standard form $mx = k$ , then solve for $x$ .

Core Rules:

Collect variable terms on one side by adding or subtracting the smaller variable coefficient from both sides.
Collect constant terms on the opposite side using inverse operations.
Simplify to isolate $x$ by dividing by the resulting coefficient.
Verify the solution in the original equation.

Consequence: This method reduces any such equation to a single-variable form, enabling straightforward solution.

Example: Solve $5x - 3 = 2x + 9$ . Subtract $2x$ : $3x - 3 = 9$ . Add 3: $3x = 12$ . Divide by 3: $x = 4$ .

EXECUTION_BLOCK

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LVL_2

EXEC: ALGORITHM

Solve the inequality: $-4x > 12$ .

DEEP_COMPUTE

ULTRA

[EXEC: MICRO_CORE]

✖️ 3. Clearing fractions and decimals from equations

🧹 Clearing Fractions and Decimals

For fractions: multiply every term by the LCM of all denominators.
For decimals: multiply every term by the appropriate power of 10.
This converts the equation to whole numbers only.
Apply the multiplication to every single term on both sides.
Then solve the simpler equation normally.

Example: Solve $\frac{x}{2} + 3 = 7$ . Multiply all terms by 2: $x + 6 = 14$ . Subtract 6: $x = 8$ .

💡 Clear the clutter first—whole numbers are easier to work with.

[EXEC: DEEP_COMPUTE]

3. Clearing fractions and decimals from equations

Clearing Fractions and Decimals by Multiplying by the LCM or Powers of 10

Equations containing fractions or decimals can be simplified by eliminating these forms through strategic multiplication. For fractions, multiply both sides by the least common multiple (LCM) of all denominators. For decimals, multiply by the appropriate power of 10.

Intuition: Clearing fractions and decimals transforms the equation into an equivalent integer-coefficient equation, which is easier to solve.

Core Rules:

For fractions: Identify the LCM of all denominators, then multiply every term by this LCM.
For decimals: Multiply every term by $10^n$ , where $n$ is the maximum number of decimal places.
Distribute carefully to avoid errors, then solve the resulting integer equation.
Simplify before solving to reduce arithmetic complexity.

Consequence: This technique preserves equality while simplifying computation.

Example: Solve $\frac{x}{3} + 2 = \frac{x}{2} - 1$ . Multiply by LCM(3,2) = 6: $2x + 12 = 3x - 6$ . Solve: $x = 18$ .

EXECUTION_BLOCK

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LVL_2

EXEC: ALGORITHM

Solve the compound inequality: $3 < 2x - 1 < 9$ .

DEEP_COMPUTE

ULTRA

[EXEC: MICRO_CORE]

✖️ 4. Identifying special cases: identities and contradictions

🔍 Special Cases: Identity vs Contradiction

Identity: variables cancel and you get a true statement like $5 = 5$ .
Identity means infinitely many solutions (every number works).
Contradiction: variables cancel and you get a false statement like $3 = 7$ .
Contradiction means no solution (no number works).
Always simplify completely to check for these cases.

Example: Solve $2x + 4 = 2x + 4$ . Subtract $2x$ : $4 = 4$ (identity, infinite solutions). Solve $x + 1 = x + 5$ . Subtract $x$ : $1 = 5$ (contradiction, no solution).

💡 If variables vanish, check what's left: truth or lie?

[EXEC: DEEP_COMPUTE]

4. Identifying special cases: identities and contradictions

Identifying Special Cases: Identities (Infinite Solutions) and Contradictions (No Solution)

Not all linear equations have exactly one solution. An identity results when simplification yields a true statement (e.g., $0 = 0$ ), indicating infinite solutions. A contradiction occurs when simplification yields a false statement (e.g., $0 = 5$ ), indicating no solution.

Intuition: If variables cancel completely, examine the remaining constant equation to determine the nature of the solution set.

Core Rules:

Identity: After simplification, both sides are identical (e.g., $3 = 3$ ). Every real number satisfies the equation.
Contradiction: After simplification, a false equality remains (e.g., $7 = -2$ ). No value of $x$ satisfies the equation.
Standard case: If a non-zero coefficient remains with $x$ , exactly one solution exists.

Consequence: Recognizing these cases prevents incorrect conclusions about solution existence.

Example: Solve $2x + 5 = 2x - 3$ . Subtract $2x$ : $5 = -3$ (contradiction). No solution exists.

EXECUTION_BLOCK

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EXEC: ALGORITHM

An elevator has a maximum payload capacity of 1000 kg. It already contains 250 kg of cargo. A person weighing $p$ kg wants to enter. Which inequality represents the allowable weight for the person?

DEEP_COMPUTE

ULTRA

[EXEC: MICRO_CORE]

✖️ 5. Applications: Calculating the exact break-even point in microeconomics

💼 Break-Even Point (Revenue = Cost)

Revenue is price per unit times quantity: $R = p \cdot q$ .
Cost is fixed cost plus variable cost per unit: $C = F + v \cdot q$ .
Set Revenue equal to Cost and solve for quantity $q$ .
The solution is the break-even quantity.
Below this quantity you lose money; above it you profit.

Example: Price is 20 dollars per unit, fixed cost is 500 dollars, variable cost is 12 dollars per unit. Set $20q = 500 + 12q$ . Subtract $12q$ : $8q = 500$ . Divide by 8: $q = 62.5$ units.

💡 Break-even is where the money coming in equals the money going out.

[EXEC: DEEP_COMPUTE]

5. Applications: Calculating the exact break-even point in microeconomics

Applications: Calculating the Exact Break-Even Point (Revenue = Cost)

In microeconomics, the break-even point occurs when total revenue equals total cost, resulting in zero profit. If revenue is $R(q) = pq$ (price times quantity) and cost is $C(q) = F + vq$ (fixed cost $F$ plus variable cost $v$ per unit), the break-even quantity $q^*$ solves $pq = F + vq$ .

Intuition: The break-even point identifies the minimum production level needed to cover all costs without loss.

Core Rules:

Set up the equation: Equate revenue and cost functions.
Isolate the variable: Collect $q$ terms on one side, constants on the other.
Solve for $q^*$ : Divide by the coefficient of $q$ .
Interpret economically: $q^*$ must be non-negative and realistic.

Consequence: This analysis informs pricing and production decisions.

Example: If $R = 15q$ and $C = 200 + 10q$ , solve $15q = 200 + 10q$ . Subtract $10q$ : $5q = 200$ . Divide: $q^* = 40$ units.

EXECUTION_BLOCK

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LVL_2

EXEC: ALGORITHM

A company has a revenue function $R = 20q$ and a cost function $C = 500 + 10q$ . Calculate the exact break-even quantity $q$ .

DEEP_COMPUTE

ULTRA

Solving linear equations with one variable

✖️ 1. Solving multi-step equations by systematically isolating the variable

🎯 Isolating the Variable Step-by-Step

1. Solving multi-step equations by systematically isolating the variable

Solving Multi-Step Equations by Systematically Isolating the Variable

✖️ 2. Handling equations with variables on both sides of the equals sign

⚖️ Variables on Both Sides

2. Handling equations with variables on both sides of the equals sign

Handling Equations with Variables on Both Sides

✖️ 3. Clearing fractions and decimals from equations

🧹 Clearing Fractions and Decimals

3. Clearing fractions and decimals from equations

Clearing Fractions and Decimals by Multiplying by the LCM or Powers of 10

✖️ 4. Identifying special cases: identities and contradictions

🔍 Special Cases: Identity vs Contradiction

4. Identifying special cases: identities and contradictions

Identifying Special Cases: Identities (Infinite Solutions) and Contradictions (No Solution)

✖️ 5. Applications: Calculating the exact break-even point in microeconomics

💼 Break-Even Point (Revenue = Cost)

5. Applications: Calculating the exact break-even point in microeconomics

Applications: Calculating the Exact Break-Even Point (Revenue = Cost)

AWAITING_CONFIRMATION