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MathChapter 2: Solving Linear Equations

Notes

Solving Equations Notes

Solve one-step through multi-step linear equations by preserving equality, simplifying structure, and checking the result.

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Solving an equation means finding every value that makes its equality statement true. The central idea is balance: each valid change must preserve equality. Reliable solvers simplify deliberately, choose inverse operations, and verify the final value in the original equation.

Learning objectives

  • Explain what a solution is and test a proposed value by substitution.
  • Use the addition, subtraction, multiplication, and division properties of equality.
  • Solve one-step, two-step, and multi-step linear equations.
  • Handle distribution, like terms, fractions, and decimals accurately.
  • Check a solution and identify unfinished or illegal algebraic steps.

What solving means

Solution of an equation
A value that makes the original equation true when substituted for its variable. For \(3x+2=17\), \(x=5\) is a solution because \(3(5)+2=17\).

The equality balance

Left side

The complete expression to the left of \(=\). Any chosen equality-preserving operation applies to this entire side.

Right side

The complete expression to the right of \(=\). Apply the same valid operation here to keep both values balanced.

Properties of equality

Properties that preserve equation solutions
PropertySymbolic ruleOriginal example
AdditionIf \(a=b\), then \(a+c=b+c\)\(x-7=12\Rightarrow x-7+7=12+7\)
SubtractionIf \(a=b\), then \(a-c=b-c\)\(y+9=20\Rightarrow y+9-9=20-9\)
MultiplicationIf \(a=b\), then \(ac=bc\)\(z/5=6\Rightarrow 5(z/5)=5(6)\)
DivisionIf \(a=b\) and \(c\ne0\), then \(a/c=b/c\)\(-4p=28\Rightarrow -4p/(-4)=28/(-4)\)

Inverse operations isolate the variable

Inverse operations undo each other: addition and subtraction are inverses, as are multiplication and division. Work in reverse order from the operations attached to the variable. In \(5x-8=27\), undo subtraction before multiplication: add \(8\), then divide by \(5\).

Worked example

Solve a two-step equation

Solve \(7x+11=-31\).

  1. Undo the constant

    Subtract \(11\) from both sides: \(7x=-42\).

  2. Undo the coefficient

    Divide both sides by \(7\): \(x=-6\).

  3. Check

    Substitute: \(7(-6)+11=-42+11=-31\), so the equality is true.

The solution is \(x=-6\).

A dependable multi-step process

Simplify → organize → isolate → verify

  1. Simplify each side

    Remove parentheses with distribution and combine like terms on each side separately.

  2. Clear awkward fractions if useful

    Multiply every term on both sides by a common denominator. This is optional, but it often reduces arithmetic errors.

  3. Undo addition or subtraction

    Move the constant attached to the variable term by applying the same inverse operation to both sides.

  4. Undo multiplication or division

    Divide or multiply both sides to make the variable's coefficient \(1\).

  5. Check the original equation

    Substitute the result into the unsimplified original. A true numerical statement confirms the solution.

Distribution and like terms

Worked example

Simplify before isolating

Solve \(4(2x-3)+5=45\).

  1. Distribute

    Multiply both terms inside the parentheses: \(8x-12+5=45\).

  2. Combine

    Combine constants on the left: \(8x-7=45\).

  3. Move the constant

    Add \(7\) to both sides: \(8x=52\).

  4. Isolate

    Divide by \(8\): \(x=\frac{52}{8}=\frac{13}{2}\).

  5. Check

    \(4(2\cdot\frac{13}{2}-3)+5=4(10)+5=45\).

The solution is \(x=\frac{13}{2}\).

Fractions and decimals

Fractions do not change the algebraic principles. You may isolate a fractional variable term directly or multiply every term by the least common denominator. For decimals, exact arithmetic is usually safest; multiplying by a power of ten can remove decimals if it is applied to every term on both sides.

Worked example

Clear denominators safely

Solve \(\frac{x-2}{3}+\frac{x}{4}=5\).

  1. Choose the LCD

    The least common denominator of \(3\) and \(4\) is \(12\).

  2. Multiply every term

    \(12\cdot\frac{x-2}{3}+12\cdot\frac{x}{4}=12\cdot5\), so \(4(x-2)+3x=60\).

  3. Simplify

    Distribute and combine: \(4x-8+3x=60\Rightarrow7x-8=60\).

  4. Isolate

    Add \(8\) and divide by \(7\): \(7x=68\Rightarrow x=\frac{68}{7}\).

  5. Check

    Substitution gives \(\frac{54/7}{3}+\frac{68/7}{4}=\frac{18}{7}+\frac{17}{7}=5\).

The solution is \(x=\frac{68}{7}\).
Worked example

Solve with decimals

Solve \(0.6x+1.7=8.9\).

  1. Remove the constant

    Subtract \(1.7\) from both sides: \(0.6x=7.2\).

  2. Isolate

    Divide by \(0.6\): \(x=12\).

  3. Check

    \(0.6(12)+1.7=7.2+1.7=8.9\).

The solution is \(x=12\).

Common mistakes and SAT strategy

Mini check

Check your understanding

Solve \(-2(3x-4)+6=26\).

  1. \(x=-2\)
  2. \(x=-1\)
  3. \(x=1\)
  4. \(x=2\)
Show answer and explanation

Answer: \(x=-2\)

Distribute: \(-6x+8+6=26\). Combine to get \(-6x+14=26\), subtract \(14\), and divide by \(-6\): \(x=-2\). Substitution returns \(26\).

Key takeaways

Key takeaways

What to remember

  • A solution makes the original equation true.
  • Preserve equality by applying the same valid operation to both complete sides.
  • Simplify grouping and like terms before isolating the variable.
  • Clear fractions or decimals only by multiplying every term on both sides.
  • Continue until the variable has coefficient \(1\), then check in the original equation.
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Put these notes into practice

Apply the ideas with SAT-style questions, then reinforce key details with flashcards.