The Central Idea

An equation is a statement that two expressions have the same value.

For example,

\[ x+5=12. \]

To solve the equation, we want to determine which value of \(x\) makes the statement true.

We can subtract \(5\) from both sides:

\[ x+5-5=12-5, \]

giving

\[ x=7. \]

The equation changed form, but the solution set did not.

This is the core of equation solving:

Apply valid transformations until the unknown is isolated, while preserving the solution set.

What Does It Mean to Solve an Equation?

To solve an equation means to find every value of the variable that makes the equation true.

For

\[ x+3=10, \]

the solution is

\[ x=7. \]

The solution set is

\[ \{7\}. \]

For an equation such as

\[ x^2=9, \]

the solution set is

\[ \{-3,3\}. \]

Equation solving is therefore a search for a set of valid values, not merely a single number.

One-Step Equations

A one-step equation can be solved with one inverse operation.

Typical forms include:

\[ x+a=b, \] \[ x-a=b, \] \[ ax=b, \]

and

\[ \frac{x}{a}=b, \qquad a\ne0. \]

The operation attached to \(x\) is undone using its inverse.

Addition Equations

Consider

\[ x+8=15. \]

Subtract \(8\) from both sides:

\[ x+8-8=15-8. \]

Then

\[ x=7. \]

The subtraction undoes the addition while the subtraction property of equality preserves the equation.

Subtraction Equations

Consider

\[ x-6=11. \]

Add \(6\) to both sides:

\[ x-6+6=11+6. \]

Therefore,

\[ x=17. \]

Multiplication Equations

Consider

\[ 5x=35. \]

Divide both sides by \(5\):

\[ \frac{5x}{5}=\frac{35}{5}. \]

So

\[ x=7. \]

Division by \(5\) is valid because \(5\ne0\).

Division Equations

Consider

\[ \frac{x}{4}=6. \]

Multiply both sides by \(4\):

\[ 4\left(\frac{x}{4}\right)=4(6). \]

Thus,

\[ x=24. \]

Inverse Operations

Operation on the Variable Inverse Operation
Addition Subtraction
Subtraction Addition
Multiplication by a nonzero quantity Division by that quantity
Division by a nonzero quantity Multiplication by that quantity
Odd power over the reals Corresponding odd root

Inverse operations are useful because they undo one another.

Two-Step Equations

A two-step equation requires two inverse operations.

Consider

\[ 3x+5=20. \]

The variable is first multiplied by \(3\), then \(5\) is added.

Undo these operations in reverse order.

Subtract \(5\):

\[ 3x=15. \]

Then divide by \(3\):

\[ x=5. \]

Why Reverse Order Works

Think of the expression

\[ 3x+5 \]

as a sequence of operations applied to \(x\):

  1. multiply by \(3\);
  2. add \(5\).

To recover \(x\), undo the outermost operation first:

  1. subtract \(5\);
  2. divide by \(3\).

This same principle scales to more complicated equations.

Multistep Equations

A multistep equation requires several transformations before the variable can be isolated.

Consider

\[ 4x+7-2x=19. \]

First combine like terms:

\[ 2x+7=19. \]

Subtract \(7\):

\[ 2x=12. \]

Divide by \(2\):

\[ x=6. \]

Different algebraic tools may be needed before inverse operations can finish the solution.

Simplify Before Solving

When possible, simplify each side of the equation before moving terms across the equality.

Consider

\[ 3x+5+2x-1=24. \]

Combine like terms:

\[ 5x+4=24. \]

Subtract \(4\):

\[ 5x=20. \]

Divide by \(5\):

\[ x=4. \]

Simplifying first reduces clutter and lowers the chance of sign errors.

Equations with Parentheses

Consider

\[ 3(x+2)=21. \]

There are at least two good approaches.

Method 1: Divide first.

\[ x+2=7, \]

so

\[ x=5. \]

Method 2: Distribute first.

\[ 3x+6=21, \]

then

\[ 3x=15, \]

so

\[ x=5. \]

Both approaches are valid. The better choice is usually the one that keeps the work simpler.

Distribution May Be Necessary

Consider

\[ 2(x+3)+x=15. \]

Distribute:

\[ 2x+6+x=15. \]

Combine like terms:

\[ 3x+6=15. \]

Subtract \(6\):

\[ 3x=9. \]

Divide by \(3\):

\[ x=3. \]

Variables on Both Sides

Consider

\[ 5x+2=3x+10. \]

We want all variable terms on one side and constants on the other.

Subtract \(3x\) from both sides:

\[ 2x+2=10. \]

Subtract \(2\):

\[ 2x=8. \]

Divide by \(2\):

\[ x=4. \]

Which Side Should Keep the Variable?

Either side is mathematically acceptable.

Consider

\[ 2x+7=5x-8. \]

We could subtract \(2x\):

\[ 7=3x-8. \]

Or subtract \(5x\):

\[ -3x+7=-8. \]

Both routes work.

In practice, choosing the subtraction that leaves a positive coefficient often makes the arithmetic easier.

Equations with Negative Coefficients

Consider

\[ -4x+3=19. \]

Subtract \(3\):

\[ -4x=16. \]

Divide by \(-4\):

\[ x=-4. \]

The sign of the coefficient must be preserved throughout the calculation.

Equations with Fractions

Consider

\[ \frac{x}{3}+\frac{1}{2}=4. \]

One approach is to subtract \(\frac12\):

\[ \frac{x}{3}=\frac{7}{2}. \]

Then multiply by \(3\):

\[ x=\frac{21}{2}. \]

Another approach is to clear the denominators at the beginning.

Clearing Numerical Denominators

For

\[ \frac{x}{3}+\frac{1}{2}=4, \]

the least common denominator is \(6\).

Multiply every term by \(6\):

\[ 6\left(\frac{x}{3}\right) + 6\left(\frac{1}{2}\right) = 6(4). \]

This gives

\[ 2x+3=24. \]

Then

\[ 2x=21 \]

and

\[ x=\frac{21}{2}. \]

Clearing denominators can make fraction equations easier to read.

Fractional Coefficients

Consider

\[ \frac{2}{5}x=8. \]

Multiply both sides by the reciprocal

\[ \frac{5}{2}. \]

Then

\[ x=8\left(\frac{5}{2}\right)=20. \]

Multiplying by the reciprocal is equivalent to dividing by the fractional coefficient.

Decimal Coefficients

Consider

\[ 0.4x+2=6. \]

Subtract \(2\):

\[ 0.4x=4. \]

Then

\[ x=10. \]

Alternatively, rewrite \(0.4\) as

\[ \frac{2}{5} \]

and solve using exact fractions.

Either representation is valid.

Equations with Several Layers

Consider

\[ 2[3(x-1)+4]=20. \]

One efficient approach is to undo the outside multiplication first.

Divide by \(2\):

\[ 3(x-1)+4=10. \]

Subtract \(4\):

\[ 3(x-1)=6. \]

Divide by \(3\):

\[ x-1=2. \]

Add \(1\):

\[ x=3. \]

We did not need to expand the nested grouping.

Choose a Useful Route

There is rarely only one valid path through an equation.

For example,

\[ 5(x-2)=3x+6. \]

Distribution is useful here because the variable appears outside the grouping on the other side.

Expand:

\[ 5x-10=3x+6. \]

Subtract \(3x\):

\[ 2x-10=6. \]

Add \(10\):

\[ 2x=16. \]

Divide by \(2\):

\[ x=8. \]

Good equation solving is strategic, not mechanical.

Checking a Solution

A solution can be checked by substituting it into the original equation.

Suppose we solved

\[ 4x-3=17 \]

and obtained

\[ x=5. \]

Substitute:

\[ 4(5)-3=20-3=17. \]

The left side equals the right side, so the solution is verified.

Why Check the Original Equation?

For ordinary reversible linear steps, checking mainly catches arithmetic mistakes.

But for equations involving radicals, rational expressions, logarithms, powers, or other restricted operations, checking the original equation can also identify extraneous solutions.

The original equation is the authoritative statement of the problem.

Identities: Infinitely Many Solutions

Sometimes solving causes the variable terms and constants to cancel completely.

Consider

\[ 2(x+3)=2x+6. \]

Distribute:

\[ 2x+6=2x+6. \]

Subtract \(2x\):

\[ 6=6. \]

This statement is always true.

Therefore, every real number satisfies the original equation.

The solution set is

\[ \mathbb{R}. \]

Contradictions: No Solution

Now consider

\[ 3(x+2)=3x+10. \]

Distribute:

\[ 3x+6=3x+10. \]

Subtract \(3x\):

\[ 6=10. \]

This is false.

Therefore, no value of \(x\) satisfies the original equation.

The solution set is

\[ \varnothing. \]

Three Outcomes for a Linear Equation

A linear equation in one variable can produce three basic outcomes:

Outcome Typical Final Form Solution Set
One solution \(x=a\) \(\{a\}\)
No solution False statement such as \(3=7\) \(\varnothing\)
Infinitely many solutions True identity such as \(5=5\) Entire intended domain

Do Not “Move” Terms Without Understanding the Operation

A shortcut such as “move the \(5\) across the equal sign and change its sign” can hide the actual mathematics.

From

\[ x+5=12, \]

the legitimate step is

\[ x+5-5=12-5. \]

Then

\[ x=7. \]

The term did not physically cross the equal sign. We applied the subtraction property of equality.

Understanding the underlying operation makes shortcuts safer.

Keep Both Sides Balanced

An equation is a relationship between two expressions.

If we apply an operation to only one side, we usually change the solution set.

For example, from

\[ x+2=7, \]

subtracting \(2\) only from the left would produce

\[ x=7, \]

which is not equivalent to the original equation.

Whatever operation is used to preserve equality must be applied to both sides.

Combining Steps

With experience, some algebraic steps may be written more compactly.

For example,

\[ 7x+5=26 \]

may quickly become

\[ 7x=21 \]

and then

\[ x=3. \]

There is no need to write every arithmetic simplification in maximum detail once the logic is understood.

However, skipped steps should never hide an invalid transformation.

Worked Example: Variables on Both Sides

Solve

\[ 7x-4=3x+20. \]

Subtract \(3x\):

\[ 4x-4=20. \]

Add \(4\):

\[ 4x=24. \]

Divide by \(4\):

\[ x=6. \]

Check:

\[ 7(6)-4=38, \]

and

\[ 3(6)+20=38. \]

The solution is correct.

Worked Example: Parentheses on Both Sides

Solve

\[ 2(x+5)=3(x-1). \]

Expand:

\[ 2x+10=3x-3. \]

Subtract \(2x\):

\[ 10=x-3. \]

Add \(3\):

\[ x=13. \]

Check:

\[ 2(18)=36, \]

and

\[ 3(12)=36. \]

Worked Example: Fraction Equation

Solve

\[ \frac{x}{4}+\frac{x}{2}=9. \]

The least common denominator is \(4\).

Multiply every term by \(4\):

\[ x+2x=36. \]

Combine:

\[ 3x=36. \]

Therefore,

\[ x=12. \]

Worked Example: Identity

Solve

\[ 4(x-2)+3=4x-5. \]

Expand:

\[ 4x-8+3=4x-5. \]

Combine:

\[ 4x-5=4x-5. \]

Subtract \(4x\):

\[ -5=-5. \]

The equation is true for every real \(x\).

Thus,

\[ \boxed{\mathbb{R}} \]

is the solution set.

Worked Example: Contradiction

Solve

\[ 5(x+1)=5x+8. \]

Expand:

\[ 5x+5=5x+8. \]

Subtract \(5x\):

\[ 5=8. \]

This is impossible.

Therefore,

\[ \boxed{\varnothing} \]

is the solution set.

A Reliable Solving Strategy

  1. Simplify each side. Distribute, remove grouping when helpful, and combine like terms.
  2. Collect variable terms. Move all terms containing the target variable to one side using properties of equality.
  3. Collect constants. Move constant terms to the other side.
  4. Isolate the variable. Divide by the remaining nonzero coefficient.
  5. Interpret the result. Decide whether the equation has one solution, no solution, or infinitely many solutions.
  6. Verify when useful. Substitute into the original equation.

Common Errors

Changing a Sign Without Performing an Operation

Signs change because an inverse operation is applied, not because a term crosses the equal sign.

Distributing to Only One Term

Incorrect:

\[ 3(x+4)=3x+4. \]

Correct:

\[ 3(x+4)=3x+12. \]

Combining Unlike Terms

The expression

\[ 3x+5 \]

cannot be simplified to

\[ 8x. \]

Dividing Only One Term

If

\[ 6x+12=24 \]

is divided by \(6\), every term must be divided by \(6\):

\[ x+2=4. \]

Stopping at a True or False Statement Without Interpreting It

A final identity such as

\[ 4=4 \]

means infinitely many solutions within the intended domain.

A contradiction such as

\[ 4=7 \]

means no solution.

Practice

1. One-Step Addition

Solve:

\[ x+9=14. \]
Check Answer \[ x=5. \]

2. One-Step Multiplication

Solve:

\[ 7x=49. \]
Check Answer \[ x=7. \]

3. Two-Step Equation

Solve:

\[ 4x+3=19. \]
Check Answer \[ 4x=16, \]

so

\[ x=4. \]

4. Combine Like Terms First

Solve:

\[ 3x+5x-6=18. \]
Check Answer \[ 8x-6=18, \] \[ 8x=24, \] \[ x=3. \]

5. Variables on Both Sides

Solve:

\[ 6x+1=4x+13. \]
Check Answer \[ 2x+1=13, \] \[ 2x=12, \] \[ x=6. \]

6. Parentheses

Solve:

\[ 3(x+2)=18. \]
Check Answer \[ x+2=6, \]

so

\[ x=4. \]

7. Identity

Solve:

\[ 2(x+4)=2x+8. \]
Check Answer

Both sides are identical for every real \(x\), so the solution set is

\[ \mathbb{R}. \]

8. Contradiction

Solve:

\[ 3(x+1)=3x+5. \]
Check Answer

Expanding gives

\[ 3x+3=3x+5, \]

so

\[ 3=5. \]

This is false, so the solution set is

\[ \varnothing. \]

Key Ideas

Where This Leads

One-step and multistep equations establish the core workflow for solving algebraic relationships by preserving equivalence.

From here, the same logic extends naturally to equations with more complex structures, inequalities, absolute values, radicals, rational expressions, systems, and other solution sets.