The Central Idea
Consider
\[ 5x+2=3x+10. \]The variable appears on both sides, so we cannot isolate \(x\) immediately.
Instead, subtract \(3x\) from both sides:
\[ 5x+2-3x=3x+10-3x. \]This gives
\[ 2x+2=10. \]Now the variable appears on only one side, and ordinary multistep solving continues:
\[ 2x=8, \] \[ x=4. \]The essential move is to collect the variable terms onto one side using the properties of equality.
Why Variables Can Be Moved to Either Side
Suppose
\[ 7x+1=4x+16. \]We could subtract \(4x\) from both sides:
\[ 3x+1=16, \]or subtract \(7x\) from both sides:
\[ 1=-3x+16. \]Both routes preserve equivalence.
The choice is strategic rather than mathematical. Usually, we prefer the route that produces simpler signs or smaller coefficients.
A Good Default Strategy
For equations with variables on both sides:
- Simplify both sides first.
- Move variable terms to one side.
- Move constant terms to the other side.
- Divide by the remaining coefficient.
- Interpret the result.
This workflow is flexible and works for most linear equations of this type.
Choose the Side That Keeps the Coefficient Positive
Consider
\[ 3x+7=8x-13. \]If we subtract \(3x\), we get
\[ 7=5x-13. \]Add \(13\):
\[ 20=5x. \]Then
\[ x=4. \]If we had subtracted \(8x\), we would instead obtain a negative coefficient. That route would still be correct, but the positive-coefficient route is slightly cleaner.
Example: Variable Terms with Different Signs
Solve
\[ 6x-5=-2x+19. \]Add \(2x\) to both sides:
\[ 8x-5=19. \]Add \(5\):
\[ 8x=24. \]Divide by \(8\):
\[ x=3. \]Example: Negative Coefficient Result
Solve
\[ 2x+9=5x+3. \]Subtract \(5x\):
\[ -3x+9=3. \]Subtract \(9\):
\[ -3x=-6. \]Divide by \(-3\):
\[ x=2. \]A negative coefficient is not a problem; it simply requires careful sign handling.
Simplify Before Collecting Variables
Consider
\[ 3x+4+2x=7x-6. \]Combine like terms on the left:
\[ 5x+4=7x-6. \]Subtract \(5x\):
\[ 4=2x-6. \]Add \(6\):
\[ 10=2x. \]Therefore,
\[ x=5. \]Simplifying first often makes the structure much easier to see.
Parentheses on One Side
Consider
\[ 4(x+2)=2x+14. \]Distribute:
\[ 4x+8=2x+14. \]Subtract \(2x\):
\[ 2x+8=14. \]Subtract \(8\):
\[ 2x=6. \]So
\[ x=3. \]Parentheses on Both Sides
Solve
\[ 3(x+4)=2(x+9). \]Expand:
\[ 3x+12=2x+18. \]Subtract \(2x\):
\[ x+12=18. \]Subtract \(12\):
\[ x=6. \]Do Not Distribute Unless It Helps
Consider
\[ 5(x-2)=5x-10. \]Expanding the left side gives
\[ 5x-10=5x-10. \]This reveals an identity immediately.
But in other equations, preserving grouping may be more efficient. The goal is not always to expand; the goal is to expose structure that helps classify or solve the equation.
Three Possible Outcomes
When variables appear on both sides, the variable terms may:
- leave a nonzero coefficient, producing one solution;
- cancel and leave a false statement, producing no solution;
- cancel and leave a true statement, producing infinitely many solutions.
These three outcomes are central to this class of equations.
One Solution
Consider
\[ 5x+1=2x+13. \]Subtract \(2x\):
\[ 3x+1=13. \]Subtract \(1\):
\[ 3x=12. \]Then
\[ x=4. \]The variable terms do not completely cancel, so one unique solution remains.
No Solution
Consider
\[ 4x+3=4x+9. \]Subtract \(4x\) from both sides:
\[ 3=9. \]This statement is false.
Therefore, no value of \(x\) can satisfy the original equation.
The solution set is
\[ \varnothing. \]Infinitely Many Solutions
Consider
\[ 6x-2=6x-2. \]Subtract \(6x\):
\[ -2=-2. \]This statement is true for every real value of \(x\).
Therefore, the equation is an identity and the solution set is
\[ \mathbb{R}. \]Identity Hidden by Parentheses
Solve
\[ 2(x+5)+3x=5x+10. \]Expand:
\[ 2x+10+3x=5x+10. \]Combine:
\[ 5x+10=5x+10. \]The equation is true for every real \(x\).
So
\[ \boxed{\mathbb{R}} \]is the solution set.
Contradiction Hidden by Parentheses
Solve
\[ 3(x+2)+x=4x+9. \]Expand:
\[ 3x+6+x=4x+9. \]Combine:
\[ 4x+6=4x+9. \]Subtract \(4x\):
\[ 6=9. \]This is impossible, so
\[ \boxed{\varnothing} \]is the solution set.
Fractions on Both Sides
Consider
\[ \frac{x}{2}+3=\frac{x}{4}+7. \]One approach is to clear denominators.
The least common denominator is \(4\).
Multiply every term by \(4\):
\[ 2x+12=x+28. \]Subtract \(x\):
\[ x+12=28. \]Subtract \(12\):
\[ x=16. \]Fractional Coefficients
Solve
\[ \frac{2}{3}x+5=\frac{1}{6}x+10. \]The least common denominator is \(6\).
Multiply the entire equation by \(6\):
\[ 4x+30=x+60. \]Subtract \(x\):
\[ 3x+30=60. \]Subtract \(30\):
\[ 3x=30. \]Therefore,
\[ x=10. \]Decimals on Both Sides
Consider
\[ 0.5x+2=0.2x+8. \]Subtract \(0.2x\):
\[ 0.3x+2=8. \]Subtract \(2\):
\[ 0.3x=6. \]Then
\[ x=20. \]If preferred, the decimals can be rewritten as fractions or cleared by multiplying the entire equation by a power of \(10\).
Variables Inside Several Groups
Solve
\[ 2(x+3)-4=3(x-1)+2. \]Expand both sides:
\[ 2x+6-4=3x-3+2. \]Simplify:
\[ 2x+2=3x-1. \]Subtract \(2x\):
\[ 2=x-1. \]Add \(1\):
\[ x=3. \]Collecting Before Expanding
Sometimes repeated structure can be combined before full expansion.
Consider
\[ 2(x+1)+3(x+1)=4x+9. \]Collect the repeated factor:
\[ 5(x+1)=4x+9. \]Expand:
\[ 5x+5=4x+9. \]Subtract \(4x\):
\[ x+5=9. \]Therefore,
\[ x=4. \]Recognizing common structure can reduce the amount of algebra required.
Symbolic Coefficients
The same structure can be described generally.
Consider
\[ ax+b=cx+d. \]Subtract \(cx\):
\[ (a-c)x+b=d. \]Subtract \(b\):
\[ (a-c)x=d-b. \]If
\[ a-c\ne0, \]then
\[ x=\frac{d-b}{a-c}. \]This formula summarizes the ordinary one-solution case.
What If the Coefficients Match?
In
\[ ax+b=cx+d, \]suppose
\[ a=c. \]Then the variable terms cancel.
The equation reduces to
\[ b=d. \]Two possibilities remain:
- If \(b=d\), the equation is true for every allowed \(x\).
- If \(b\ne d\), the equation has no solution.
This explains the three-outcome structure in general form.
General Classification
For
\[ ax+b=cx+d, \]we can classify the solution set:
| Condition | Outcome |
|---|---|
| \(a\ne c\) | One solution: \(\displaystyle x=\frac{d-b}{a-c}\) |
| \(a=c\) and \(b=d\) | Infinitely many solutions |
| \(a=c\) and \(b\ne d\) | No solution |
Why Classification Matters
The final result of an equation is not always a number.
For example, a solving process may end with
\[ 0=0. \]This is not a failure. It tells us that the original equation represented an identity.
Likewise,
\[ 0=5 \]reveals that the original equation was inconsistent.
Solving means identifying the solution set, whatever form that set takes.
Check by Substitution
Suppose we solve
\[ 8x-3=5x+12 \]and obtain
\[ x=5. \]Check the original equation.
Left side:
\[ 8(5)-3=37. \]Right side:
\[ 5(5)+12=37. \]Both sides agree, so the solution is verified.
Graphical Interpretation
An equation with variables on both sides can be interpreted as asking where two functions have equal outputs.
For example,
\[ 2x+3=5x-6 \]compares the functions
\[ y=2x+3 \]and
\[ y=5x-6. \]The solution is the \(x\)-coordinate where the two lines intersect.
Solving algebraically:
\[ 2x+3=5x-6 \] \[ 9=3x \] \[ x=3. \]At \(x=3\), both sides equal \(9\), so the lines intersect at
\[ (3,9). \]No Solution as Parallel Lines
Consider
\[ 2x+1=2x+5. \]The corresponding lines are
\[ y=2x+1 \]and
\[ y=2x+5. \]They have the same slope but different intercepts, so they are parallel and never intersect.
Algebraically, the equation reduces to
\[ 1=5. \]This geometric interpretation matches the empty solution set.
Infinitely Many Solutions as the Same Line
Consider
\[ 2(x+3)=2x+6. \]Both sides represent the same linear function.
Graphically, the two lines coincide.
Algebraically, the equation reduces to a true identity.
Every \(x\) in the domain gives equal outputs, so there are infinitely many solutions.
Worked Example: Strategic Side Choice
Solve
\[ 9x-4=3x+20. \]Subtract \(3x\) to keep the variable coefficient positive:
\[ 6x-4=20. \]Add \(4\):
\[ 6x=24. \]Divide by \(6\):
\[ x=4. \]Worked Example: Fractions and Parentheses
Solve
\[ \frac{1}{2}(x+4)=\frac{1}{4}(x+12). \]Multiply the entire equation by \(4\):
\[ 2(x+4)=x+12. \]Expand:
\[ 2x+8=x+12. \]Subtract \(x\):
\[ x+8=12. \]Therefore,
\[ x=4. \]Worked Example: No Solution
Solve
\[ 3(x+2)=3x+10. \]Expand:
\[ 3x+6=3x+10. \]Subtract \(3x\):
\[ 6=10. \]This is false.
Therefore,
\[ \boxed{\varnothing}. \]Worked Example: Infinitely Many Solutions
Solve
\[ 4(x-1)+2=4x-2. \]Expand:
\[ 4x-4+2=4x-2. \]Simplify:
\[ 4x-2=4x-2. \]This is true for every real \(x\).
Therefore,
\[ \boxed{\mathbb{R}}. \]Common Errors
Changing the Sign Without Applying an Operation
From
\[ 5x+2=3x+10, \]the \(3x\) does not simply “become negative.” We subtract \(3x\) from both sides.
Combining Terms Across the Equal Sign
The terms \(5x\) and \(3x\) cannot be directly combined while they remain on opposite sides. First use a property of equality to place them on the same side.
Forgetting to Simplify Both Sides
Parentheses and like terms can hide identities or contradictions.
Misreading a Final Constant Statement
A true statement means infinitely many solutions; a false statement means no solution.
Assuming Every Equation Must Have One Numerical Answer
Equations can have one solution, no solution, or every value in the intended domain as a solution.
Practice
1. One Solution
Solve:
\[ 7x+2=4x+20. \]Check Answer
\[ 3x+2=20, \] \[ 3x=18, \] \[ x=6. \]2. Negative Coefficients
Solve:
\[ -2x+9=4x-3. \]Check Answer
Add \(2x\):
\[ 9=6x-3. \]Add \(3\):
\[ 12=6x. \]Therefore,
\[ x=2. \]3. Parentheses
Solve:
\[ 5(x-1)=3x+7. \]Check Answer
\[ 5x-5=3x+7, \] \[ 2x-5=7, \] \[ 2x=12, \] \[ x=6. \]4. Fractions
Solve:
\[ \frac{x}{3}+2=\frac{x}{6}+5. \]Check Answer
Multiply by \(6\):
\[ 2x+12=x+30. \]Then
\[ x=18. \]5. No Solution
Solve:
\[ 2x+5=2x-1. \]Check Answer
Subtract \(2x\):
\[ 5=-1. \]This is false, so
\[ \varnothing \]is the solution set.
6. Infinitely Many Solutions
Solve:
\[ 3(x+4)=3x+12. \]Check Answer
Expanding gives
\[ 3x+12=3x+12. \]This is an identity, so the solution set is
\[ \mathbb{R}. \]Key Ideas
- When variables appear on both sides, collect all variable terms on one side using the properties of equality.
- Either side may hold the variable; choose the route that keeps the algebra simple.
- Simplify each side before collecting variable terms when possible.
- Parentheses, fractions, decimals, and repeated groups may need to be simplified first.
- If a nonzero variable coefficient remains, the equation has one solution.
- If the variable terms cancel and leave a false statement, the equation has no solution.
- If the variable terms cancel and leave a true statement, the equation has infinitely many solutions in the intended domain.
- The general equation \(ax+b=cx+d\) has one solution when \(a\ne c\).
- When \(a=c\), the constants determine whether the equation is an identity or a contradiction.
- Graphically, solving an equation with variables on both sides means finding where two expressions have equal outputs.
Where This Leads
Equations with variables on both sides deepen the idea that solving is really the controlled comparison of two expressions.
Next lessons can extend the same framework to equations with fractions, absolute values, powers, radicals, and inequalities, where domain restrictions and case structure become increasingly important.