Algebraic Problem Solving

Introduction

Many algebraic problems cannot be solved using just one operation. Instead, they require several connected steps, such as simplifying expressions, expanding brackets, collecting like terms, rearranging equations, factoring, and checking solutions.

These are called multi-step algebraic problems.

For example, consider:

3(2x − 4) + 5 = 29

We cannot immediately determine x. First, we must expand the brackets, simplify the equation, and isolate the variable.

3(2x − 4) + 5 = 29

6x − 12 + 5 = 29

6x − 7 = 29

6x = 36

x = 6

We can verify the solution by substituting x = 6 into the original equation:

3(2 × 6 − 4) + 5 = 29

3(8) + 5 = 29

24 + 5 = 29

29 = 29 ✓

This example demonstrates an important principle: complex algebraic problems become manageable when they are divided into smaller, logical steps.

In this topic, we will explore how to choose efficient solution strategies, simplify complicated expressions, organise calculations, solve equations involving several operations, and communicate mathematical reasoning clearly.

Understanding Multi-Step Algebraic Problems

A multi-step algebraic problem requires two or more mathematical operations to reach a solution.

These problems may involve:

• Brackets and the distributive property.

• Like terms.

• Fractions and rational expressions.

• Variables on both sides of an equation.

• Powers and roots.

• Quadratic expressions.

• Systems of equations.

• Word problems and mathematical modelling.

The challenge is not simply knowing how to perform each operation. It is knowing which operation to use, when to use it, and why it is appropriate.

Expressions and equations

An algebraic expression contains numbers, variables, and operations.

Example:

4x² − 3x + 7

An equation states that two expressions are equal.

Example:

4x² − 3x + 7 = 0

We can simplify an expression, but we solve an equation.

This distinction helps us select the correct mathematical strategy.

Choosing an Appropriate Solution Strategy

Before beginning a complicated problem, examine its structure.

Ask yourself:

1. Are there brackets that need expanding?

2. Can any like terms be combined?

3. Are there fractions that can be eliminated?

4. Does the variable appear on both sides?

5. Is the variable squared or raised to a higher power?

6. Would factoring or substitution simplify the problem?

7. Are there restrictions on the possible solutions?

Different problem structures suggest different strategies.

A useful general approach is:

Understand → Plan → Simplify → Solve → Verify → Communicate

Simplifying Complex Algebraic Expressions

Simplification is often the first step in solving a complicated algebraic problem.

An expression is simplified by applying algebraic rules to produce an equivalent expression that is easier to work with.

Collecting Like Terms

Like terms have identical variable parts, including the same exponents.

For example:

3x + 5x = 8x

4x² − x² = 3x²

However:

3x + 4x²

cannot be combined because x and x² are different variable parts.

Worked Example: Multiple terms

Simplify:

7x − 3 + 4x + 8 − 2x

Group like terms:

7x + 4x − 2x − 3 + 8

Combine:

9x + 5

Final answer: 9x + 5

Worked Example: Several variables

Simplify:

5a + 3b − 2a + 7b − 4

Group like terms:

5a − 2a + 3b + 7b − 4

Combine:

3a + 10b − 4

Notice that a and b terms cannot be combined with each other.

Expanding and Simplifying Brackets

The distributive property states:

a(b + c) = ab + ac

This means that every term inside a bracket must be multiplied by the factor outside it.

Worked Example: Single bracket

Simplify:

4(3x − 2) + 5x

Expand:

12x − 8 + 5x

Collect like terms:

17x − 8

Worked Example: Multiple brackets

Simplify:

3(2x + 5) − 2(x − 4) + 7

Expand carefully:

6x + 15 − 2x + 8 + 7

Collect variable terms:

6x − 2x = 4x

Collect constants:

15 + 8 + 7 = 30

Final answer: 4x + 30

The negative sign before the second bracket is especially important.

−2(x − 4) = −2x + 8

A common error is to write −2x − 8.

Worked Example: Products of binomials

Simplify:

(2x + 3)(x − 5)

Multiply each term in the first bracket by each term in the second:

2x(x) + 2x(−5) + 3(x) + 3(−5)

= 2x² − 10x + 3x − 15

Combine like terms:

2x² − 7x − 15

This method is sometimes called double distribution or the FOIL method when multiplying two binomials.

Solving Multi-Step Linear Equations

A linear equation contains a variable to the first power.

The goal is to isolate the variable while preserving equality.

Worked Example: Brackets and constants

Solve:

5(2x − 3) + 4 = 39

Expand:

10x − 15 + 4 = 39

Simplify:

10x − 11 = 39

Add 11:

10x = 50

Divide by 10:

x = 5

Verification:

5(2 × 5 − 3) + 4

= 5(7) + 4

= 39 ✓

Worked Example: Variables on both sides

Solve:

4(2x − 1) = 3x + 21

Expand:

8x − 4 = 3x + 21

Subtract 3x:

5x − 4 = 21

Add 4:

5x = 25

Divide by 5:

x = 5

Worked Example: Multiple brackets on both sides

Solve:

3(2x + 4) − 5 = 2(x + 9) + 1

Expand:

6x + 12 − 5 = 2x + 18 + 1

Simplify:

6x + 7 = 2x + 19

Subtract 2x:

4x + 7 = 19

Subtract 7:

4x = 12

x = 3

Verification:

Left side:

3(2 × 3 + 4) − 5 = 25

Right side:

2(3 + 9) + 1 = 25

Both sides are equal.

Solving Equations Containing Fractions

Fractions can make algebraic equations appear more complicated.

A useful strategy is to multiply every term by the lowest common denominator (LCD).

Worked Example: Two fractions

Solve:

x/3 + x/4 = 7

The LCD of 3 and 4 is 12.

Multiply every term by 12:

12(x/3) + 12(x/4) = 12(7)

Simplify:

4x + 3x = 84

7x = 84

x = 12

Verification:

12/3 + 12/4 = 4 + 3 = 7 ✓

Worked Example: Fractions and brackets

Solve:

(x − 2)/3 + (x + 1)/2 = 6

The LCD is 6.

Multiply every term by 6:

2(x − 2) + 3(x + 1) = 36

Expand:

2x − 4 + 3x + 3 = 36

Simplify:

5x − 1 = 36

Add 1:

5x = 37

x = 37/5 = 7.4

Verification:

(7.4 − 2)/3 + (7.4 + 1)/2

= 5.4/3 + 8.4/2

= 1.8 + 4.2

= 6 ✓

Important: Variables in denominators

Consider:

1/(x − 2) = 3

The denominator cannot equal zero.

Therefore:

x ≠ 2

Multiply both sides by x − 2:

1 = 3(x − 2)

1 = 3x − 6

7 = 3x

x = 7/3

The solution is valid because it does not make the denominator zero.

When multiplying by expressions containing variables, always consider restrictions and check the final solution.

Solving Problems Involving Powers and Roots

Some multi-step equations involve squared variables, square roots, or higher powers.

These problems require us to combine ordinary algebraic operations with inverse powers and roots.

Worked Example: Squared expression

Solve:

2(x − 3)² + 5 = 55

Subtract 5:

2(x − 3)² = 50

Divide by 2:

(x − 3)² = 25

Take square roots:

x − 3 = ±5

Solve both cases:

x − 3 = 5 → x = 8

x − 3 = −5 → x = −2

Final answer: x = 8 or x = −2

Verification:

For x = 8:

2(8 − 3)² + 5 = 2(25) + 5 = 55

For x = −2:

2(−2 − 3)² + 5 = 2(25) + 5 = 55

Both are correct.

Worked Example: Square root equation

Solve:

√(2x + 9) = x

Square both sides:

2x + 9 = x²

Rearrange:

x² − 2x − 9 = 0

Use the quadratic formula:

x = (2 ± √(4 + 36))/2

x = (2 ± √40)/2

x = 1 ± √10

Because the original equation requires x ≥ 0, the candidate x = 1 − √10 is invalid.

The remaining solution is:

x = 1 + √10 ≈ 4.162

Verification:

For x = 1 + √10:

2x + 9 = 11 + 2√10

Also:

(1 + √10)² = 11 + 2√10

Therefore √(2x + 9) = 1 + √10, as required.

This example demonstrates why solutions obtained after squaring must be checked.

Solving Quadratic Equations

A quadratic equation contains a squared variable and can be written in the form:

ax² + bx + c = 0

where a ≠ 0.

There are several possible strategies:

• Factoring.

• Completing the square.

• Using the quadratic formula.

• Taking square roots when the squared expression is isolated.

Worked Example: Factoring

Solve:

x² + 7x + 12 = 0

Find two numbers that multiply to 12 and add to 7.

The numbers are 3 and 4.

Factor:

(x + 3)(x + 4) = 0

Using the zero-product property:

x + 3 = 0 or x + 4 = 0

x = −3 or x = −4

Worked Example: Quadratic formula

Solve:

2x² + 3x − 4 = 0

The quadratic formula is:

x = (−b ± √(b² − 4ac))/(2a)

Here:

a = 2, b = 3, c = −4

Substitute:

x = (−3 ± √(3² − 4 × 2 × (−4)))/(2 × 2)

x = (−3 ± √41)/4

Therefore:

x ≈ 0.851 or x ≈ −2.351

The quadratic formula is particularly useful when a quadratic does not factor conveniently into rational factors.

Organising Long Algebraic Solutions

One of the most important skills in advanced algebra is presenting calculations in a clear, logical order.

Consider:

2(3x − 4) + (x + 5)/2 = 17

A disorganised solution may contain several operations on the same line, making errors difficult to identify.

A more effective method is to simplify systematically.

Step 1: Eliminate the fraction

Multiply every term by 2:

4(3x − 4) + (x + 5) = 34

Step 2: Expand brackets

12x − 16 + x + 5 = 34

Step 3: Collect like terms

13x − 11 = 34

Step 4: Isolate the variable

13x = 45

Step 5: Solve

x = 45/13

Step 6: Verify

Substituting x = 45/13 into the original equation gives 17 on the left-hand side.

The solution is correct.

Why clear organisation matters

Organised algebra helps us:

• Recognise mistakes before they become larger problems.

• Explain why each operation is valid.

• Keep track of negative signs and fractions.

• Check intermediate calculations.

• Communicate solutions to other people.

In examinations, clearly written working can also demonstrate mathematical understanding even if a final arithmetic error occurs.

Solving Systems of Equations

Some multi-step problems involve two or more unknown variables.

A system of equations contains multiple equations that must be satisfied simultaneously.

Worked Example: Substitution method

Solve:

2x + y = 11

x − y = 1

First rearrange the second equation:

y = x − 1

Substitute into the first equation:

2x + (x − 1) = 11

3x − 1 = 11

3x = 12

x = 4

Now calculate y:

y = 4 − 1

y = 3

Final answer: x = 4, y = 3

Verification:

2(4) + 3 = 11

4 − 3 = 1

Both equations are satisfied.

Worked Example: Elimination method

Solve:

3x + 2y = 16

2x − 2y = 4

Add the equations:

5x = 20

x = 4

Substitute into the second equation:

2(4) − 2y = 4

8 − 2y = 4

−2y = −4

y = 2

Final answer: x = 4, y = 2

Elimination is especially efficient when the coefficients of one variable are equal or opposites.

Translating Word Problems into Algebra

Real-world problems are often presented using words rather than equations.

The first challenge is translating the information into mathematical relationships.

A useful approach is:

1. Identify the unknown quantity.

2. Assign a variable.

3. Express other quantities using that variable.

4. Construct an equation.

5. Solve systematically.

6. Interpret the result in context.

Worked Example: Ticket sales

A school sells adult tickets for $12 and student tickets for $7.

A total of 80 tickets are sold, producing $760 in revenue.

How many of each type were sold?

Let:

a = number of adult tickets

s = number of student tickets

Total tickets:

a + s = 80

Total revenue:

12a + 7s = 760

Rearrange the first equation:

s = 80 − a

Substitute:

12a + 7(80 − a) = 760

Expand:

12a + 560 − 7a = 760

Simplify:

5a + 560 = 760

5a = 200

a = 40

Then:

s = 80 − 40

s = 40

Answer: 40 adult tickets and 40 student tickets.

Verification:

40 + 40 = 80 tickets

12(40) + 7(40) = 480 + 280 = $760

Both conditions are satisfied.

Real-World Applications

Engineering: Calculating Material Dimensions

A rectangular metal plate has an area of 240 cm². Its length is 4 cm greater than its width.

Find the dimensions.

Let the width be x cm.

Length = x + 4

Area = length × width

240 = x(x + 4)

Expand:

240 = x² + 4x

Rearrange:

x² + 4x − 240 = 0

Factor:

(x + 20)(x − 12) = 0

x = −20 or x = 12

A negative width is not physically meaningful.

Therefore:

Width = 12 cm

Length = 16 cm

Verification:

12 × 16 = 192 cm²

This does not match the given area, so the proposed factorisation must be incorrect.

The correct factorisation is not (x + 20)(x − 12), because that expands to x² + 8x − 240.

Instead, solve:

x² + 4x − 240 = 0

Using the quadratic formula:

x = (−4 ± √(16 + 960))/2

x = (−4 ± √976)/2

x = −2 ± 2√61

The positive solution is:

x = −2 + 2√61 ≈ 13.62 cm

Length = x + 4 ≈ 17.62 cm

Final dimensions: approximately 13.62 cm by 17.62 cm.

The exact dimensions are:

Width = 2√61 − 2 cm

Length = 2√61 + 2 cm

Verification using exact values:

(2√61 − 2)(2√61 + 2)

= (2√61)² − 2²

= 244 − 4

= 240 cm² ✓

This example illustrates why verification is important: it can reveal an error in an earlier algebraic step.

Physics: Force and Motion

A 10 kg object experiences a forward force of 80 N and a resistive force of 25 N.

Calculate its acceleration.

Resultant force:

Fₙₑₜ = 80 − 25

Fₙₑₜ = 55 N

Newton's Second Law:

Fₙₑₜ = ma

Rearrange:

a = Fₙₑₜ/m

a = 55/10

a = 5.5 m/s²

The object accelerates forward at 5.5 m/s².

Finance: Comparing Two Pricing Plans

A service offers two subscription plans.

Plan A charges $15 per month plus a one-time $40 setup fee.

Plan B charges $23 per month with no setup fee.

After how many months will the total costs be equal?

Let m = number of months.

Plan A:

C = 15m + 40

Plan B:

C = 23m

Set the costs equal:

15m + 40 = 23m

Subtract 15m:

40 = 8m

m = 5

Answer: The plans cost the same after 5 months.

Verification:

Plan A: 15(5) + 40 = $115

Plan B: 23(5) = $115

For fewer than five months, Plan B is cheaper. For more than five months, Plan A is cheaper, assuming the stated charges remain unchanged.

Recognising and Correcting Algebraic Errors

Even when we understand algebraic techniques, mistakes can occur during long calculations.

The ability to recognise and correct errors is an important mathematical skill.

Error Example: Incorrect distribution

A student writes:

3(2x − 5) = 6x − 5

This is incorrect because the 3 must multiply both terms inside the brackets.

Correct:

3(2x − 5) = 6x − 15

Error Example: Incorrect handling of negative signs

A student writes:

−2(x − 4) = −2x − 8

Correct:

−2(x − 4) = −2x + 8

The product of two negative numbers is positive.

Error Example: Dividing by a variable

Consider:

x² = 5x

A student divides both sides by x:

x = 5

However, dividing by x assumes x ≠ 0 and loses a possible solution.

Instead, rearrange:

x² − 5x = 0

Factor:

x(x − 5) = 0

Therefore:

x = 0 or x = 5

Both values satisfy the original equation.

Error Example: Squaring a binomial

Incorrect:

(x + 4)² = x² + 16

Correct:

(x + 4)² = x² + 8x + 16

The middle term must be included.

Verifying Solutions

Verification means checking whether the proposed answer satisfies the original equation.

There are several useful methods.

Substitution: Replace the variable with the proposed solution and evaluate both sides.

Reverse operations: Work backwards from the solution to check the original relationship.

Graphical verification: When appropriate, graph both sides of an equation and identify their intersections.

Estimation: Decide whether the result is reasonable before accepting it.

Checking restrictions: Confirm that solutions do not cause division by zero, invalid real square roots, or other domain violations.

Worked Example: Verification by substitution

Solve:

4(x − 2) + 3x = 27

Expand:

4x − 8 + 3x = 27

7x − 8 = 27

7x = 35

x = 5

Check:

4(5 − 2) + 3(5)

= 12 + 15

= 27 ✓

The solution is valid.

Communicating Mathematical Reasoning

A complete algebraic solution should show not only the final answer but also the reasoning used to obtain it.

A useful format is:

• Define: Identify the unknown quantities.

• Formulate: Write the equation or equations.

• Simplify: Expand, collect like terms, or eliminate fractions.

• Solve: Isolate the variable using valid operations.

• Verify: Check the answer against the original conditions.

• Interpret: Explain the result in the context of the problem.

Worked Example: A Complete Mathematical Explanation

A rectangle has a perimeter of 54 cm. Its length is 3 cm more than twice its width. Find its dimensions.

Let w = width in centimetres.

Then:

Length = 2w + 3

The perimeter formula is:

P = 2l + 2w

Substitute the known information:

54 = 2(2w + 3) + 2w

Expand:

54 = 4w + 6 + 2w

Combine like terms:

54 = 6w + 6

Subtract 6:

48 = 6w

Divide by 6:

w = 8

Calculate the length:

l = 2(8) + 3

l = 19

Verification:

P = 2(19) + 2(8)

P = 38 + 16

P = 54 cm ✓

Final answer: The rectangle is 19 cm long and 8 cm wide.

The answer includes the equation, algebraic steps, verification, units, and interpretation.

Common Mistakes

Expanding brackets incorrectly

Remember to multiply every term inside a bracket by the outside factor.

Combining unlike terms

Terms such as 3x and 5x² cannot be combined through addition because their variable parts are different.

Ignoring the order of operations

Brackets, exponents, multiplication, division, addition, and subtraction must be handled correctly.

Performing an operation on only one side of an equation

An equation must remain balanced. Any operation used to isolate a variable must preserve equality.

Cancelling terms incorrectly

For example:

(x + 3)/x ≠ 3, in general.

Cancellation applies to common factors, not arbitrary terms in a sum.

Dividing by an expression that could equal zero

Dividing by x can eliminate the solution x = 0. Check restrictions before dividing.

Forgetting negative solutions

When solving equations involving even powers, consider both positive and negative possibilities where appropriate.

Failing to verify answers

An algebraic manipulation can introduce invalid solutions or hide earlier errors. Substitute final answers into the original equation.

Giving answers without context

In word problems, include the meaning of the variable, appropriate units, and a clear concluding statement.

Key Terms

• Algebraic expression: A mathematical combination of numbers, variables, and operations.

• Equation: A mathematical statement that two expressions are equal.

• Variable: A symbol representing an unknown or changing quantity.

• Coefficient: A number multiplying a variable.

• Constant: A term without a variable.

• Like terms: Terms with identical variable parts and exponents.

• Distributive property: A rule allowing multiplication across terms inside brackets.

• Simplification: Rewriting an expression in an equivalent, more manageable form.

• Equivalent expressions: Expressions that have the same value for all allowed values of their variables.

• Inverse operations: Operations that reverse one another.

• Linear equation: An equation involving variables to the first power.

• Quadratic equation: An equation that can be written as ax² + bx + c = 0, where a ≠ 0.

• Factorisation: Expressing an expression as a product of factors.

• Substitution: Replacing a variable with a value or equivalent expression.

• Elimination: Removing a variable by combining equations.

• Lowest common denominator: The smallest positive denominator that is a multiple of all denominators being considered.

• Extraneous solution: A proposed solution that does not satisfy the original equation.

• Verification: Checking whether a result satisfies the original conditions.

• Mathematical modelling: Representing real-world situations using mathematical relationships.

• Solution strategy: A planned sequence of mathematical operations used to solve a problem.

Key Takeaways

• Multi-step algebraic problems require a planned sequence of operations.

• Understanding the structure of a problem helps identify the most efficient strategy.

• Simplification often involves expanding brackets and collecting like terms.

• Fractions can frequently be eliminated using a common denominator.

• Variables appearing on both sides of an equation should be collected systematically.

• Quadratic equations may require factoring, completing the square, or the quadratic formula.

• Systems of equations can be solved using substitution or elimination.

• Word problems must be translated into mathematical relationships before solving.

• Clear organisation makes errors easier to identify and correct.

• Every final solution should be verified using the original equation or conditions.

• Mathematical communication should include reasoning, calculations, appropriate units, and interpretation.

Check Your Understanding

Recall and Understanding

1. Explain the difference between simplifying an expression and solving an equation.

2. Why is it important to expand brackets carefully before collecting like terms?

3. Describe two methods for solving simultaneous linear equations.

4. Why should a solution be substituted into the original equation?

Application

5. Simplify:

4(3x − 2) − 2(x + 5) + 7

6. Solve:

5(2x − 3) + 4 = 3x + 24

7. Solve:

(x + 2)/3 + (x − 1)/4 = 5

8. Solve:

3(x − 2)² − 12 = 63

9. Solve the simultaneous equations:

2x + 3y = 19

x + y = 7

10. Solve the quadratic equation:

x² − 5x − 24 = 0

Analysis and Challenge

11. A rectangle has a perimeter of 70 cm. Its length is 5 cm greater than twice its width. Calculate its dimensions.

12. A company charges a fixed delivery fee of $25 plus $8 per kilogram of goods. Another company charges $13 per kilogram with no fixed fee. At what mass will the two delivery costs be equal?

13. Solve:

√(x + 10) = x − 2

Identify and reject any extraneous solutions.

14. A student solves the equation:

2(x + 4) − 3(x − 2) = 12

and obtains x = 2.

Determine whether the answer is correct. If it is incorrect, identify the mistake and find the correct solution.

15. A school purchases 45 notebooks and pens altogether. Each notebook costs $4 and each pen costs $2. The total cost is $130. Determine how many notebooks and pens were purchased.

Answers to Check Your Understanding

Final Summary

Multi-step algebraic problems develop more than calculation skills. They require logical thinking, strategic planning, accuracy, and the ability to explain mathematical ideas.

The most effective approach is to understand the problem, select a suitable strategy, simplify systematically, solve carefully, and verify the final answer.

With practice, complicated equations become a sequence of manageable operations rather than a single difficult task.

Successful algebra is not just about finding an answer. It is about understanding why the answer is correct and being able to explain the reasoning clearly.