Algebra Guide: Equations, Factoring, and Systems of Equations

Algebra is the foundation of nearly every math course you will take from middle school through college. Whether you are solving for x in a simple equation or working through a system of equations on a timed exam, having a solid grasp of algebra makes everything else easier. This guide covers the major topics you need to know, with example problems and clear explanations for each one.

If you ever get stuck on a problem, you can screenshot it and get a full step-by-step breakdown using Math.Photos. But the goal here is to help you build real understanding so you can work through problems on your own.

Solving Linear Equations

A linear equation is any equation where the variable has an exponent of 1. These are the building blocks of algebra, and you will see them constantly.

The Basic Approach

The goal is always the same: isolate the variable on one side of the equation. You do this by performing the same operation on both sides.

Example: Solve 3x + 7 = 22

1. Subtract 7 from both sides: 3x = 15
2. Divide both sides by 3: x = 5

Check: 3(5) + 7 = 15 + 7 = 22. Correct.

Equations With Variables on Both Sides

When the variable appears on both sides, collect all variable terms on one side and all constants on the other.

Example: Solve 5x - 3 = 2x + 9

1. Subtract 2x from both sides: 3x - 3 = 9
2. Add 3 to both sides: 3x = 12
3. Divide both sides by 3: x = 4

Equations With Fractions

Fractions make equations look harder than they are. The trick is to multiply every term by the least common denominator (LCD) to clear the fractions first.

Example: Solve (x/2) + (x/3) = 10

1. The LCD of 2 and 3 is 6
2. Multiply every term by 6: 3x + 2x = 60
3. Combine like terms: 5x = 60
4. Divide by 5: x = 12

A common mistake is forgetting to multiply every term by the LCD, including the constant on the right side. Always apply it to the entire equation.

Try it yourself: Screenshot any algebra problem and get step-by-step solutions with Math.Photos — free browser extension.

Factoring

Factoring is the process of breaking an expression into a product of simpler expressions. It shows up everywhere in algebra, from simplifying fractions to solving quadratic equations. For a deeper dive, check out our guide on how to factor polynomials step by step.

Greatest Common Factor (GCF)

Always check for a GCF first. Pull out the largest factor that every term shares.

Example: Factor 12x³ + 18x²

1. The GCF of 12 and 18 is 6. Both terms have at least x².
2. Factor out 6x²: 6x²(2x + 3)

Factoring Trinomials (x² + bx + c)

For trinomials where the leading coefficient is 1, find two numbers that multiply to c and add to b.

Example: Factor x² + 7x + 12

1. Find two numbers that multiply to 12 and add to 7
2. Those numbers are 3 and 4
3. Factor: (x + 3)(x + 4)

Verification: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12. Checks out.

Factoring When the Leading Coefficient Is Not 1

When you have something like 2x² + 7x + 3, the process requires more care. One reliable method is the AC method:

1. Multiply a and c: 2 * 3 = 6
2. Find two numbers that multiply to 6 and add to 7: those are 6 and 1
3. Rewrite the middle term: 2x² + 6x + 1x + 3
4. Factor by grouping: 2x(x + 3) + 1(x + 3)
5. Factor out the common binomial: (2x + 1)(x + 3)

Difference of Squares

Any expression in the form a² - b² factors as (a + b)(a - b). Recognize this pattern and factoring becomes instant.

Example: Factor 25x² - 49

1. 25x² = (5x)² and 49 = 7²
2. Factor: (5x + 7)(5x - 7)

This pattern does not work for a sum of squares. The expression x² + 9 does not factor over the real numbers.

Try it yourself: Screenshot any algebra problem and get step-by-step solutions with Math.Photos — free browser extension.

The Quadratic Formula

When factoring is difficult or the trinomial does not factor neatly, the quadratic formula always works. For any equation ax² + bx + c = 0:

x = (-b ± sqrt(b² - 4ac)) / (2a)

For worked examples with different types of solutions, see our post on quadratic formula examples.

Using the Quadratic Formula Step by Step

Example: Solve 2x² - 5x - 3 = 0

1. Identify a = 2, b = -5, c = -3
2. Calculate the discriminant: b² - 4ac = (-5)² - 4(2)(-3) = 25 + 24 = 49
3. Apply the formula: x = (5 ± sqrt(49)) / 4 = (5 ± 7) / 4
4. Two solutions: x = (5 + 7)/4 = 3 and x = (5 - 7)/4 = -1/2

What the Discriminant Tells You

The discriminant (b² - 4ac) reveals the nature of the solutions before you finish solving:

• Positive and a perfect square: Two rational solutions (the equation factors nicely)
• Positive but not a perfect square: Two irrational solutions (expect square roots in the answer)
• Zero: Exactly one repeated solution
• Negative: No real solutions; the solutions are complex numbers

This is a useful shortcut on exams. If a question asks “how many real solutions does 3x² + 2x + 5 = 0 have?” you can compute the discriminant (4 - 60 = -56, which is negative) and answer zero without solving the whole thing.

Completing the Square

Completing the square is another method for solving quadratics, and it is how the quadratic formula itself is derived. It is also essential for converting quadratic functions to vertex form.

Example: Solve x² + 6x + 2 = 0 by completing the square

1. Move the constant: x² + 6x = -2
2. Take half the coefficient of x (which is 3), square it (which is 9), and add to both sides: x² + 6x + 9 = 7
3. Factor the left side: (x + 3)² = 7
4. Take the square root: x + 3 = ± sqrt(7)
5. Solve: x = -3 ± sqrt(7)

Try it yourself: Screenshot any algebra problem and get step-by-step solutions with Math.Photos — free browser extension.

Systems of Equations

A system of equations involves two or more equations with two or more variables. The solution is the set of values that satisfies all equations simultaneously. We have a dedicated walkthrough on how to solve systems of equations if you want to go deeper.

Substitution Method

Best when one equation already has a variable isolated or when it is easy to isolate one.

Example: Solve the system:

• y = 2x + 1
• 3x + y = 11

Since y is already isolated in the first equation, substitute it into the second:

1. 3x + (2x + 1) = 11
2. 5x + 1 = 11
3. 5x = 10
4. x = 2
5. Substitute back: y = 2(2) + 1 = 5

Solution: (2, 5)

Elimination Method

Best when both equations are in standard form and you can easily cancel a variable by adding or subtracting.

Example: Solve the system:

• 2x + 3y = 12
• 4x - 3y = 6

The y terms are already opposites (3y and -3y), so add the equations:

1. (2x + 3y) + (4x - 3y) = 12 + 6
2. 6x = 18
3. x = 3
4. Substitute into the first equation: 2(3) + 3y = 12, so 3y = 6, so y = 2

Solution: (3, 2)

Systems With No Solution or Infinite Solutions

Not every system has a single solution.

• No solution (inconsistent): The equations represent parallel lines. Example: y = 2x + 1 and y = 2x + 5. Same slope, different intercepts. They never intersect.
• Infinite solutions (dependent): The equations represent the same line. Example: y = 2x + 1 and 2y = 4x + 2. The second equation is just the first one multiplied by 2.

When you attempt to solve an inconsistent system, you will reach a contradiction like 0 = 4. When you solve a dependent system, you will get a tautology like 0 = 0.

Try it yourself: Screenshot any algebra problem and get step-by-step solutions with Math.Photos — free browser extension.

Polynomials

Polynomials are expressions with one or more terms, where each term has a variable raised to a non-negative integer power. Understanding how to work with them is essential for algebra and everything that comes after it.

Adding and Subtracting Polynomials

Combine like terms. Like terms have the same variable raised to the same power.

Example: (3x² + 5x - 2) + (x² - 3x + 7)

1. Group like terms: (3x² + x²) + (5x - 3x) + (-2 + 7)
2. Simplify: 4x² + 2x + 5

When subtracting, distribute the negative sign first.

Example: (3x² + 5x - 2) - (x² - 3x + 7)

1. Distribute the negative: 3x² + 5x - 2 - x² + 3x - 7
2. Combine: 2x² + 8x - 9

A frequent error is forgetting to distribute the negative to every term in the second polynomial. Watch out for that minus sign in front of the parentheses.

Multiplying Polynomials

Multiply each term in the first polynomial by each term in the second, then combine like terms.

Example: (2x + 3)(x² - x + 4)

1. 2x * x² = 2x³
2. 2x * (-x) = -2x²
3. 2x * 4 = 8x
4. 3 * x² = 3x²
5. 3 * (-x) = -3x
6. 3 * 4 = 12
7. Combine: 2x³ + (-2x² + 3x²) + (8x - 3x) + 12 = 2x³ + x² + 5x + 12

Polynomial Long Division

When dividing polynomials, the process mirrors long division with numbers.

Example: Divide (2x³ + 3x² - 5x + 1) by (x + 2)

1. Divide 2x³ by x to get 2x². Multiply (x + 2) by 2x² to get 2x³ + 4x². Subtract to get -x² - 5x + 1.
2. Divide -x² by x to get -x. Multiply (x + 2) by -x to get -x² - 2x. Subtract to get -3x + 1.
3. Divide -3x by x to get -3. Multiply (x + 2) by -3 to get -3x - 6. Subtract to get 7.

Result: 2x² - x - 3 with a remainder of 7, or equivalently: 2x² - x - 3 + 7/(x + 2).

If you find polynomial division confusing on paper, try screenshotting the problem to see each step broken down visually.

Try it yourself: Screenshot any algebra problem and get step-by-step solutions with Math.Photos — free browser extension.

Inequalities

Inequalities work similarly to equations, with one critical difference: when you multiply or divide both sides by a negative number, you must flip the inequality sign.

Linear Inequalities

Example: Solve 3x - 7 >5

1. Add 7 to both sides: 3x >12
2. Divide by 3: x >4

The solution is all real numbers greater than 4. On a number line, you would use an open circle at 4 and shade to the right. In interval notation: (4, infinity).

Compound Inequalities

A compound inequality combines two inequalities. “And” means both conditions must hold; “or” means at least one must hold.

Example: Solve -2 <3x + 1 <10

1. Subtract 1 from all three parts: -3 <3x <9
2. Divide all parts by 3: -1 <x <3

The solution is the interval (-1, 3).

Quadratic Inequalities

For quadratic inequalities, first solve the corresponding equation, then test intervals.

Example: Solve x² - 5x + 6 <0

1. Factor: (x - 2)(x - 3) <0
2. The roots are x = 2 and x = 3, which divide the number line into three intervals: (-infinity, 2), (2, 3), and (3, infinity)
3. Test a value from each interval:
   • x = 0: (0-2)(0-3) = 6 >0. Does not satisfy.
   • x = 2.5: (0.5)(-0.5) = -0.25 <0. Satisfies.
   • x = 4: (2)(1) = 2 >0. Does not satisfy.
4. Solution: 2 <x <3, or in interval notation, (2, 3).

Absolute Value Inequalities

Absolute value inequalities split into two cases.

For |expression| <k (where k >0): -k <expression <k

For |expression| >k (where k >0): expression <-k or expression >k

Example: Solve |2x - 3| <7

1. Write as a compound inequality: -7 <2x - 3 <7
2. Add 3 to all parts: -4 <2x <10
3. Divide by 2: -2 <x <5

Try it yourself: Screenshot any algebra problem and get step-by-step solutions with Math.Photos — free browser extension.

Tips for Studying Algebra Effectively

Algebra is a skill, and like any skill, it improves with practice. Here are a few strategies that actually help.

Work problems by hand. Reading through solutions feels productive, but it is not the same as solving problems yourself. Write out each step. When you get stuck, that is where the real learning happens.

Check your answers by substituting back. After solving an equation, plug your answer back into the original equation. This catches arithmetic errors and builds confidence in your process.

Identify your weak spots. If you consistently struggle with factoring trinomials or setting up word problems, spend extra time on those topics specifically. A tool like the free math checker on Math.Photos can help you verify your work and pinpoint where mistakes are happening.

Understand why, not just how. Memorizing steps will get you through some problems, but understanding the reasoning behind each step makes you faster and more flexible. When you encounter a problem you have never seen before, conceptual understanding is what carries you.

Use technology as a learning tool, not a crutch. Tools like Math.Photos are most effective when you attempt the problem first, then check your work against the step-by-step solution. Compare your approach to the AI’s approach and note where they differ. This is far more effective than just copying answers. Read about the difference between learning and copying in our Math.Photos vs Photomath comparison.

Frequently Asked Questions About Algebra

What is the most important concept in algebra?

Understanding how to manipulate equations by performing the same operation on both sides is arguably the single most important concept. Nearly everything in algebra, from solving linear equations to working with systems of equations, relies on this principle. Once this clicks, the rest of algebra becomes a series of variations on the same idea.

How do I know which method to use for solving quadratics?

Start by checking if the equation factors easily. If you can spot the factors within a few seconds, factoring is the fastest route. If not, use the quadratic formula, which works for every quadratic equation regardless of whether it factors. Completing the square is most useful when you need vertex form or when the problem specifically asks for it.

Why do I keep getting the wrong answer on algebra problems?

The most common sources of error are sign mistakes (especially when distributing negatives or subtracting polynomials), forgetting to apply an operation to every term, and dropping terms when combining like terms. Get in the habit of writing every step clearly and checking your work by substitution. If you want to identify exactly where you went wrong, screenshot your work with Math.Photos and compare your steps to the AI solution.

Is algebra used in real life?

Yes, constantly, though not always in an obvious way. Algebra is the language behind budgeting, cooking conversions, calculating interest rates, programming, data analysis, engineering, and any field that involves modeling relationships between quantities. Even when you are not writing equations on paper, the logical thinking that algebra builds is used in problem-solving across virtually every profession.

How long does it take to learn algebra?

Most students spend one to two academic years covering Algebra 1 and Algebra 2. However, the time needed varies widely depending on your background and how much you practice. If you are reviewing algebra as an adult or preparing for a standardized test, focused study of two to four weeks can cover the essentials if you practice daily. The key is consistent practice rather than marathon study sessions.

What comes after algebra?

After Algebra 1, most students take Geometry, then Algebra 2. From there, the typical path is Precalculus (which includes trigonometry) and then Calculus. Algebra skills are used heavily in all of these courses, so gaps in your algebra knowledge will show up repeatedly. Investing time in algebra now pays dividends throughout your entire math education.