Algebra Fundamentals

deck is a great starting point if you are new to algebra or feel shaky on the foundational vocabulary. It walks you through the building blocks first, such as variables, coefficients, constants, and algebraic expressions, before moving on to how to evaluate and simplify them. By the time you finish, you will have practiced the distributive property, combining like terms, and solving linear equations, including ones that contain fractions or turn out to have no solution or infinitely many solutions.

It is designed for middle or high school students working through their first algebra course, for adults returning to math after a break, or for anyone who wants a refresher before tackling more advanced topics. The cards mix short conceptual questions, like defining a variable or explaining the distributive property, with hands-on practice problems where you actually solve for x. This balance helps you check both your understanding of the language of algebra and your ability to apply it.

To get the most out of the deck, try to spend a little time with it every day rather than cramming everything into one long session. When you hit a practice problem, work it out on paper before flipping the card, since doing the algebra by hand is where real learning happens. If you get a card wrong, read the answer carefully and then try to redo a similar problem from memory a day or two later to strengthen your recall.

Pay special attention to the cards about what it means for an equation to have no solution or infinitely many solutions, because these ideas often trip students up but come up again and again in later algebra. Treat any card you miss as a signal to revisit the earlier cards on simpler ideas, since the harder questions usually depend on them. With steady, focused practice, the patterns will start to feel familiar and solving equations will become second nature.

Algebra is built on the idea of using symbols to represent unknown or changeable quantities. A variable is a symbol, usually a letter such as \(x\) or \(y\), that stands in for a value we may not yet know or that may vary depending on context. When a variable is combined with constants (fixed numbers) and operations like addition, subtraction, multiplication, or division, the result is an algebraic expression. For example, \(3x + 5\) is an algebraic expression because it combines the variable \(x\), the constant \(5\), and the operations of multiplication and addition, all without an equals sign.

Within any expression, each term has its own structure. The numerical factor multiplied by a variable in a term is called a coefficient; in the term \(7x\), the coefficient is \(7\). A constant, on the other hand, is a number standing on its own whose value does not change; in \(2x + 9\), the constant is \(9\). To evaluate an expression means to substitute specific numerical values for each variable and then simplify using the order of operations to obtain a single numerical answer.

Two terms are called like terms when they contain the same variable raised to the same power, such as \(3x^2\) and \(-5x^2\). Like terms can be combined by adding or subtracting their coefficients. The distributive property, which states that \(a(b + c) = ab + ac\), is another essential tool: it lets you multiply a single term by every term inside a set of parentheses. For instance, to simplify \(4(2x - 3) + 5x\), you first distribute to get \(8x - 12 + 5x\), and then combine the like terms \(8x\) and \(5x\) to obtain \(13x - 12\).

A linear equation is one in which the highest power of the variable is 1; its graph is a straight line, and its general form is \(ax + b = 0\). The goal when solving such an equation is to isolate the variable using inverse operations. For example, in \(3x + 7 = 22\), you subtract \(7\) from both sides to get \(3x = 15\), and then divide both sides by \(3\) to find \(x = 5\). When the equation contains parentheses, the distributive property is used first; in \(5(x - 2) = 3x + 4\), distributing gives \(5x - 10 = 3x + 4\), after which subtracting \(3x\) and adding \(10\) leads to \(2x = 14\) and finally \(x = 7\).

When fractions appear in a linear equation, multiplying every term by the least common denominator clears them so that standard solving methods can be applied. For instance, in \(\frac{x}{3} + 2 = 5\), you can subtract \(2\) from both sides to obtain \(\frac{x}{3} = 3\), and then multiply both sides by \(3\) to get \(x = 9\). Not every linear equation has exactly one solution, however. An equation such as \(2(x + 3) = 2x + 6\) simplifies to \(0 = 0\), a statement that is always true regardless of the value of \(x\); such an equation is said to have infinitely many solutions. By contrast, simplifying some equations leads to a contradiction like \(0 = 5\), which is never true and indicates that the equation has no solution.

Inequalities extend the idea of equations by comparing quantities with symbols such as \(<\), \(>\), \(\leq\), or \(\geq\). The symbols \(<\) and \(>\) represent strict inequalities, in which the boundary value is not included in the solution set, while \(\leq\) and \(\geq\) are non-strict and do include the boundary. Most inequality-solving steps mirror those used for equations, with one crucial exception: multiplying or dividing both sides by a negative number reverses the direction of the inequality sign. So if \(-2x > 6\), dividing by \(-2\) yields \(x < -3\). A compound inequality like \(-2 < 3x + 1 \leq 10\) can be solved by performing the same operation on all three parts at once, leading to \(-1 < x \leq 3\).

Solution sets are commonly written in interval notation, where parentheses indicate an excluded endpoint and brackets indicate an included endpoint. For example, the solution to \(3x - 4 \leq 11\) is \(x \leq 5\), which in interval notation is \((-\infty, 5]\). The solution to \(2x + 5 > 3x - 1\) is \(x < 6\), written as \((-\infty, 6)\). Absolute value equations and inequalities are another important family of problems. Since the absolute value \(|a|\) represents the distance of \(a\) from zero, it is always non-negative. The equation \(|x| = a\) for \(a \geq 0\) splits into two cases, \(x = a\) or \(x = -a\), while \(|x| < a\) becomes \(-a < x < a\) and \(|x| > a\) becomes \(x > a\) or \(x < -a\). Applied to \(|2x - 3| = 7\), for instance, the two cases give \(x = 5\) and \(x = -2\).

A polynomial is an expression made up of variables and coefficients combined using addition, subtraction, and multiplication, where every exponent on a variable is a non-negative integer; an example is \(4x^3 - 2x + 7\). The degree of a polynomial is the highest power of the variable that appears, so \(5x^4 + 3x^2 - 1\) has degree \(4\). Polynomials are added or subtracted by combining like terms, and they are multiplied by distributing each term of one polynomial across every term of the other. For two binomials, a convenient shortcut is the FOIL method (First, Outer, Inner, Last), which expands \((a + b)(c + d)\) into \(ac + ad + bc + bd\). Applied to \((2x + 3)(x - 4)\), FOIL gives \(2x^2 - 8x + 3x - 12\), which combines to \(2x^2 - 5x - 12\).

Factoring reverses the multiplication process and is one of the most useful skills in algebra. The first step in factoring any polynomial is to identify and pull out the greatest common factor, or GCF. For example, the GCF of \(6x^3 + 9x^2\) is \(3x^2\), and factoring it out gives \(3x^2(2x + 3)\). Several recognizable patterns make factoring faster: the difference of squares pattern \(a^2 - b^2 = (a + b)(a - b)\) applies to expressions like \(x^2 - 16 = (x + 4)(x - 4)\); perfect square trinomials follow \(a^2 + 2ab + b^2 = (a + b)^2\) or \(a^2 - 2ab + b^2 = (a - b)^2\), as in \(x^2 + 10x + 25 = (x + 5)^2\); and a trinomial of the form \(x^2 + bx + c\) factors as \((x + p)(x + q)\) where \(p\) and \(q\) are two numbers whose product is \(c\) and whose sum is \(b\). The sum and difference of cubes formulas, \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\) and \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), can be remembered with the mnemonic SOAP for the signs (Same, Opposite, Always Positive).

A quadratic equation is a polynomial equation of degree 2, written in standard form as \(ax^2 + bx + c = 0\) with \(a \neq 0\). The expression \(b^2 - 4ac\) beneath the square root in the quadratic formula is called the discriminant, and it reveals the nature of the solutions: if the discriminant is positive, there are two distinct real solutions; if it is zero, there is exactly one real solution (a repeated root); and if it is negative, there are two complex conjugate solutions. The quadratic formula itself provides the solutions to any quadratic equation in standard form. Applied to \(2x^2 + 3x - 2 = 0\), the discriminant is \(9 + 16 = 25\), and the formula gives \(x = \frac{-3 \pm 5}{4}\), producing the solutions \(x = \frac{1}{2}\) and \(x = -2\).

Two other important techniques for working with quadratics are completing the square and writing in vertex form. Completing the square rewrites a quadratic as \(a(x - h)^2 + k\) by taking half of the coefficient of \(x\), squaring it, and adding and subtracting that value. For example, to solve \(x^2 + 6x + 5 = 0\), you first rewrite it as \(x^2 + 6x = -5\) and add \((\frac{6}{2})^2 = 9\) to both sides to obtain \((x + 3)^2 = 4\); taking square roots then gives \(x + 3 = \pm 2\), so \(x = -1\) or \(x = -5\). The vertex form \(y = a(x - h)^2 + k\) has its vertex at the point \((h, k)\), and the sign of \(a\) tells you whether the parabola opens upward (if \(a > 0\)) or downward (if \(a < 0\)). By Vieta's formulas, the sum of the roots of \(ax^2 + bx + c = 0\) is \(-\frac{b}{a}\) and the product of the roots is \(\frac{c}{a}\), a quick way to check your answers. Factoring is often the simplest method when applicable, as with \(x^2 - 5x + 6 = 0\), which becomes \((x - 2)(x - 3) = 0\) and yields \(x = 2\) or \(x = 3\). The Remainder Theorem adds another tool: when a polynomial \(f(x)\) is divided by \((x - c)\), the remainder equals \(f(c)\), so if \(f(c) = 0\), then \((x - c)\) is a factor.

A system of equations is a collection of two or more equations that share the same variables, and its solution is the set of values that satisfies every equation in the system at the same time. There are three principal methods for solving a system of two linear equations. The graphing method finds the point where the two lines intersect on a coordinate plane. The substitution method solves one equation for one variable in terms of the others and then plugs that expression into the remaining equation. The elimination method adds or subtracts the equations from each other in order to cancel one of the variables.

The substitution method works well when one equation is already solved, or easily solved, for a variable. Given the system \(y = 2x + 1\) and \(3x + y = 11\), you substitute \(2x + 1\) for \(y\) in the second equation to obtain \(3x + (2x + 1) = 11\), which simplifies to \(5x + 1 = 11\), so \(x = 2\). Substituting back gives \(y = 2(2) + 1 = 5\), and the solution is the ordered pair \((2, 5)\). The elimination method, by contrast, is most efficient when the coefficients of one variable are opposites or can be made so. In the system \(2x + 3y = 12\) and \(4x - 3y = 6\), adding the equations eliminates \(y\) and yields \(6x = 18\), so \(x = 3\). Substituting back gives \(2(3) + 3y = 12\), or \(3y = 6\), and therefore \(y = 2\); the solution is \((3, 2)\).

Not every system has a unique solution. A system is called inconsistent when it has no solution, which graphically corresponds to two parallel lines that never cross; algebraically, this situation arises when the equations reduce to a contradiction. A system is called dependent when it has infinitely many solutions, which happens when the two equations actually describe the same line so that every point on the line is a solution. You can predict the number of solutions for a \(2 \times 2\) linear system by comparing the slopes and y-intercepts of its lines: different slopes give one solution, equal slopes with different intercepts give no solution, and equal slopes with equal intercepts give infinitely many solutions.

For a quick algebraic approach to a \(2 \times 2\) system, Cramer's Rule uses determinants. For the system \(ax + by = e\) and \(cx + dy = f\), the solutions are \(x = \frac{ed - bf}{ad - bc}\) and \(y = \frac{af - ce}{ad - bc}\), provided the determinant \(ad - bc\) is not zero. A zero determinant signals either no solution or infinitely many solutions, mirroring the parallel-line and same-line cases.

Exponents obey a handful of rules that simplify many calculations. The product rule states that when powers share the same base, their exponents are added: \(a^m \cdot a^n = a^{m+n}\). The power rule says that raising a power to another power multiplies the exponents: \((a^m)^n = a^{mn}\). The quotient rule says that dividing powers with the same base subtracts the exponents: \(\frac{a^m}{a^n} = a^{m-n}\) (with \(a \neq 0\)). From the quotient rule, any nonzero number raised to the zero power equals \(1\), because \(a^n / a^n = a^{n-n} = a^0 = 1\). A negative exponent, meanwhile, signals a reciprocal: \(a^{-n} = \frac{1}{a^n}\), so \(2^{-3} = \frac{1}{2^3} = \frac{1}{8}\).

Radicals are closely tied to fractional exponents. The \(n\)th root of \(a\) can be written as \(a^{1/n}\), and more generally \(\sqrt[n]{a^m} = a^{m/n}\). This connection lets you convert freely between radical and exponential form. To simplify a radical, factor the radicand to expose perfect powers, as in \(\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}\). When a radical appears in a denominator, it is standard practice to rationalize the denominator by multiplying both numerator and denominator by the radical, so that \(\frac{1}{\sqrt{3}}\) becomes \(\frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{3}\).

A rational expression is a fraction whose numerator and denominator are both polynomials, such as \(\frac{x^2 + 1}{x - 3}\). Such an expression is undefined wherever the denominator equals zero, so the first step in working with rational expressions is to identify and state the restrictions on the variable. To simplify, factor the numerator and denominator completely and cancel any common factors. Two rational expressions are multiplied by multiplying numerators together and denominators together, and they are divided by multiplying the first expression by the reciprocal of the second. To add or subtract rational expressions with different denominators, find the least common denominator, rewrite each fraction with that LCD, and then combine the numerators before simplifying the result.

A function is a relation that assigns exactly one output value to every input value. If \(f(x) = x^2\), then each value of \(x\) produces exactly one value of \(f(x)\). The vertical line test captures this idea geometrically: a graph represents a function if and only if every vertical line crosses the graph at most once; if any vertical line meets the graph in two or more points, the relation is not a function. Two sets describe a function's behavior: the domain is the set of all permissible input values, and the range is the set of all possible output values. Common restrictions on the domain come from division by zero and from taking even roots of negative numbers. For \(f(x) = \frac{1}{x - 3}\), the denominator cannot be zero, so \(x \neq 3\) and the domain is \((-\infty, 3) \cup (3, \infty)\). For \(f(x) = \sqrt{x - 2}\), the radicand must be non-negative, so \(x \geq 2\) and the domain is \([2, \infty)\).

Function composition builds new functions by feeding the output of one function into another. The composition \((f \circ g)(x)\) equals \(f(g(x))\): you first evaluate \(g(x)\) and then apply \(f\) to that result. For example, with \(f(x) = 2x + 1\) and \(g(x) = x^2\), the value of \((f \circ g)(3)\) is found by first computing \(g(3) = 9\) and then \(f(9) = 2(9) + 1 = 19\). Compositions are particularly useful for chaining operations and for describing transformations in stages.

An inverse function reverses the action of the original function: if \(f(a) = b\), then \(f^{-1}(b) = a\). Graphically, the graph of \(f^{-1}\) is the reflection of the graph of \(f\) across the line \(y = x\). To find an inverse algebraically, replace \(f(x)\) with \(y\), swap \(x\) and \(y\) throughout the equation, and then solve for \(y\); the resulting expression is \(f^{-1}(x)\). A useful check is to verify that \(f(f^{-1}(x)) = x\). Some functions are not defined by a single rule over their entire domain; a piecewise function uses different expressions for different intervals, such as \(f(x) = x^2\) when \(x \geq 0\) and \(f(x) = -x\) when \(x < 0\).

A logarithm is the inverse of exponentiation. The statement \(\log_b(x) = y\) means exactly the same thing as \(b^y = x\), answering the question: to what power must the base \(b\) be raised to produce \(x\)? Two specific bases are so common that they have their own notation: the common logarithm uses base \(10\) and is written as \(\log(x)\), so \(\log(1000) = 3\) because \(10^3 = 1000\); the natural logarithm uses base \(e\) (approximately \(2.718\)) and is written as \(\ln(x)\). The natural logarithm appears constantly in calculus and in models of exponential growth and decay. To solve an equation like \(\log_2(x) = 5\), you convert from logarithmic to exponential form, obtaining \(x = 2^5 = 32\).

Logarithms obey three key rules that mirror the rules for exponents. The product rule states that \(\log_b(MN) = \log_b(M) + \log_b(N)\); the quotient rule states that \(\log_b(M / N) = \log_b(M) - \log_b(N)\); and the power rule states that \(\log_b(M^n) = n \cdot \log_b(M)\). These rules allow complicated logarithmic expressions to be expanded or condensed. The change of base formula, \(\log_b(x) = \frac{\log_a(x)}{\log_a(b)}\), makes it possible to evaluate any logarithm on a calculator that only has common or natural log keys, since \(\log_b(x) = \frac{\ln(x)}{\ln(b)}\).

Sequences are ordered lists of numbers that often follow a pattern. An arithmetic sequence is one in which each term differs from the previous term by a constant called the common difference \(d\); the \(n\)th term is given by \(a_n = a_1 + (n - 1)d\). For example, the sequence \(3, 7, 11, 15, \ldots\) has common difference \(4\), so its tenth term is \(a_{10} = 3 + 9 \cdot 4 = 39\). The sum of the first \(n\) terms of an arithmetic series is \(S_n = \frac{n(a_1 + a_n)}{2}\), or equivalently \(S_n = \frac{n}{2}[2a_1 + (n - 1)d]\), which can be remembered as the number of terms times the average of the first and last terms.

A geometric sequence, by contrast, is one in which each term is obtained by multiplying the previous term by a constant called the common ratio \(r\); the \(n\)th term is \(a_n = a_1 \cdot r^{n-1}\). The sum of the first \(n\) terms of a geometric series is \(S_n = \frac{a_1(1 - r^n)}{1 - r}\), provided \(r \neq 1\). When the absolute value of the common ratio is less than one, the series converges as more and more terms are added, and the infinite geometric series has a finite sum given by \(S = \frac{a_1}{1 - r}\). For the series \(8 + 4 + 2 + 1 + \ldots\), the first term is \(8\) and the ratio is \(\frac{1}{2}\), so the sum is \(\frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16\).