Algebra often feels like a puzzle where your goal is to find the value of an unknown, usually called $x$. The key is to keep the equation balanced. Whatever you do to one side of the equals sign, you must do to the other.

Let's start with a basic example: $2x + 5 = 11$. Our mission is to get $x$ by itself. First, we'll undo the addition. To get rid of the $+5$, we subtract 5 from both sides.

Now, $x$ is being multiplied by 2. To undo that, we divide both sides by 2.

The same principle applies when the variable appears on both sides. Just gather all the $x$ terms on one side and all the constant numbers on the other. For an equation like $4x - 3 = 2x + 7$, we can start by subtracting $2x$ from both sides.

Now it's a simple two-step equation. Add 3 to both sides, then divide by 2, and you'll find that $x = 5$. Sometimes, you'll need to use the distributive property first, which means multiplying a term outside a parenthesis by everything inside it. For $3(x - 2) = 9$, we distribute the 3:

From here, add 6 to both sides and then divide by 3 to get $x = 5$.

Inequalities work almost exactly like equations. You can add, subtract, multiply, and divide both sides to isolate the variable. There's just one special rule to remember: if you multiply or divide both sides by a negative number, you must flip the inequality sign.

Let's solve $-2x + 4 < 10$. First, subtract 4 from both sides.

Now, we need to divide by -2. Because we're dividing by a negative number, we flip the $<$ sign to a $>$ sign.

Many algebra problems on the SAT involve linear functions, which create straight lines when graphed. The most common form you'll see is the slope-intercept form:

In this equation:

• $m$ is the slope of the line. It tells you how steep the line is and in which direction it goes. Think of it as "rise over run"—how much the line goes up (or down) for every unit it moves to the right.
• $b$ is the y-intercept. This is the point where the line crosses the vertical y-axis. Its coordinates are always $(0, b)$.

You might also see function notation, like $f(x) = mx + b$. Don't let this intimidate you. The notation $f(x)$ is just a fancy way of saying $y$. It's useful because it tells us the name of the function (in this case, $f$) and what the input variable is ($x$).

For example, if you have the function $f(x) = 2x + 1$ and are asked to find $f(3)$, you simply substitute 3 for every $x$ in the equation.

This means that when the input is 3, the output is 7. On the graph of the line, this corresponds to the point $(3, 7)$. Let's visualize the function $f(x) = 2x + 1$. Its y-intercept ($b$) is 1, and its slope ($m$) is 2, which we can write as $\frac{2}{1}$. This means for every 1 unit we move to the right, we move 2 units up.

Sometimes you'll be given two linear equations with two variables, like $x$ and $y$. This is called a system of equations. The solution is the one pair of $(x, y)$ values that makes both equations true. Graphically, this is the point where the two lines intersect. There are two main algebraic methods for solving these systems.

1. Substitution Method This method is best when one variable is already isolated in one of the equations.

Consider this system:

The first equation tells us that $y$ is the same as $2x + 1$. We can substitute this expression for $y$ into the second equation:

Now we have a single equation with only one variable, which we can solve. Distribute the 2, combine the $x$ terms, and solve to find $x = 2$. To find $y$, plug $x=2$ back into either of the original equations. The first one is easier: $y = 2(2) + 1 = 5$. The solution is $(2, 5)$.

2. Elimination Method This method is useful when the equations are both in the form $Ax + By = C$. The goal is to add or subtract the equations to eliminate one variable.

Here's a new system:

Notice that the coefficients of $y$ are opposites ($+3y$ and $-3y$). If we add the two equations together, the $y$ terms will cancel out.

This simplifies to $6x = 18$, so $x = 3$. Now substitute $x=3$ into either original equation to find $y$. Using the first one: $2(3) + 3y = 8$, which simplifies to $6 + 3y = 8$. Solving this gives $3y=2$, or $y = \frac{2}{3}$. The solution is $(3, \frac{2}{3})$.

Sometimes you might need to multiply one or both equations by a number first to get the coefficients to be opposites before you can eliminate a variable.

Ready to test your algebra skills? Let's try a few questions.

Mastering these foundational skills will set you up for success on a huge portion of the SAT math section.