Updated September 1, 2026

SAT Algebra: The 22 Skills You Need for a Perfect Score

According to the College Board, 35% of the questions on the SAT math section fall under the category “Algebra.” Another 35% fall under “Advanced Math,” which expands on the skills tested in the former. 

We’ve compiled a list of 22 essential algebra concepts that students need to know on test day. This post focuses on the core skills that students must grasp to score well on SAT math, and at the end, we’ll provide tips about applying these skills to SAT questions.

In this post, we use simple examples to illustrate the key algebra skills needed for the SAT. See which skills you’ve already mastered and which ones need extra practice. If you would benefit from 1-1 support from an expert tutor, get in touch with our team to find the right match.

Jump to section:
Algebra Skills (1-7)
Algebra Skills: (8-14)
Advanced Math Skills: (15-22)
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Algebra Skills

According to the College Board, about one-third of the questions on the SAT math section are categorized as “Algebra.” This umbrella mostly includes skills for solving linear equations and inequalities. 
Most of these Algebra questions test skills for solving linear and nonlinear equations. 

We’ll break this down further into the concrete skills that students need to know to master the questions in the “Algebra” category.

1. Combining Like Terms

Much of the SAT’s “Algebra” section involves knowing how to solve linear equations. In these questions, you’re usually working with a variable (like x). To solve an equation, you need to figure out which value for x would make the equation true.

For instance, you might be given an equation like 3x + 4x + 5 = 19.

To approach these kinds of questions, one of the most fundamental skills to understand is how to combine like terms. 

Like terms have the same variable raised to the same power. The terms in an equation are separated by an operator, like an addition sign or subtraction sign. In the example above, 3x and 4x both have the variable x raised to the same power: 1. This means we can directly add their coefficients (the numbers in front of the variable) together, to get 7x . This simplifies our equation, making it the first step in solving it.

The same logic applies when combining like terms for variables that are raised to higher powers. 

For example, take the expression 2x2 + x2. There are two terms, 2x2 and x2. In this case, we see that they both contain the same variable raised to the same power: x2. That means we can directly add the coefficients to get (2+1)x2 = 3x2. 

2. Isolating Variables

After simplifying an equation by combining all the like terms, we need to isolate the variable we want to solve. This means moving the variable we’re interested in solving on one side of the equal sign, and moving everything else to the other side.

For example, when solving an equation like 2x + 3 = 7, the goal is to put x alone on one side of the equal sign. 

To start, we want to move the 3 to the other side. On the left side, we can “undo” the addition of 3 by subtracting 3. Any operation you do on one side must be done on the other side as well. So we subtract 3 from both sides:

2x + 3 – 3 = 7 – 3
2x = 4

Now, to get x by itself, we have to deal with the 2. On the left side, because x is being multiplied by 2, we can “undo” this by dividing it by 2. And we do the same on the right.

(2x) / 2 = 4/2
x = 2

Now we have isolated the x and found that it equals 2.

3. Finding the Greatest Common Factor

One key strategy in solving linear equations is to find the greatest common factor (GCF), which helps us simplify an equation and make it easier to solve. 

For example, if we have the linear equation 8x + 4 = 0, we can simplify it by noticing that every term in the equation can be divided by 4. In this case, 4 is the largest factor that satisfies this condition. (In other words, even though each term is divisible by 2, it’s not the greatest common factor.) 

We can pull out this common factor, writing the equation as 4(2x+1) = 0. From there, we can divide both sides by 4, getting 2x+1=0. This makes our equation much easier to tackle. Using our skills in isolating the variable, we solve the equation:

2x + 1 – 1 = 0 – 1
2x = -1
x = -1/2

Thus, our solution is x = -1/2 .

4. Solving Linear Inequalities

Unlike equations, where students need to find one value for a variable, inequalities describe a range of values for a variable.

Still, solving inequalities is a lot like solving equations. First, we isolate the variable on one side of the inequality sign. To do so, add or subtract terms to move constants to the other side. Multiply or divide by a constant to get x alone. 

Note: If you multiply or divide by a negative number, flip the inequality sign. This is a key difference from equations.

Once the variable is isolated, you can solve for x.

Example:

3x + 5 ≤ 11

Step 1: Subtract 5 from both sides:

3x ≤ 6

Step 2: Divide both sides by 3:

x ≤ 2

Thus the solution is x ≤ 2, meaning x can be any number less than or equal to 2.

5. Solving Linear Systems Using Substitution

For the SAT “Algebra” section, students not only need to understand how to solve one linear equation. They also need to know how to solve systems of multiple linear equations.

One key skill in solving linear systems is to use substitution. To solve systems of equations using this method, you solve one of the equations (usually the simpler one) for one variable and plug the result into the other equation. For example, let’s say we have the following system of two equations:

x + y = 5
2x – y = 4

We can take the first equation, x + y = 5, and solve for x. This gives us x = 5 – y. If we know that x = 5 – y, then we can take that and plug it in for x in the second equation, 2x – y = 4. This gets us:

2(5 – y) – y = 4

Then we solve for y:

10 – 2y – y = 4
10 – 3y = 4
-3y = -6
y = 2

Finally, we can plug our solution for y back into the first equation to solve for x:

x + 2 = 5
x = 3

Thus, our solutions to this system are x = 3 and y = 2.

6. Solving Linear Systems Using Elimination

You can also solve linear systems using a method called elimination. Using this method, instead of solving for one equation first, you combine the two equations together and solve that. 

This is done by adding or subtracting the equations together to eliminate one of the variables. 

Let’s say we have the system:

3x + 2y = 16
2x – 3y = -2

We can choose to eliminate x. To do that, we need to make sure the coefficients in front of x in both equations are the same. That way, we can cleanly subtract them from each other with nothing left over. In the first equation, the x coefficient is 3, while in the second, it’s 2. 

Let’s make them both the same number by multiplying the top equation by 2 and the bottom one by 3. Remember, whatever you do to one side of the equation, you must do to the other.

2(3x + 2y) = 2(16)
3(2x-3y) = 3(-2)  

This gives us:

6x + 4y = 32
6x – 9y = -6

Now, let’s eliminate the x terms by subtracting the bottom equation from the top:

6x + 4y = 32
-(6x – 9y = -6)
_________________
13y = 26

And solve to get y=2. Now, plug the solution for y back into either one of the original equations to find x.

3x + 2(2) = 16
3x + 4 = 16
3x = 12
x = 4

The solution to this system of equations is thus x = 4 and y = 2.

7. Solving Linear Systems With No Solution or Infinite Solutions

Sometimes we get systems of equations that have no solution or infinitely many solutions.

In linear systems with no solution, the two equations represent parallel lines that never meet. When simplified, the variables cancel out and you get a false statement like 0=5. This means there is no solution, or no points at which the two lines equal each other. 

In linear systems with infinitely many solutions, the two equations represent the same line. When simplified, the variables cancel out and you get a true statement like 0 = 0. 

This means there are infinitely many solutions, or infinitely many points at which the two lines equal each other. 

8. Finding a Line’s Slope and Y-Intercept

When working with linear equations, students also need to understand how these equations are visualized as lines on a graph.

The point-slope form for a line is y = mx + b , where

m = slope 
b = y-intercept 
y = y-coordinate 
x = x-coordinate

To find the slope, denoted as m, between two points on a line, assign one point to be (x1,y1) and the other as (x2, y2) and then plug the values into the slope formula:

m = (y1 – y2) / (x1 – x2 ) = “rise” / “run”

Equation of a line: y = mx + b 
m = slope, b = y-intercept
Slope: m = (y1 – y2) / (x1 – x2 ) = “rise” / “run”
Midpoint: ((x1 – x2 / 2), (y1 – y2 / 2))

9. Finding The Point of Intersection Between Two Points

To find the point of intersection between two points on a line, figure out where they cross by setting the two equations equal to each other.

Example: Find the point of intersection between y = 2x +3  and y = -x + 1

Step 1: Make sure both lines are in point-slope form:

Line 1: y = 2x +3  
Line 2: y = -x + 1

Step 2: Find where they intersect on the y axis by setting the two y’s equal to each other

2x + 3 = -x + 1 

Step 3: Solve for x to find the x-coordinate of the intersection point

3x  = -2 
x = -⅔    

Step 4: Substitute the x-coordinate into one of the equations to find the y-coordinate

y = 2(-⅔)+3
y = -5/3

The point at which the two lines intersect is (-⅔ , 5/3).

10. Finding The Equations of Parallel and Perpendicular Lines

Parallel lines are lines that never intersect. Two lines are parallel when the slope m of both lines are equal, but the y-intercepts are different. 

For instance, lines y = 4/5x + 3 and  y = 4/5x are parallel because they have the same slope and different y-intercepts 

Example: Find the equation of the line that is parallel to y = 2x+1 that goes through point (2,2) 

We can use point-intercept form to solve this question.

Step 1: Parallel lines have the same slope, m. According to point-intercept form, In the equation y = 2x + 1, m = 2  

Step 2: Plug point (2,2) and m = 2 into y = mx + b to find b

2 = 2(2) + b
2 = 4 + b
-2 = b

Step 3: Plug it all back into point-slope form to identify the final equation.

y = mx + b 
m = 2
b = -2
y = 2x – 2 

11. Finding the Midpoint of Two Points on a Line

The midpoint of a line is the point that lies exactly halfway between two endpoints. Questions asking for the midpoint will usually provide two endpoints, and the midpoint can be found using the midpoint formula:

Midpoint of a line: ((x1 – x2 / 2), (y1 – y2 / 2))

Example problem: find the midpoint between endpoints (1,5) and (3, 11) 

midpoint = ((x1 – x2 / 2), (y1 – y2 / 2))
midpoint = ((1+3)/2, (5+11)/2)
midpoint = (2, 8)

The midpoint between endpoints endpoints (1,5) and (3, 11)  is (2, 8)

12. Finding the Distance between Two Points on a Line

he distance formula finds the distance between two points (x₁, y₁) and (x₂, y₂), on the graph. 

d= √(x₂ -x₁)2+(y₂ -y₁) / 2

d = distance
(x₁, y₁) = coordinates of the first point 
(x₂, y₂) = coordinate of the second point 

Example: Find the distance between point A (1, 2) and B (9, 17) 

Step 1: Write out the distance (d) formula

d= √(x₂ -x₁)2+(y₂ -y₁)2

Step 2: Plug in the coordinates where point A stands for (x₁, y₁) and B stands for (x₂, y₂)

d= √(9 -1)2+(17 -2)2

Step 3: Solve 

d= (8)2+(15)2
d= 64 + 225
d= 289 
d= 17

The distance between A and B is 17.

13. Evaluating Linear Functions

A function is a rule that defines a relationship between one variable (the independent variable) and another variable (the dependent variable). Functions can be represented by an equation that represents the relationship between an independent and dependent variable.

Function notation: a function is represented by f(x) , where f(x) is determined by any given x value. 

Think of a function f(x) as a machine, where every x-value is an input that when inserted into the machine leads to a result: f(x). For example: if the x-value of a point is 5 and f(x)=x+5 , then f(5) = 10.

14. Evaluating Composite Functions

Composite functions are expressed in the form (f∘g)(x) = f(g(x)), where one function is defined by another function.

Example: solve (f∘g)(x)=f(g(x)) when f(x) = x + 2 and g(x) = 2x  at x = 3

First, Plug x = 3 into g(x) = 2x 

Hint: order of operations matters. (f∘g)(x)=f(g(x)) is different than (g∘f)(x)=g(f(x)) and will lead to a different result.

g(3) = 2*3
g(3) = 6

Next, plug the g(3) into (f∘g)(x)=f(g(x))

(f∘g)(x) = f(g(3))
(f∘g)(x) = f(6)
(f∘g)(x) = 6 + 2
(f∘g)(x) = 8

Advanced Math

The College Board estimates that another one-third of SAT Math questions are in the category of  “Advanced Math.” This includes more high-level skills for solving linear equations as well as skills for solving nonlinear equations and inequalities. 

We’ll break this down further into the skills that students need to know to master the questions in the “Advanced Math” category.

15. Evaluating Exponents and Radicals

Nonlinear equations are equations that contain variables raised to a power greater than 1. On a graph, this might look like a curve or a parabola. These equations will often have more than one value for x. The “solutions” simply tell you where the curve crosses the x axis on a graph. 

Because these equations are nonlinear, many will contain a variable raised to a power. To solve them, students need to understand how to evaluate exponents and radicals. 

Let’s start with the basics. An exponent tells us how many times to multiply a value by itself. For example, 23 = 2 * 2 * 2=8. Here, the 2 is called the “base” and the 3 is called the “exponent.” By the same token, x4 = x * x * x * x. 

A radical is the “reverse” of an exponent. It is the value that, when multiplied by itself some number of times, gives you the value under the root symbol. For instance, to simplify ∛8 , ask: which number, multiplied by itself 3 times (the “degree”), would give us 8 (the “radicand”)? As we found above, the answer is 2. 

Radicals can be written as exponents, and vice versa. For example, the expression ∛8 can be written using an exponent as 81/3.

16. Understanding Exponent Rules

To begin solving equations with exponents and radicals, it’s important to follow the exponent rules:

These rules guide us in working with exponents and radicals. 

For example, to simplify an expression like 3(5x-3x)-2, we first combine the like terms within the parentheses to get 3(2x)-2. Then we evaluate the exponent. Using the negative exponent rule and remembering to distribute the exponent to all the items inside the parentheses, we get 3(1/4x2).  Finally, we evaluate the multiplication to get 3/4x2 as our final solution.

17. Solving Quadratic Equations

Once we understand what exponents and radicals are and what rules we need to follow when working with them, we can start to solve nonlinear equations. 

A common form of nonlinear equation is a quadratic equation, which contains a variable raised to the power of 2. Quadratic equations follow the basic format ax2 + bx + c and a ≠ 0. When graphed, quadratic equations form the shape of a parabola.

A parabola is a curve that is approximately u-shaped. Because the parabola is symmetrical, there is a single point called a vertex at the deepest point of the “curve” where the parabola begins to change direction. The vertex is located on the axis of symmetry, a line that cuts the parabola in half. 

X-value of vertex (“axis of symmetry”):  x = -b / 2a

To find the vertex of a parabola, use the formula x = -b / 2a to find the x-coordinate of the vertex, then plug that value back into the original equation of the parabola to find the y-coordinate.

Vertex form:  f(x) = a(x-h)2 + k
(h, k) is the vertex, axis of symmetry is x=h 
a indicates the direction and width of the graph

The points at which the quadratic crosses the x-axis are called the solutions or roots of the quadratic. Find these points by setting the parabola equation to 0. 

18. Factoring Quadratics

A key skill in solving quadratics is factoring. To factor quadratic equations of the form ax2 + bx + c, look for two numbers that multiply to ac and add to b. 

For example, to factor x2 + 5x + 6, look for two numbers that multiply to 6 and add to 5. These numbers would be 2 and 3. Then you can rewrite x2 + 5x + 6 = (x+2)(x+3).  Test this out the distributive property to see how both sides of the equal sign are equivalent to each other.

19. Factoring by Completing the Square

If a quadratic equation isn’t factorable by the method above, there’s still another way of factoring: by completing the square. To do this, we rely on two rules: 

Sum of squares: (a + b)2 = a2 + 2ab + b2
Difference of squares: (a-b)2 = (a+b)(a-b)

Example 1:

f(x) = 9x2 – 4 (difference of 2 perfect squares)

9x2 – 4 = (3x)2 + (2)2

This is of the form (a – b)2 = (a+b)(a-b).

Hence we factor the equation 9x2-4 = (3x+2)(3x-2)

Example 2:

f(x) = 4x2 + 12x + 9 (sum of 2 perfect squares)

4x2 + 12x + 9 = 2x2+ 2(2x)(3) + 32

This is of the form (a + b)2 = a2 + 2ab + b2:

(2x)2 + 2(2x)(3) + (3)2 = (2x+3)2

Hence, (2x + 3)(2x + 3) is the factored form of f(x) = 4x2 + 12x + 9.

20. Evaluating Quadratics using the Quadratic Formula

If neither of the two methods above work, we can solve any quadratic equation using the quadratic formula.

Quadratic formula:

We can also use the quadratic formula to determine how many solutions there are for a given equation. Certain quadratic equations have no solutions (equation does not pass x = 0). 

The discriminant, b2 – 4ac, the part of the quadratic formula underneath the root, determines the number of solutions:

b2-4ac > 0: Two real solutions.
b2-4ac = 0: One real solution.
b2-4ac < 0: No real solutions (complex numbers)

21. Solving Systems of Quadratic Equations

In a quadratic system with one solution, the two quadratic curves intersect at exactly one point. 

Example: 

y = x2 + 2
y=(x – 1)2

In a quadratic system with two solutions, the two quadratic curves intersect at exactly two points. 

Example: 

y = x2 + 2
y=(2x – 1)2

In a quadratic system with infinitely many solutions, the two quadratic curves are identical.

Example: 

y = x2 + 4x + 4
y=(x + 2)2

In a quadratic system with no solution, the two quadratic curves do not intersect.

Example: 

y = x2 + 2
y= x2 – 3

22. Solving Systems of Quadratic Equations

Solving nonlinear inequalities is similar to solving linear ones, except you might get more than one answer for x. As a result, you need to identify the intervals on the number line broken up by each of the values for x and test each interval to see if the inequality holds true in that range.

Let’s say we have the following quadratic inequality: 

x2 – 5x + 6 ≤ 0

First, let’s factor the left hand side:

(x-2)(x-3) ≤ 0

“Solving” it, we get:

x1 = 2, x2 = 3  

In this case, x1 and x2 break up the number line into 3 intervals: −∞ to 2, 2 to 3, and 2 to ∞. You can represent the intervals like this: ( −∞,2], (2,3), [3,∞). The “(“ means the number itself is not included, while the “]” indicates that it is included.

Step 2: Pick an easy number in each interval to test. For the interval ( −∞,2], we can pick 0. For (2,3), we can pick 5/2. For [3,∞), we can pick 6. Plug in each number for x and see if the inequality holds.

x = 0: 0 – 0 + 6 ≤ 0 → FALSE
x = 5/2: 25/4 – 25/2 + 6 = -5/4 ≤ 0 → TRUE
x = 6: 36 – 30 + 6 ≤ 0 → FALSE

So we know the inequality is true in the interval between (2,3). Then, make sure to test the x1 and x2 themselves to see if the inequality is true for those numbers.

Solution: the inequality is true for interval [2,3].

Next Steps

Understanding these 22 core skills for SAT Algebra will give you the foundation you need for a top score. Once you master these skills and learn how to apply them, you should be able to walk into test day with confidence.

If you’re aiming for a top SAT Math score but find yourself stuck, personalized guidance can make all the difference. Our SAT Prep coaches specialize in helping students master the test through targeted strategies and a proven curriculum designed by test experts. Reach out to us if you’d like tailored 1-1 support.

At the end of the day, SAT success comes down to not just working hard, but working smart. As you start preparing, be sure to make a study plan that hones in on all of the key topics that you will be tested on during the exam. 

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