# Advanced SAT Math Section Techniques for High-Scoring Students

Whether you have been studying for the SAT for some time or are just starting with your prep and looking for a high SAT math score, you have come to the right place! While there are a number of strategies for high scorers, this article will focus on advanced SAT math section techniques for high-scoring students. More specifically, we will discuss many granular advanced SAT math strategies across various math topics.

## A Quick Overview of the SAT Math Topics

SAT math encompasses virtually every math topic you have ever learned, and even a few you haven’t! SAT advanced math covers the tough questions on those topics, so it’s important to be familiar with the 20 topics that are covered. They’re categorized within the College Board framework as follows:

• Heart of Algebra – Solving Linear Equations, Inequalities and Absolute Value, Coordinate Geometry, Linear Functions, Systems of Linear Equations
• Problem Solving and Data Analysis – General Word Problems, Rates, Unit Conversions, Ratios, Statistics, Percentages
• Passport to Advanced Math SAT – Exponents, Roots, Quadratic Equations, Functions, Graph Interpretation, Table Data
• Additional Topics in Math SAT – Geometry, Trigonometry, Complex Numbers

Now that we know what we’ll face on the exam, let’s look at some of the optimal strategies for SAT Math section preparation for high-scorers.

## Technique # 1: Get Rid of Fractions in Equations

When thinking about the advanced SAT math tips you’d learn in this article, you probably did not think you’d be starting with fractions in equations. However, not all high-achiever SAT math tactics are based on complicated formulas or concepts. In fact, the secret of top-scorers is that they execute simple things flawlessly and efficiently, a major key to achieving top scores in SAT math.

With this in mind, a simple tactic you need to know for improving your score and time-management is how to quickly eliminate fractions in equations.

TTP PRO TIP:

Always try to get rid of fractions in equations.

For example, let’s say you had to solve for x in the following equation:

1/2 + x = 3x + 3/8

While we could get the least common denominator for the fractions and combine them, we can also multiply the entire equation by that same common denominator to eliminate the fractions.

Since the least common denominator is 8, we can multiply the entire equation by 8, so we have:

8/2 + 8x = 24x + 24/8

4 + 8x = 24x + 3

1 = 16x

1/16 = x

Keep in mind that the process above can be applied to math problems of all question types, in both the calculator section and the non-calculator section, and even when dealing with advanced math SAT questions. Let’s practice with an example.

### Example: Getting Rid of Fractions in an Equation

If 3/y + 1/3 = 3/4, then what is the value of y?

• 3.6
• 5.4
• 6.8
• 7.2

#### Solution:

To start, we can focus on the fractions 1/3 and 3/4. We should see that the least common denominator of those fractions is 12. However, we also have a fraction with a denominator of y. So, to clear all the fractions from the equation, we must multiply the entire equation by the LCD of 12 and y, which is simply 12y. Multiplying the equation by 12y gives us:

36y/y + 12y/3 = 36y/4

36 + 4y = 9y

36 = 5y

36/5 = y

Notice that solving this question by clearing the fractions first avoids all kinds of problems! The problem becomes easily solvable, and we need only 3 steps to solve it. This technique is a lifesaver!

## Technique # 2: Solving Equations With No Solution

Early on in your SAT prep, even if you are on a path to advanced SAT math preparation, you will learn about systems of equations. The easy questions will ask you to solve for a single variable, but as these questions get more advanced, you’ll have to deal with equations that produce no solution. In general, we know that we have two equations with no solution when, after we manipulate them, the variable coefficients are the same, but the constant terms are not.

TTP PRO TIP:

A system of linear equations will have no solution if the equations are expressed such that the coefficients of the variables of both equations are equal but the constant terms are not.

For example, let’s say we have the following system of equations:

2x + y = 10

4x + 2y = 36

If we multiply the first equation by 2, we have:

4x + 2y = 20

Now our two equations are 4x + 2y = 36 and 4x + 2y = 20. We should see that the coefficients of x and y, which are 4 and 2, respectively, are the same, while the constants of 36 and 20 differ. Thus, the equation has no solution.

Let’s practice with a real example that adds even more difficulty using extra variables in the equations.

### Example: System of Linear Equations With No Solution

ax – 4y = 5 (equation 1)

2x – 8y = 12 (equation 2)

If a is a constant in the above system of equations, and the system has no solution, what is the value of a?

• 1
• -1
• 2
• -2

#### Solution:

We learned that we can determine that a system of equations has no solution when the variable coefficients of the variables in each of the equations are equal, but the constant terms are not equal.

Thus, we’ll manipulate equation (1) so that its x- and y-coefficients match those in equation (2). To accomplish this, we can multiply equation (1) by 2, obtaining:

2ax – 8y = 10

So, our two equations are now:

2ax – 8y = 10 (equation 3)

2x – 8y = 12 (equation 4)

Thus, when a = 1, we see that the x- and y-coefficients in equation (3) are equal to those in equation (4), yet the constant terms differ. Therefore, the system will produce no solution.

## Technique # 3: Use Discriminant Analysis

When discussing high-scoring SAT math techniques, we must include discriminant analysis, a topic that is included in the Passport to Advanced Math category. Using the discriminant, we can easily determine the number and type of roots of a quadratic equation. The discriminant is part of the quadratic formula (not covered in this article), and specifically, for a quadratic equation in standard form (ax^2 + bx + c = 0), the discriminant is b2 – 4ac.

TTP PRO TIP:

Using the discriminant, we can determine the number and type of roots of a quadratic equation.

The great thing about the discriminant is that we can memorize a few simple rules to know the number of roots based on the result of b2 – 4ac. Let’s cover those rules now.

• If the value of the discriminant b2 – 4ac is positive, there are two real roots.
• If the value of the discriminant b2 – 4ac is zero, there is one real (repeated) root.
• If the value of the discriminant b2 – 4ac is negative, there are no real roots. Instead, there are 2 complex roots.

So, for example, let’s say we have the quadratic equation 10x^2 – 4x + 2 = 0. We could determine the number of roots by substituting 10 for a, -4 for b, and 2 for c:

b^2 – 4ac = (-10)^2 – 4(10)(2)

b^2 – 4ac = 100 – 80 = 20

Since the value of the discriminant is positive, we have two real roots. Note that we didn’t have to waste time factoring the trinomial at all. We simply had to substitute the values of a, b, and c and evaluate the discriminant.

Let’s now practice a full example with an added layer of difficulty.

### Example: Discriminant Analysis

If the equation 9x2 + 3x + k = 0 has only one solution, and k is a constant, what is the value of k?

• 4
• -4
• 1/4
• -1/4

#### Solution:

First, we know that since the equation has just one solution, the discriminant must be equal to zero. We also see that a = 9, b = 3, and c = k. So, we have:

b^2 – 4ac = 0

(3)^2 – 4(9)(k) = 0

9 – 36k = 0

9 = 36k

1/4 = k

Next, let’s take the knowledge learned here and discuss an additional high-score technique, that of dealing with a system of equations when the system contains one linear equation and one quadratic equation.

## Technique # 4: Solving a System of Equations With One Linear Equation and One Quadratic Equation

When you first start with your SAT studying, you likely will encounter an SAT math question or two that deals with systems of equations. Generally, in those questions, you’ll deal with two linear equations. However, as these questions become more advanced, the SAT will throw at you a system of equations with one linear equation and one quadratic equation. So, dealing with this question type is an important aspect of mastering the SAT math sections.

TTP PRO TIP:

Be prepared for questions dealing with one linear equation and one quadratic equation.

The good news is that this question type is quite predictable. Generally, you will be asked to solve for a specific variable or set of variables, or you’ll be asked to determine the number of solutions the equations produce. To come up with an answer to these questions, you may need to rely on your skills of substitution, factoring, the quadratic formula, or discriminant analysis.

For example, let’s say we are given the following system of equations (one linear and one quadratic):

y = 3x + 31

y = x^2 + 3

Since both equations equal y, we can set them equal to each other and solve for x.

x^2 + 3 = 3x + 31

x^2 – 3x – 28 = 0

(x – 7)(x + 4) = 0

Thus, x = 7 or x = -4.

TTP PRO TIP:

When dealing with one linear and quadratic equation, be prepared to substitute, factor quadratics, use the quadratic formula, and use your knowledge of discriminant analysis.

Next, let’s practice a full example.

### Example: One Linear and One Quadratic Equation

y = x2 + 5

y = x

How many real solutions are there to the system of equations given above?

• 0
• 1
• 2
• infinitely many

#### Solution:

Since we have y on the left side of each equation, we can set the two expressions for y equal to each other, and then simplify:

x^2 + 5 = x

x^2 – x + 5 = 0

Since we cannot easily factor, we can use the discriminant. Notice that for this quadratic equation, a = 1, b = -1, and c = 5: Thus, we have:

b^2 – 4ac = (-1)^2 – 4(1)(5)

b^2 – 4ac = 1 – 20 = -19

Since the value of the discriminant is negative, we have no real solutions to this system of equations. (Note: Because the value of the discriminant is negative, we know that we have 2 complex roots.)

## Technique # 5: Add Inequalities When the Signs Face in Opposite Directions

In any advanced study guide for the SAT, you will learn a lot about dealing with systems of inequalities. For example, one of the more advanced techniques for high SAT math scores is adding inequalities when the signs face in opposite directions.

You may remember learning how to add inequalities. You may have learned that you can add them only if the signs are facing in the same direction. However, adding the two inequalities is also possible when their signs face in opposite directions — with one caveat. Before adding them together, you must multiply one inequality by -1 (or any negative number, depending on the question). Recall that multiplying an inequality by a negative number requires you to reverse the inequality sign. The result here is that your two inequalities will now have their signs facing the same direction. From there, you can add the two inequalities. This rule is something we teach SAT students who are going for high scores.

TTP PRO TIP:

When adding inequalities whose signs face opposite directions, you must first multiply one inequality by a negative number.

For example, let’s say we needed to add x + 2y > 2 and 3x + 2y < 8, with the goal of isolating variable x. We should see that the first inequality has a greater than sign and the second has a less than sign. Thus, we can multiply the first inequality by -1 to flip the sign, giving us:

-1(x + 2y > 2) = -x – 2y < -2

With the signs facing the same direction, we can now add our inequalities. Summing -x – 2y < -2 and 3x + 2y < 8 give us:

2x < 6

x < 3

Now, let’s practice a full example.

### Example: Adding Inequalities When the Signs Face Opposite Directions

7y – x > 12

7y < 3x

Given the system of inequalities above, what is the smallest integer value of x?

#### Solution:

Here we see a grid-in question rather than a multiple-choice problem-solving question.

As we learned above, we must get the signs facing the same direction before we can add these inequalities. Thus, we can multiply the second equation by -1 and flip the sign, giving us -7y > -3x.

Next, we can sum 7y – x > 12 and -7y > -3x, giving us:

-x > 12 – 3x

2x > 12

x > 6

Thus, the least integer value of x is 7.

## Technique # 6: Solving Absolute Value Equations

Absolute values can show up in several ways on the SAT. One such way is in the form of equations. While absolute value equations can seem intimidating, there is a systematic process you can follow to solve them. We solve an absolute value equation for two cases: one when the expression inside the absolute value bars is positive and one when it’s negative.

For example, let’s say we need to determine possible values of x when given the equation |x| – 4 = 10. Notice that only the variable x is enclosed inside the absolute value bars. We would do the following:

Case 1: When x is positive

x – 4 = 10

x = 14

Case 2: When x is negative

-x – 4 = 10

-x = 14

x = -14

Thus, x can be 14 or -14.

Let’s practice with a more complicated example.

### Example: Absolute Value Equations

If |4x + 8| = 20, then what could be a value of x?

#### Solution:

We should notice that this problem has been made extra-complicated because it’s a grid-in question. In other words, we would not have the option to find an answer by plugging in answer choices if we did not know how to solve the problem. The good news is that we do know how to solve it!

So, let’s test our cases. Notice that the entire expression (4x + 8) is enclosed inside the absolute value bars.

Case 1: When (4x + 8) is positive

4x + 8 = 20

4x = 12

x = 3

Case 2: When (4x + 8) is negative

-1(4x + 8) = 20

-4x – 8 = 20

-4x = 28

x = -7

Thus, the correct answers for x are 3 or -7. However, since the answer to a grid-in question cannot be negative, the only answer that works is 3.

## Technique # 7: Know the Equation of a Circle in the Coordinate Plane

As you know, there are a variety of equations and formulas you need to memorize when dealing with Coordinate Geometry topics. One such formula is the equation of a circle in the coordinate plane. One great thing about learning this formula is that you can easily use it even in the most seemingly complicated questions once you have it down.

So, for any circle in the coordinate plane with a center at the point (a, b) and radius r, the standard form of the circle equation is the following:

(x – a)^2 + (y – b)^2 = r^2

For example, the equation of a circle with a radius of 3 and a center located at point (3, -2) would be the following:

(x – 3)^2 + [y -(- 2)]^2 = 3^2

(x – 3)^2 + (y + 2)^2 = 9

Notice that in order to express the information in standard form, we initially had to subtract -2 in the expression [y -(- 2)]^2, since the standard form requires a negative sign. Then, in the final step, we simplified the expression as (y + 2)^2.

KEY FACT:

For any circle in the coordinate plane with a center at the point (a, b) and radius r, the standard form of the equation is (x – a)^2 + (y – b)^2 = r^2.

Let’s now practice with an example.

### Example: Equation of a Circle

(x + 3)^2 + (y – 2)^2 = 5

The equation above defines a circle in the xy-plane. What are the coordinates of the center and the length of the radius of the circle?

• Center: (3,-2), Radius = 5
• Center: (-3,2), Radius = 5
• Center: (-3,2), Radius = √5
• Center: (3,-2), Radius = √5

#### Solution:

To start, we must recall that the standard form of the equation of a circle is (x – a)^2 + (y – b)^2 = r^2. Therefore, we must have:

x – a = x + 3

-a = 3

a = -3

and

y – b = y – 2

-b = -2

b = 2

and

r^2 = 5

r = √5

Thus, the center of the circle is (-3,2) and the radius is √5.

Alternate Solution:

We could first express the equation in standard form, as follows:

(x + 3)^2 + (y – 2)^2 = 5

[x -(-3)]^2 + (y – 2)^2 = (√5)^2

Now we can read the values of a, b, and r from the standard form as a = -3, b = 2, and r = √5. Thus, the center of the circle is (-3,2) and the radius is √5.

## Technique # 8: Memorize the Formula for the Area of an Equilateral Triangle

Many question types are associated with equilateral triangles on the SAT. For example, besides the fact that an equilateral triangle has all equal sides and all equal angles, when a perpendicular line is dropped from the vertex to the base, it creates two equal 30-60-90 right triangles. Knowing those relationships allows us to answer various questions. However, sometimes, using those relationships can be time-consuming.

With that in mind, we can refer to a useful formula when asked about the area of an equilateral triangle. That formula is area = [side^2(√3)]/4. This formula will allow you to answer equilateral area questions in seconds rather than minutes!

KEY FACT:

The formula for the area of an equilateral triangle is [side^2(√3)]/4.

Let’s practice with an example.

### Example: Using the Formula for the Area of an Equilateral Triangle

Example: Using the Formula for the Area of an Equilateral Triangle

If the side of an equilateral triangle is 8 units long, what is the area of the triangle?

• 4√3
• 8√3
• 24
• 64

#### Solution:

We can find the area of this equilateral triangle using the formula area = [side^2(√3)]/4 by substituting 4 for side:

area = [8^2(√3)]/4

area = [64(√3)]/4

area = 8√3

## Technique #9: Memorize “SOHCAHTOA”

If you have been studying for the SAT for some time, you likely know that some of the more challenging questions on the exam involve trigonometry. Luckily, we can use one simple technique to answer many trigonometry questions.

Most trigonometry questions force us to use Sine, Cosine, and Tangent. When solving questions using those, we can create the following fractions:

• Sine: Opposite/Hypotenuse

To help remember the definitions above, we can use the acronym “SOHCAHTOA.”

Let’s put this acronym to work in an example.

### Example: Memorize “SOHCAHTOA”

According to the figure above, what is the value of cos x?

• 1/2
• 1/5
• 3/5
• 4/5

#### Solution:

To answer this question, we need to focus on the “CAH” portion of the acronym of SOHCAHTOA:

Cosine of x = adjacent / hypotenuse

We do not initially know the length of the side adjacent to angle x, but since we know the lengths of the other two sides of this right triangle, we can calculate the length of the third side using the Pythagorean theorem. We can let y equal the side adjacent to angle x. Then, we have:

y^2 + 8^2 = 10^2

y^2 + 64 = 100

y^2 = 36

y = 6

Thus, cos x = 6/10 = 3/5.

## Technique # 10: When a Constant Value Is Added to or Subtracted From Each Data Point in a Set

Statistics is quite an important topic on the SAT, so there is much to learn on that topic. Furthermore, if you are pushing to be an advanced SAT test-taker, then even the most seemingly minor rules need to be memorized. One such rule is how adding a constant to or subtracting a constant from each number in a data set affects the average (arithmetic mean), median, range, and standard deviation. Let’s review each below.

When a constant x is added to or subtracted from each term of a data set:

• The average will increase by x or decrease by x, respectively.
• The median will increase by x or decrease by x, respectively.
• The range will be unchanged.
• The standard deviation will be unchanged.

Let’s practice an example, using the rules above.

### Example: When a Constant Value Is Added to or Subtracted From Each Data Point

Mr. Smith’s students earned the following scores on a math test:

92, 87, 95, 98, 77

However, after finishing grading, he realized that he accidentally awarded 5 extra points to each test. Mr. Smith calculated the average, median, range, and standard deviation of both the original and corrected test scores. He then calculated the positive difference for each measure between its original and corrected values. What is the sum of those positive differences?

• 5
• 10
• 15
• 20

#### Solution:

Using the rules above, we see that the average will decrease by 5, the median will decrease by 5, the range will not change, and the standard deviation will not change. So, the sum of the positive differences is 5 + 5 = 10.

## In Summary

SAT math tests you on 20 topics. In this article, we identified 10 high-value tips and techniques that can help you if you are motivated to score at or near 800 on the math section:

1. Get Rid of Fractions in Equations
2. Solving Equations With No Solution
3. Use Discriminant Analysis
4. Solving a System of Equations with One Linear Equation and One Quadratic Equation
5. Add Inequalities When the Signs Face in Opposite Directions
6. Solving Absolute Value Equations
7. Know the Equation of a Circle in the Coordinate Plane
8. Memorize the Formula for the Area of an Equilateral Triangle
9. Memorize “SOHCAHTOA”
10. When a Constant Is Added to or Subtracted From Each Data Point in a Set

Knowing these techniques will definitely give you a leg up when you encounter challenging SAT math questions. Take a look at our SAT Math Prep Course to review and learn every topic tested on the SAT in depth!  You can’t go wrong!

### What are some advanced techniques high-scoring students use for the SAT math section?

There are a variety of techniques that students use for the SAT math section. We have covered many in this article, but keep in mind that there are hundreds of specific techniques and strategies spanning the various SAT math topics. Learning these techniques and formulas will help you eliminate careless mistakes and improve your SAT math skills. To learn so many strategies, take a topical approach to studying, so you can focus on one SAT math topic at a time.

### How can I improve my SAT math score with advanced strategies?

A great way to improve your math score with advanced strategies is to practice these strategies in topical practice. By practicing just one topic at a time, you’ll ensure that you master and execute all strategies for one topic before moving to the next one.