Best SAT Math Prep: Linear Equations in One Variable

Frontier Lesson: "Linear Equations in One Variable" for the Digital SAT

"Mastering linear equations is your golden ticket to boosting your SAT Math score—let's unlock it step by step!"

Why This Skill Matters on the SAT

Easy Points: These problems are low-hanging fruit! They typically appear in the first half a single Math module, so nailing them builds confidence and saves time for harder questions.
Gateway to Advanced Math: Many word problems, systems of equations, and even graphing questions depend on solving linear equations first.

How the SAT Tests "Linear Equations in One Variable"

Expect questions in two formats:

Multiple Choice (MCQ):

Tip: Plug answer choices back into the equation (works great for harder problems too!).

Student-Produced Response (SPR, "grid-in"):

Crucial: Double-check your arithmetic—no answer choices to guide you!

Understand "Linear Relationships"

1. Constant rate of change:

For every unit change in $x$ , $y$ changes by a fixed amount.

2. Proportionality:

The relationship can be represented as $y = m x + b$ , where:

$m$ = slope (rate of change),
$b$ = y-intercept (value when $x = 0$ ).

➢ SAT Tip: Word problems often describe linear relationships (e.g., "a car drives at 60 mph" $⟹$ Distance = 60 $\times$ Time.)

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What are "Linear Equations in One Variable"?

➢ A linear equation in one variable is an equation that can be written in the form:

$a x + b = 0$

Where:

$x$ is the variable (the unknown we're solving for)
a and b are constants (known numbers)
a ≠ 0 (if a=0, it's no longer a linear equation)

➢ Key Characteristics:

Only one variable (usually $x$ ) is present
The variable has an exponent of 1 (no $x^{2}$ , $x^{3}$ , etc.)
The graph would be a straight line (that's why it's called "linear")

➢ Example Equations:

Simple: $2 x + 3 = 7$
With fractions: $\frac{1}{2} x - 4 = 0$
With parentheses: $3 (x - 2) = 2 x + 1$

Key Points for Solving "Linear Equations in One Variable" Questions

1. Moving Terms (Changing Signs)

When you move a term to the other side of the equation, change its sign.

Example:
$x + 5 = 9 \Rightarrow x = 9 - 5$
The "+5" becomes "-5" when moved right.

2. Handling Fractions

To eliminate fractions, multiply all terms by the LCD (Least Common Denominator).

Example:
$\frac{x}{2} + \frac{1}{3} = \frac{x}{4}$
LCD is 12 → Multiply every term by 12:
$6 x + 4 = 3 x$

3. Keeping x on One Side

Always aim to have all x terms on one side and constants on the other.

Example:
$2 x - 3 = x + 4 \Rightarrow 2 x - x = 4 + 3 \Rightarrow x = 7$

4. x on Both Sides

When x appears on both sides, move all x terms to one side.

Example:
$3 x + 2 = 2 x - 5 \Rightarrow 3 x - 2 x = - 5 - 2 \Rightarrow x = - 7$

5. Parentheses Around x Terms

Use the distributive property to eliminate parentheses first.

Example:
$\frac{1}{2} (x - 1) = 3$
First distribute: $\frac{1}{2} x - \frac{1}{2} = 3$
Then solve normally.

6. Cross Multiplication

A method can be used to solve equations involving fractions.

Let's say we have two fractions in an equation: $\frac{a}{b} = \frac{c}{d}$
We can always have another equation: $a \times d = b \times c$ , quickly remove the denominator.

7. Special Cases

No solution: When all variables cancel out and you get a false statement (e.g., $0 = 5$ )
Infinite solutions: When all variables cancel out and you get a true statement (e.g., $3 = 3$ )

🎯 "Linear equations in one variable" are a common topic on the SAT. In the next part, we'll learn how they will be tested in real Digital SAT.

Common SAT "Linear Equations in One Variable" Question Types

Example 1. Solving for a Single Variable

These questions typically provide a simple linear equation and ask you to solve for $x$ .

Example: What is the solution to the equation $2 x + 3 = - 3$ ?

A. $1$

B. $1.5$

C. $- 3$

D. $0$

Approach: Isolate $x$ on one side by undoing any additions, subtractions, multiplications, or divisions.

Example 2. Substitute to Find Another Expression

You might be asked to compute the value of another related expression.

Example: If $5 (2 x + 3) + 2 = 2 x + 3$ , what is the value of $2 x + 3$ ?

Approach: Solve for $x$ , then substitute the value into the new expression; or take " $2 x + 3$ " as a whole and isolate it on one side to find its value.

Example 3. Equations with Equivalent Representations

These questions ask you to identify the equation that is equivalent to the original one, and it may phrase the question differently.

Example: $\frac{1}{2} (3 x + 3) = 84$

Which equation has the same solution as the given equation?

A. $x + 1 = 56$

B. $x + 1 = \frac{84}{3}$

C. $x - 1 = 42$

D. $x + 57 = 0$

Approach: Simplify the given equation logically; look for equivalent transformations like distributing or combining like terms.

Example 4. Identifying the Equation that Has the Same Solution

You may be asked to identify the equation that has the same solution as the given equation.

Example: Which of the following is equivalent to $\frac{3}{2} + \frac{x}{2} = 7$ ?

A. $3 + 2 x = 14$

B. $6 + 2 x = 42$

C. $3 + x = 14$

D. $6 + x = 42$

Approach: Solve the given equation and substitute the value of $x$ into the four options and calculate which one is correct. Or solve the value of $x$ to the given equation and the four equations of options and match them.

Example 5. Number of Solutions Determination

You will be given an quation and asked to determine how many solutions it has.

Example: How many solutions does the equation $10 (13 x - 9) = - 13 (8 - 10 x)$ have?

A. Exactly one

B. Exactly two

C. Infinitely many

D. Zero

Approach: Simplify both sides. Look for contradictions (no solution), tautologies (infinite solutions), or just one solution.

Example 6. Variables Representing Constants

These problems provide additional variables/constants, such as $b$ , $p$ , $k$ , $w$ , etc. and ask you to determine their values based on properties like "no solutions" or "infinitely many solutions."

Example: $(b - 9) x = 8$

In the given equation, $b$ is a constant. If the equation has no solution, what is the value of $b$ ?

A. $10$

B. $17$

C. $0$

D. $9$

Approach: For an equation of the form $a \cdot x = c$ , you can determine the nature of the solutions as follows:

One unique solution: When $a \neq = 0$ , there is a unique solution $x = \frac{c}{a}$
No solution: When $a = 0$ and $c \neq = 0$ .
Infinite solutions: When $a = 0$ and $c = 0$

Example 7. Word Problems Leading to Equations

The question is a text description of an equation, and you need to "translate" it into the equation form.

Example: 2 more than $4$ times a number $x$ is equal to $- 38$ . Which equation represents this situation?

A. $(2) (4) x = - 38$

B. $4 x = - 38 + 2$

C. $2 x + 4 = - 38$

D. $4 x + 2 = - 38$

Approach: Carefully define the variables from the problem and write the equation step-by-step.

Example 8. Geometric Situations Leading to Equations

Questions might involve geometric problems (e.g., perimeter, sum of angles) where their relationship is exactly linear.

Example: The perimeter of an isosceles triangle is $78$ inches. Each of the two congruent sides of the triangle has a length of $20$ inches. What is the length, in inches, of the third side?

Approach: Write an expression for the perimeter or relevant geometric property, substituting known side lengths and solving for the missing side.

Example 9. Real-World Context with Proportions and Rates

Questions test your ability to calculate a certain value such as the rate/speed/price/quantity/etc. in a real-life scenario.

Example:

John used a tool called an auger to remove corn from a storage bin at a constant rate. The bin contained $25, 000$ bushels of corn when John began to use the auger. After $4$ hours of using the auger, $21, 360$ bushels of corn remained in the bin. If the auger continues to remove corn at this rate, what is the total number of hours John will have been using the auger when $13, 170$ bushels of corn remain in the bin?

A. $15$

B. $10$

C. $9$

D. $13$

Approach: Interpret the linear relationship, write an expression to find the value required.

Example 10. Interpreting the Components of an Equation

You might be required to interpret the components of an equation in a real-world context.

Example: $1.2 n + 5 = 17$

A tree had a height of 5 feet when it was planted. The equation above can be used to find how many years $n$ it took the tree to reach a height of $17$ feet. Which of the following is the best interpretation of the number $1.2$ in this context?

A. The number of years it took the tree to double its height

B. The average number of feet that the tree grew per year

C. The average number of years it takes similar trees to grow $17$ feet

D. The height, in feet, of the tree when the tree was $1$ year old

Approach: Analyze the relationship and the role of each term in the equation to understand its real-world meaning.

Example 11. Building Equations from Word Problems

These involve creating an equation from a description of costs, totals, or other quantities.

Example:

John paid a total of $ $174$ for an electron microscope by making a down payment of $ $42$ plus $w$ monthly payments of $ $12$ each. Which of the following equations represents this situation?

A. $12 w - 42 = 174$

B. $42 w + 12 = 174$

C. $12 w + 42 = 174$

D. $42 w - 12 = 174$

Approach: Similar to previous word problems types, you need to write down the constants and variables one by one according to the meaning of the text, and list the equations according to their linear relationship so that the left and right sides of the equations have the same meaning but are expressed in different ways.

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Common Mistakes in Real Digital SAT

When solving linear equations, students often make these mistakes:

1. Forget to apply operations to both sides:

Remember, whatever operation you perform on one side of the equation must be done to the other side as well.

2. Sign errors when moving terms:

When moving a term from one side to another, you need to change its sign (addition becomes subtraction and vice versa).

Incorrect: $x + 5 = 10$ → $x = 10 + 5$
Correct: $x + 5 = 10$ → $x = 10 - 5$

3. Distribute incorrectly:

When applying the distributive property, make sure to multiply each term inside the parentheses.

Incorrect: $3 (x + 2) = 3 x + 2$
Correct: $3 (x + 2) = 3 x + 6$

4. Divide by the coefficient incorrectly:

When isolating the variable, divide all terms on both sides by the coefficient.

Incorrect: $4 x + 2 = 14$ → $x + 2 = 14/4$ or $4 x + 2 = 14$ → $2 x + 2 = 14/2$
Correct: $4 x + 2 = 14$ → Divide both sides of the equation by 2: $2 x + 1 = 7$ → $2 x = 7 - 1$ → $2 x = 6$ → $x = 3$

5. Misread the question to ask for a value of $x$ or a formula containing $x$ :

Sometimes, the question will ask you to solve for the value of $x$ , but sometimes, it's asking what's the value of " $6 x + 2$ ", " $2 x$ ", etc. Read the question carefully and calculate the required value.

Quick Practice: Test Your Skills!

Question 1

If $3 (2 x - 4) = 18$ , what is the value of $x$ ?

A. $5$

B. $6$

C. $7$

D. $8$

Solution:

To solve the equation $3 (2 x - 4) = 18$ :

Divide both sides by 3:

$\frac{3 ( 2 x - 4 )}{3} = \frac{18}{3} ⟹ 2 x - 4 = 6$

Move 4 to the right side:

$2 x = 6 + 4 ⟹ 2 x = 10$

Divide both sides by 2:

$\frac{2 x}{2} = \frac{10}{2} ⟹ x = 5$

Final Answer: Option A

Question 2

How many solutions does the equation $\frac{x - 3}{2} = \frac{x + 5}{4}$ have?

A. Zero

B. Infinitely many

C. Exactly one

D. Exactly two

Solution:

To find the number of solutions for $\frac{x - 3}{2} = \frac{x + 5}{4}$ :

Cross-multiply:

$4 (x - 3) = 2 (x + 5)$

Expand:

$4 x - 12 = 2 x + 10$

Subtract $2 x$ :

$2 x - 12 = 10$

Move 12 to the right side:

$2 x = 10 + 12 ⟹ 2 x = 22$

Divide both sides by 2:

$x = 11$

Thus, the given equation has one unique answer, which matches Option C.

Question 3

Terry is a car salesperson. His total monthly income consists of a base salary of $ $2, 300$ and a bonus for each car he sells. If Terry's total monthly income is $ $5, 270$ and he sells $18$ cars, which equation can be used to calculate Terry's bonus $b$ per car in dollars?

A. $2, 300 + 18 + b = 5, 270$

B. $2, 300 + 18 b = 5, 270$

C. $5, 270 + 18 b = 2, 300$

D. $5, 270 + 18 + b = 2, 300$

Solution:

Understand the components of Terry's income:
- Base salary: $2,300
- Bonus: $b$ per car
- Total cars sold: 18
- Total Bonus: 18 $\times$ $b$
- Total income: $5,270
Set up the equation:

Total income = Base salary + (Bonus per car × Number of cars sold)

$5, 270 = 2, 300 + 18 b$
Match with the given options:

The correct equation is B $2, 300 + 18 b = 5, 270$ .

Answer: B

Question 4

A store sells both large and small packages of pens. The large package contains $60$ pens, which is $4$ less than $2$ times the number of pens in the small package. How many pens are in the small package?

Solution:

Let $s$ be the number of pens in the small package.
The problem states: 60 pens = 2 times the small package minus 4.
Translate this into an equation: $60 = 2 s - 4$
Add 4 to both sides: $60 + 4 = 2 s$
Simplify: $64 = 2 s ⟹ s = 32$

Answer: There are $32$ pens in the small package.

Question 5

$(p + 4) \times x + \frac{3}{2} = 0$

In the given equation, $p$ is a constant. If the equation has no solution, what is the value of $p$ ?

A. $- \frac{3}{2}$

B. $0$

C. $- \frac{1}{2}$

D. $- 4$

Solution:

A linear equation has no solution only if:
- The coefficient of $x$ is 0 (so the equation simplifies to a false statement).
- The constant term is non-zero (e.g., $0 = non-zero$ ).
Set the coefficient of $x$ to 0:

$p + 4 = 0 ⟹ p = - 4$
Check the constant term when $p = - 4$ :

$\frac{3}{2} \neq = 0$

This confirms the equation becomes $0 = \frac{3}{2}$ , which is unsolvable.

Answer: The correct value is Options D: p = -4

Your Turn! Realistic "Linear Equations in One Variable" Questions for DSAT Success

Question 1

Difficulty level: Easy

$\frac{j}{8} = k + 11 m$

The given equation relates the distinct positive numbers $j$ , $k$ , and $m$ . Which equation correctly expresses $j$ in terms of $k$ and $m$ ?

A. $j = k + 8 (11 m)$

B. $j = 8 (k + 11 m)$

C. $j = 8 k + 11 m$

D. $j = (k + 11 m) /8$

Question 2

Difficulty level: Medium

A chemist combines water and isopropanol to make a mixture with a volume of $52 milliliters (mL)$ . The volume of isopropanol in the mixture is $10 mL$ . What is the volume of water, in $mL$ , in the mixture? (Assume that the volume of the mixture is the sum of the volumes of water and isopropanol before they were mixed.)

Answer:________

Question 3

Difficulty level: Hard

$a (4 x - 8) + 5 a = 4 (a x - 2 a) - 20$

In the given equation, $a$ is a constant. The equation has infinitely many solutions. What are all possible values of $a$ ?

A. $- 4$ only

B. $0$ only

C. Any real number

D. No real number

"Linear Equations in One Variable" Learning Checklist

🔘 Know the two types of "Linear Equations in One Variable" Questions in Digital SAT:
- Multiple Choice (MCQ)
- Student-Produced Response ("grid-in")
🔘 Understand the "Constant rate of change" and "Proportionality" of Linear Relationships
🔘 Be able to recognize a "linear equation in one variable": $a x + b = 0$
🔘 Master 7 key points for solving "Linear Equations in One Variable" Questions.
- Moving Terms (Changing Signs)
- Handling Fractions
- Keeping $x$ on One Side
- $x$ on Both Sides
- Parentheses Around $x$ Terms
- Cross Multiplication
- Special Cases(No solution & Infinite solutions)
🔘 Master 11 common question types in real SAT: none-word questions & word questions
🔘 Know how to avoid 5 common mistakes of "Linear Equations in One Variable" Questions.

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Best SAT Math Prep: Linear Equations in One Variable

Frontier Lesson: "Linear Equations in One Variable" for the Digital SAT

Why This Skill Matters on the SAT

How the SAT Tests "Linear Equations in One Variable"

Understand "Linear Relationships"

1. Constant rate of change:

2. Proportionality:

Free Full Length SAT Tests withOfficial-StyleQuestions

What are "Linear Equations in One Variable"?

Key Points for Solving "Linear Equations in One Variable" Questions

1. Moving Terms (Changing Signs)

2. Handling Fractions

3. Keeping x on One Side

4. x on Both Sides

5. Parentheses Around x Terms

6. Cross Multiplication

7. Special Cases

Common SAT "Linear Equations in One Variable" Question Types

Example 1. Solving for a Single Variable

Example 2. Substitute to Find Another Expression

Example 3. Equations with Equivalent Representations

Example 4. Identifying the Equation that Has the Same Solution

Example 5. Number of Solutions Determination

Example 6. Variables Representing Constants

Example 7. Word Problems Leading to Equations

Example 8. Geometric Situations Leading to Equations

Example 9. Real-World Context with Proportions and Rates

Example 10. Interpreting the Components of an Equation

Example 11. Building Equations from Word Problems

Free Full Length SAT Tests withOfficial-StyleQuestions

Common Mistakes in Real Digital SAT

1. Forget to apply operations to both sides:

2. Sign errors when moving terms:

3. Distribute incorrectly:

4. Divide by the coefficient incorrectly:

5. Misread the question to ask for a value of x or a formula containing x:

Quick Practice: Test Your Skills!

Question 1

Question 2

Question 3

Question 4

Question 5

Your Turn! Realistic "Linear Equations in One Variable" Questions for DSAT Success

Question 1

Question 2

Question 3

"Linear Equations in One Variable" Learning Checklist

Essential SAT Prep Tools

Personalized Study Planner

Expert-Curated Question Bank

Smart Flashcards

Score Calculator

SAT Skills Lessons

Full-Length Practice Tests

Pro Tip

5. Misread the question to ask for a value of $x$ or a formula containing $x$ :