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# SAT Math: Linear Equations 📏

### SAT🎓

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## SAT Math: Linear Equations 📏

### Overview

The SAT Math section contains approximately 60 questions, separated into:
• No-Calculator Section
• 15 Multiple Choice
• 5 Grid-Ins
• 25 Minutes
• Calculator Section
• 30 Multiple Choice
• 8 Grid-Ins
• 55 Minutes
• Total Time: 80 Minutes
These questions (both with-calculator and no-calculator) test on four major content areas: the Heart of Algebra, Problem Solving, Data Analysis, and the Passport to Advanced Math.

## Resources:

Linear Equations fall under the Heart of Algebra content section.

Topics Related to “Linear Equations”

• Systems of Linear Equations
• Linear Functions
• Linear Inequalities
• Graphical & Algebraic Manipulation
• Linear & Algebraic Reasoning

What You Have to Do

• Convey a thorough understanding of critical thinking, and multiple core concepts in Algebra, specifically:
• Linear Equations
• Systems of Linear Equations
• Linear Functions
• Demonstrate that you can convert between graphical representations of linear equations and their algebraic representations (and vice versa)
• Analyze word problems to manipulate and create linear equations to solve for different variables
• Be able to define and isolate key variables
• Be able to write expressions, equations, and inequalities
• Be able to solve linear equations and interpret the solution to apply it to the context of the problem

### Sample Problem Section 📚

Non-calculator practice problems 🚫🖩

General directions for this section you should expect to see:
• The use of a calculator is not permitted.
• All variables and expressions used represent real numbers unless otherwise indicated.
• Figures provided in this test are drawn to scale unless otherwise indicated.
• All figures lie in a plane unless otherwise indicated.

1. In 2014, County X had 783 miles of paved roads. Starting in 2015, the county has been building 8 miles of new paved roads each year. At this rate, how many miles of paved road will County X have in 2030? (Assume that no paved roads go out of service.)
1. 546
2. 911
3. 780
4. 612
• Let’s break down this linear equation-based word problem and start with what we know!
• Our first step is to conclude which variables we need to define (also referred to as the variable we need to solve for!). Since this problem asks how many miles of paved road the county will have in 2030, our unknown variable will be the number of years after 2014!
• Our second step is to write our linear equation.
Let’s Break it Down!
⇒ Each year after 2014 (where the number of miles of paved roads has already been determined) can be represented with the variable x.
⇒ Since we already have 783 miles, the number of miles of paved roads in 2030 can be represented by 783 + 8*x, but what is x?
⇒ Glad you asked! Since x is the number of years after 2014, a quick calculation of 2030 (the year of interest) - 2014 (the current year), tells us that x = 16.
⇒ From here, we can plug this value into our linear expression!
⇒ 783 + 8(16) = 783 + 128 = 911 miles of paved roads in 2030.

Feeling good? Let’s try another one!
2. In 2014, County X had 783 miles of paved roads. Starting in 2015, the county has been building 8 miles of new paved roads each year. At this rate, in which year will County X have at least 1,000 miles of paved roads? (Assume that no paved roads go out of service.)
1. 2025
2. 2058
3. 2042
4. 2029

AP College Board SAT Practice Problems

• Let’s break down this linear equation-based word problem and start off with what we know!
• Before we get started, let’s recognize a keyword-phrase hint: “At least.”
• Typically, the phrase “at least” indicates that you’ll be utilizing an inequality rather than an expression or equation.
• This question tests your ability to write linear expressions, differentiate between equations and inequalities, and translate word problems into equations.
• Just like in Equation 1, let’s let x represent the number of years after 2014.
⇒ Since we start with 783 miles, x is the number of years after 2014, and the rate of paving is 8 miles a year, our expression can be represented as 783 + 8n
⇒ This question asks when there'll be at least 1,000 miles of total paved roads in County X (not when an additional 1,000 miles will be paved).
⇒ Our inequality can be written as 783 + 8n >= 1000
⇒ Why isn’t it <, <=, or >? We aren’t looking for when the amount of total miles will be less than 1,000, and at least indicates that we can include 1,000; therefore it’s >=
⇒ 783 + 8n >=1000
⇒ 8n >= 217
⇒ n >= 27.125
⇒ Once we add 27.125 to 2014, we get 2041.125, which does not make sense in the context of the question. In the year 2041, only 999 miles of paved road exist in the county. Therefore, we need to round up to 2042 to have at least 1,000 miles of paved roads total!
Onto Question 3!
3. To edit a manuscript, Miguel charges \$50 for the first 2 hours and \$20 per hour after the first 2 hours. Which of the following expresses the amount, C, in dollars, Miguel charges if it takes him x hours to edit a manuscript, where x>2?
1. C = 20x
2. C = 20x + 10
3. C = 20x + 50
4. C = 20x + 90

AP College Board SAT Practice Problems

• Let’s break down this linear equation-based word problem and start with what we know!
• Miguel charges \$50 for the first 2 hours and \$20 only after the first 2 hours.
• Therefore, the remaining time can be represented by x-2 (since the first two hours are \$50)
• Therefore, our equation is C = 50 + 20(x-2)
⇒ C = 50 + 20(x-2)
⇒ C = 50 + 20x - 40
⇒ C = 10 + 20x
⇒ C = 20x + 10, which is option

Let’s try one more!
1. -2x = 4y + 6
2(2y + 3) = 3x - 5
What is the solution (x,y) to the system of equations above?
1. (1,2)
2. (1,-2)
3. (-1,-1)
4. (-1,1)

AP College Board SAT Practice Problems

• Let’s break down this system of linear equations and start off with what we know!
• This system of linear equations can be solved via substitution.
• Substitution is a method of solutions through isolating for one variable and plugging that answer into another equation
• Let’s start off with -2x = 4y + 6
⇒ -2x = 4y + 6 ⇒ let’s simplify this and divide all terms by 2!
⇒ -x = 2y + 3
⇒ x = -2y - 3
⇒ Now that we’ve isolated for x, we can plug this into the second equation and solve!
⇒ 2(2y + 3) = 3x - 5
⇒ 2(2y + 3) = 3(-2y - 3) - 5 ⇒ Since x = -2y - 3, we plugged in -2y - 3 in place of x
⇒ 4y + 6 = -6y - 9 - 5
⇒ 4y + 6 = -6y - 14
⇒ 10y = - 20
⇒ y = -2
⇒ Now that we’ve solved for y, we can plug this back into our first equation!
⇒ -2x = 4y + 6
⇒ -2x = 4(-2) + 6
⇒ -2x = -8 + 6
⇒ -2x = -2
⇒ x = 1
⇒ Since x = 1 and y = -2, our answer is (1, -2)
TIP: Always plug your answer back into either the first or second equation (or both) to make sure they work! If your solution is correct, both sides of the equation should be equal
Excellent work!

### Closing 🤩

Congratulations! You’ve made it to the end of this prep activity 🙌 You learned about “SAT Math: Linear Equations ” You should have a better understanding of the Math sections for the SAT© , topic highlights, what you will have to be able to do in order to succeed, as well as have seen some practice questions that put the concepts in action. Good luck studying for the SAT Math section 👏
Need more resources? Check out our complete SAT Math Study Guide w/ Practice Problems. Pressed on time? Access our SAT cram sessions and watch the Night Before the SAT cram session. You got this 🥳.

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