Before starting this unit, and the rest of your sat prep, there is an INCREDIBLY important skill you must possess: isolating variables and numbers.
This is commonly taught in either pre-algebra or algebra 1 classes. If you need a refresher or you're starting from the beginning, these videos do a great job covering the skill of isolating variables.
Linear equations | Khan Academy
Rearrange formulas to isolate specific variables | Khan Academy
Isolating Variables | SAT Math
A linear relationship is a constant relationship between any two variables that creates a line when graphed. Let's look at an example to make that make a bit more sense
Let's say Jackson wants to rent out a kayak which costs $20 an hour. The total cost that Jackson has to pay will then be
\[\text{Cost} = \text{Hourly rate} \times \text{amount of hours kayaking}\]
We can represent the amount of hours kayaking using a variable, such as h, and the total cost as another variable, such as C.
Replacing our text with variables, we get the linear equation
\[
c = 20h
\]
Where 20h is the total amount that Jackson has to pay.
In another town, if Jackson wants to rent a kayak he has to pay an initial fee of $30 and an hourly rate of $10. No matter the duration that Jackson decides to rent the kayak, he has to pay the fee of $30 dollars. This can now be written as
\[
\text{Cost} = (\text{Hourly rate} \times \text{amount of hours}) + \text{initial fee}
\]
If we use the same variables as earlier, the new equation becomes
\[
c = 10h + 30
\]
Where 10h is the amount that Jackson spends every hour, and 30 is the initial fee he has to pay.
Let's go over some real questions you’ll see on the test.
It costs Sean 11 dollars every day to rent out the tent, meaning that the amount Sean has to pay every day is
\[
11 \times \text{amount of days}
\]
The variable given for days is d, So we can simplify this part of the equation as \(11d\)
On top of the daily fee, Sean must pay a one-time fee of $10. Since he only has to pay it once, there’s no variable attached to it, so we can add it to the end of our equation to get
\[
11d + 10
\]
Meaning choice C is the correct answer.
This question is a bit different as we’re already given the total cost. Though, this doesn't change anything. We’re simply replacing our variable with a number
John has to pay \(p\) monthly payments of 16 dollars each, meaning the amount spent on monthly payments is \(16p\). He is also paying a one-time down payment of 37 dollars. So the total cost would be equal to \(16p +37\).
Since were already given the total cost, our full equation comes out to be
\[
16p + 37 = 165
\]
Making choice C the correct answer.
While many questions will simply ask you to write a linear equation, some questions require you to write AND solve for a number. Refer to the example below.
We can start by writing our equation. The website costs us d dollars each month plus an initial 350 dollar fee, giving us...
\[
C = dm + 350
\]
Where \(C\) is total cost, \(d\) is the amount of dollars per month, and \(m\) is the amount of months.
Our final objective is to solve for \(d\), so plug in all inormation given and solve for \(d\)
\[C = dm + 350\]
\[1010 = d(12) + 350\]
\[660 = d(12)\]
\[55 = d\]
Making choice D the correct answer.
Whether you’ve noticed or not, all the equations we’ve written are in slope-intercept form
\[ y=mx+b\]
Where \(m\) is the slope and \(b\) is the y-intercept.
Slope is the name given to a number that dictates how steep a line is. Commonly, the slope is found/referred to as rise/run. Take a look at the example below
\[
\text{Slope} = m = \frac{\text{rise}}{\text{run}}
\]
This means that slope is defined as the distance in the y-direction over the distance in the x-direction of 2 points.
Let's choose \( (1, 2) \) and \( (2, 4) \)
To find the slope of the graph you can count how far the points are from each other in the y-direction first
Starting at \( y = 2 \)and counting up to the next point, the distance between the two is 2 units, meaning our rise is 2.
Counting over from our stopping point, the distance between the points in the x-direction is 1 unit, meaning our run is 1. Putting this together we get a slope of
\[
m = \frac{\text{rise}}{\text{run}} = \frac{2}{1} = 2
\]
NOTE: always count from left to right
The slope will not always be positive. A negative slope indicates a negative association between \(x\) and \(y\), and has a downward trend going from left to right.
When counting from left to right, the rise ends up being negative, meaning we count down 2 and move left 1, making our slope - \( \frac{2}{1} \) or simply \( -2 \)
Watch this for more on finding slope from a graph.
Two parallel lines are lines that never intersect
Two perpendicular lines are lines that create a 90-degree angle when they intersect
In terms of slope-intercept form, two parallel lines share the same exact slope. Since they have the exact rise/run, they will never run into each other.
The rule for perpendicular lines may be a bit harder to remember, but is still relatively simple.
Two perpendicular lines, the slope of each line is the negative reciprocal of the other line.
The reciprocal is found by swapping the numerator and denominator; to then find the negative value, multiply the reciprocal by -1.
The negative reciprocal of \( \frac{3}{2} \)would be found by first swapping the numerator and denominator, giving us \( \frac{2}{3} \), then multiplying it by \(-1\), giving us a final answer of -\( \frac{2}{3} \) This means that two lines with slope of \( \frac{3}{2} \) and - \( \frac{2}{3} \)are perpendicular.
NOTE: the reciprocal of any whole number is 1 over the given number.
The reciprocal of 2 is \( \frac{1}{2} \)
The reciprocal of 4 is \( \frac{1}{4} \)
Oftentimes, you won't actually be given a graph of a line, instead you’ll be asked to find slope using either two points or a table.
The equation to find slope is
\[m = \frac{y_2 - y_1}{x_2 - x_1}\]
Where \( (y_1) \)is the y-value of the first point and \( (y_1) \) is the y-value for the second point, same for the x-values.
Let's try to find the slope between the two points \( (5, 3) \) and \( (7, 6) \). Ill assign \( x_1 \) and \( y_1 \)to \( (5, 3) \) and \( x_2 \)and \( y_2 \) to \( (7, 6) \)
Plug each value into the equation to get slope
\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 3}{7 - 5} = \frac{3}{2}\]
NOTE: Whenever finding the slope between two points, I highly recommend labeling each number either \( (x_1, x_2, y_1, and y_2) \) before you plug it into the equation. Many students will often plug in the numbers in the wrong order which ends up giving a false slope.
Watch this for more. on finding the slope from two points.
This can all be done using Desmos too! check out our Desmos course to learn how.
The y-intercept is where a graph intercepts the y-axis, which can also be defined as when \( x \)is equal to zero
Take a look at the graph of the equation \( y= 2x + 3 \)
In the graph above, the line crosses through the y-axis at \((0,3)\), making 3 the y-intercept. This is because when 0 is plugged into x in the original equation, the y-value is 3.
In slope intercept form, the y-intercept is defined by the constant b
\( y = x + b \)
Let's look at a couple of real SAT problems
NOTE: in this case, you can treat n as the x-values, and \( f(n) \) as y-values - check out our functions page for more explanation on the notation.
The first step to any one of these questions is finding the slope. You can choose any two points, for this example ill choose \( (2, 1) \) and \( (3,4) \)
2 will be by \( x_1 \)value
1 will be my \( y_1 \) value
3 will be my \( x_2 \) value
4 will be my \( y_2 \)value
You could now plug these into the slope equation.
\[
\frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 1}{3 - 2} = \frac{3}{1} = 3
\]
Giving us a slope of 3.
Once you have the slope though, how do you find the y-intercept? Many tables won't actually give you the value for when x equals zero, that's too easy. What you have to do is plug in your slope, and an \( x \) & \( y \) value.
The slope-intercept equation is \( y = mx + b \). We have the slope, so we can substitute that in for \( m \), giving us a new equation of \( y = 3x + b \). To find the \( b \) value, simply choose any point on the line and plug in the \( (x, y) \) values for \( x \) and \( y \) in the equation. I’ll choose \( (3, 1) \) to do this
\[ y = 3x + b \]
\[1 = 3(2) + b\]
\[ 1 = 6 + b \]
\[ -5 = b \]
Giving us a \( b \) value of -5. Putting it all together into an equation we get \( y = 3x -5 \), making answer choice C correct.
NOTE: if the slopes of each answer choice are different, you wouldn't actually have to find the y-intercept. In this particular question you could have found that the slope is 3 and immediately choose answer choice C, as its the only choice with a slope of 3.
A large subset of linear equation questions involve word problems, where you have to understand what each part of an equation represents in real life.
The y-intercept (b-value) is often called the initial value. This is your ‘starting point’ in any real-life situation. If my bank account starts off with $100, and I make $20 a day, then my y-intercept/initial value would be 100: it's where I started at day zero.
The slope (m-value) is often called the rate of change. It's how quickly something changes, usually over a set amount of time. If I make 20 dollars per day, then my rate of change is 20 dollars/1 day, or a slope of 20.
Before even looking at the answer choices, you can identify that 331.4 is the y-intercept/initial value. This means that the speed of sound is 331.4 when \(T\) is equal to zero. Now scan all the answers to see which one lines up the best.
Both C and D refer to slope, as it refers to the increase/decrease of something.
Looking at both A and B, A is the answer choice that defines it as the speed when T is equal to 0, making choice A the correct answer.
The equation models the number of students per classroom starting from the year 2000, meaning that during the year 2000, there were approximately 27.2 students per room (initial value), and each year it increases by an average number of 0.56. Since the slope is associated with an increase of a number, 0.56 will be the slope.
We can now scan the answer choices to see which one matches up with this definition.
A says that it would be the total number of students in the school in 2000. The equation gives us the number of students per classroom, not the whole school.
B, The average number of students per classroom in 2000 would be the y-intercept of the equation, since 2000 is our starting point/when x equals zero. This is not what we are looking for.
This leads us to C, the estimated increase in the average number of students per classroom each year. Everything here matches up to our previous definition, making choice C the correct answer.
Most problems that will test you on standard form are word problems, similar to the ones above. The concepts and ways of writing these equations is incredibly similar to slope-intercept, and most times its even easier due to the way information is presented
The standard form of a linear equation is
\[Ax + By = C\]
For example, Let’s say William opens a fruit stand. He sells \(x\) apples for $2 each, and \(y\) pears for $3 each. By the end of the day, William wants to make $46. How would you write an equation for this scenario?
Since he sells x apples for $2 each, the total amount of money he makes off of apples will be \(2x\).
Since he sells y pears for $3 each, the total amount of money he makes off of pears will be \(3y\).
This means the total amount of money made off both will end up being \( 2x + 3y \). Since we want the total amount of money made to be $46, we have to set this expression equal to 46, giving us a final equation of \( 2x + 3y = 46 \)
We’re firstly given that the amount of filler in Fertilizer A contains 60% filler. 60% is equivalent to 0.6 (if you're confused on how we got to that, check out the percentages page) The total amount of filler in fertilizer A would then be \(0.6x\).
B contains 40% filler, or 0.4, so our total amount of filler within Fertilizer B is 0.4y
We know that the sum will equal our total amount of filler in both fertilizers, which is 240 pounds. So our final equation will be
\[ 0.6x + 0.4y = 240 \]
Making choice B the correct answer.
Not all standard-form questions will be word problems though. Sometimes you will be given a graph and asked which standard form equation best represents the line, or be asked to find a parallel/perpendicular line to another given equation. In this case, The best thing you could do is convert the standard form equations into slope-intercept form, as slope-intercept directly tells you both the slope of the line and the y-intercept, making it much easier to match up with a graph or compare slopes.
Converting from standard to slope-intercept form is simply isolating \(y\). Take the equation
\[ 3x + 2y = 14 \]
In order to convert to slope-intercept, get y by itself on one side, and everything on the other.
\[ 3x + 2y = 14 \]
\[ 2y = -3x + 14 \]
\[y = -\frac{3}{2}x + 7\]
Watch this for more on converting from standard form to slope-intercept form.
As we went over earlier, for two lines to be parallel, they must share the same slope. So, we have to find which one of these equations has a slope of -3.
Starting of with answer choice A, when converted into slope-intercept form we get
\[ 6x + 2y = 15 \]
\[ 2y = -6x +15 \]
\[ y = -3x + \frac{15}{2} \]
This equation has a slope of negative -3, making it parallel to the original like, meaning choice A is the correct answer.
The best way to check your answer is to simply plot the original given equation, and your answer choice in standard form into desmos, if the lines never intersect you are correct.
While you could plot every single line in Desmos and see which are parallel, it's almost always more efficient to manually find the slope of each equation, and if you haven' noticed already, there's actually a shortcut to finding slope from a standard form equation.
In any equation in standard form \(Ax + By = C\), the slope of that line is \( \frac{-A}{B} \).
If given simple enough numbers, finding the slope of all 4 answer choices could be done within less than 10 seconds.
For an example of writing linear equations from word problems
For more on lines and their equations
For an overall review of the lesson