Best Digital SAT Math Prep: "Systems of Two Linear Equations in Two Variables"

On the Digital SAT, questions about "Systems of Two Linear Equations in Two Variables" fall under the "Algebra" content domain.

You'll encounter these problems in two formats: multiple-choice questions and student-produced responses (grid-in questions). The difficulty of this specific concept doesn't noticeably lean towards any one level of easy, medium, or hard. The good news is that it's a relatively straightforward skill, and with a strong grasp of the methods, you can definitely find these questions to be manageable, offering a great opportunity to rack up points!

The SAT tests this topic in a combined way, building on earlier concepts related to linear equations. The methods and thought processes you rely on here—such as substituting, eliminating, and analyzing the relationships between lines—aren't particularly "new." With a bit of practice, success is within reach.

🎯 The SAT often tests this concept in three main ways:

1. Solving the system algebraically using substitution or elimination.

2. Determining the number of solutions—one, none, or infinitely many—both algebraically and graphically.

3. Applying systems of equations to word problems that are connected to real-world scenarios.

In this part, we will learn how to solve systems of two linear equations with two variables. A "system of equations" is a set of equations that we solve simultaneously to find values for the variables that satisfy all the equations in the system. There are two commonly used methods for solving these systems: Substitution and Elimination. Let's explore them step by step.

📚 What is a System of Equations?

In Digital SAT, a system of equations consists of two linear equations. Each equation typically contains one or two variables. In the context of a system of two linear equations with two variables, we are solving for values of two variables, such as $x$ and $y$, that make both equations true at the same time.

For example:

Here, $x$ and $y$ are the variables. Our goal is to determine the specific values of $x$ and $y$ that satisfy both equations.

💡 Note: A single equation on its own, like $-2x+5y=65$, is not sufficient to determine the exact values of $x$ and $y$. Multiple solutions exist if you only solve one equation.

📚 Substitution Method

The Substitution Method involves isolating one variable in one equation and substituting its expression into the other equation. This method works especially well when one of the equations is already solved for one variable in terms of the other, or can easily be rearranged to this form.

➥ Steps to Solve Using Substitution:

#1. Solve for one variable in one of the equations.

• Rearrange one equation to express one variable (e.g., $x$) in terms of the other (e.g., $y$).

#2. Substitute this expression into the other equation.

• Replace the variable in the second equation with its equivalent expression from Step 1.

• This will reduce the system to a single equation with one variable.

#3. Solve for the variable.

#4. Substitute back the value found in Step 3 into the expression from Step 1 to find the value of the other variable.

➥ When to Use Substitution:

• One of the equations is already in the form $x=$ or $y=$, or can be easily rearranged into this form.

• The coefficients of one variable make isolation easy.

➥ See an Example:

Solve the following system using Substitution:

Step 1: Solve for one variable (already done in Equation 1: $y=2x+1$).

Step 2: Substitute $y=2x+1$ into Equation 2.

Step 3: Simplify and solve for $x$.

Step 4: Substitute $x=2$ into Equation 1 to find $y$.

Final Answer: $x=2,y=5$

📚 Elimination Method

The Elimination Method involves adding or subtracting the equations to eliminate one of the variables. This method is particularly useful when the coefficients of one variable are the same (or can be made the same through multiplication).

➥ Steps to Solve Using Elimination:

#1. Align the equations so that like terms (e.g., $x$, $y$, and constants) are in columns.

#2. Multiply one or both equations (if necessary) so that the coefficients of one variable are identical or opposites.

#3. Add or subtract the equations to eliminate the chosen variable.

• Addition eliminates a variable when the coefficients are opposites (e.g., $3x$ and $-3x$).

• Subtraction eliminates a variable when the coefficients are the same (e.g., $4y-4y=0$).

#4. Solve for the remaining variable.

#5. Substitute the found value into one of the original equations to find the value of the other variable.

➥ When to Use Elimination:

• The variables have coefficients that are already the same, opposite, or can easily be made so.

• Substitution would be messy or overly complicated.

➥ See an Example:

Solve the following system using Elimination:

Step 1: Align the equations. (already aligned)

Step 2: Add the equations to eliminate $y$.

Step 3: Solve for $x$.

Step 4: Substitute $x=3$ into one of the original equations to find $y$.

Using $2x+y=7$:

Final Answer: $x=3,y=1$

By practicing both methods (Substitution and Elimination), you can choose the one that's most efficient depending on the structure of the system of equations. Identifying which method to use and applying it correctly can make you solve in a quick and wise way!

For systems of two linear equations in two variables, there are only three possibilities for the number of solutions on the Digital SAT: one solution, no solution, or infinitely many solutions. Understanding these scenarios both graphically and algebraically is crucial for solving these problems efficiently.

📚 One Solution

➥ Graphically:

If there is one solution, it means that the two lines represented by the equations intersect at exactly one point on the graph. That intersection point is the solution to the system because it satisfies both equations.

➥ Algebraically:

• This happens when the two lines have different slopes. The $y$-intercepts don't matter in this case—differing slopes guarantee that the lines will meet.

• You can determine if there is one solution by first rewriting each equation in slope-intercept form $(y=mx+b)$, where $m$ is the slope and $b$ is the y-intercept. If the slopes ($m$) are different, the system has exactly one solution.

➥ Example:

$2x+3y=5$

$2x-y=-2$

How many solutions does the given system of equations have?

Step 1: Rewrite both equations in slope-intercept form.

1. $2x+3y=5⟹y=-\frac{2}{3}x+\frac{5}{3}$ (Slope $m=-\frac{2}{3}$)

2. $2x-y=-2⟹y=2x+2$ (Slope $m=2$)

Step 2: Compare the slopes.

• Equation 1 has a slope of $-\frac{2}{3}$, and Equation 2 has a slope of $2$.- Since the slopes are different, the lines intersect at one point, so there is exactly one solution.

➥ Some Graphic Examples:

📚 No Solution

➥ Graphically:

If the system has no solution, it means that the two lines are parallel and will never intersect.

➥ Algebraically:

When the two equations are written in slope-intercept form, if the slopes ($m$) are equal, but the y-intercepts ($b$) are different, the system has no solution.

➥ Example:

$2x+y=4$

$px-y=10$

The given system of equations has no solution, and $p$ is a constant. What's the value of $p$?

Step 1: Rewrite both equations in slope-intercept form.

1. $2x+y=4⟹y=-2x+4$ (Slope $m=-2$, $y$-intercept $b=4$)

2. $px-y=10⟹y=px-10$ (Slope $m=p$, y-intercept $b=-10$)

Step 2: Compare the slopes and y-intercepts.

• Sine there is no solution, the lines are parallel, which means the two equations have the same slope but different $y$-intercepts.

• Therefore, $p=-2$.

➥ Some Graphic Examples:

📚 Infinitely Many Solutions

➥ Graphically:

If the system has infinitely many solutions, it means that the two lines are actually the same line. This means every point on one line is also a point on the other line.

➥ Algebraically:

• When both equations are rewritten in slope-intercept form, if the slopes ($m$) and the $y$-intercepts ($b$) are identical, the system has infinitely many solutions.

• While the equations might initially look different, you can transform one into the other by simplifying or rearranging.

➥ Example:

$2x+y=6$

$4x=12-2y$

How many solutions does the given system of equations have?

Step 1: Rewrite both equations in slope-intercept form.

1. $2x+y=6⟹y=-2x+6$ (Slope $m=-2$, $y$-intercept $b=6$)

2. $4x=12-2y⟹2y=-4x+12⟹y=-2x+6$ (Slope $m=-2$, $y$-intercept $b=6$)

Step 2: Compare the slopes and $y$-intercepts.

• Both equations have the same slope ($-2$) and the same $y$-intercept ($6$).

• Therefore, the lines are identical, and the system has infinitely many solutions.

➥ Some Graphic Examples:

💡 Tips for Success

1. Always rewrite both equations in slope-intercept form ($y=mx+b$) to easily compare slopes and $y$-intercepts.

2. When solving algebraically, you can double-check your math by substituting the solution back into both original equations.

3. Memorize the summary table:

On the Digital SAT, word problems involving linear relationships may seem intimidating at first, but the process for solving them is always the same. The challenge lies in translating the word problems into mathematical equations.

There are three essential steps: identify the variables, build the equations, and solve the system of equations. Let's break it down.

📚 #1: Identify the Variables

The first step in solving any word problem is identifying what the variables represent. This requires basic reading comprehension skills to determine which parts of the problem are "unknowns" that can change. Variables typically represent quantities like quantities, prices, weights, temperatures, hours worked, distances traveled, etc. These are the elements you are solving for.

It's a good idea to use the highlight tool in the Bluebook testing platform to underline or mark the key pieces of information that will point to your variables. Moreover, since the SAT consistently tests systems of equations with two equations, you can confidently expect two variables in these problems.

Example 1: Identifying Variables

"A farmer sells apples and oranges at a market. Apples cost $3 per pound and oranges cost $2 per pound. On a certain day, the farmer sold a total of 50 pounds of apples and oranges combined for $120. How many pounds of each fruit did the farmer sell?"

Solution:

You need to identify the variables:

• Let $a$ = the number of pounds of apples sold.

• Let $o$ = the number of pounds of oranges sold.

Choosing clear labels for your variables helps keep track of them during the problem-solving process.

📚 #2: Build the Equations

The next step is translating the information in the problem into useful equations. Start by identifying totals or "sums" mentioned in the problem. These sums often relate to:

• Total quantity (e.g., pounds, number of items),

• Total price or total cost, or

• Other totals like time, distance, or amounts.

In addition, you can also explore other relationships between the variables, such as quantities, multiples, and so on. Once you've identified the relationships, use relationships like "add", "subtract" or "multiply" to construct the equations. Remember, since the problem involves a system of two equations, you will always have enough information to build two separate equations with your variables.

Example 2: Building Equations

Using the previous example:

• Total pounds equation (quantity):

This equation represents the total number of pounds sold, which is the sum of the pounds of apples and oranges.

• Total cost equation (price):

This equation represents the total cost, where $3a$ is the cost of apples sold ($3 per pound times the pounds of apples) and $2o$ is the cost of oranges sold ($2 per pound times the pounds of oranges).

Now you have the system of equations:

📚 #3: Solve the System of Equations

After establishing the system of equations, use either the Substitution or Elimination method to solve. Choose the method that is easiest based on the structure of the equations.

Example 3: Solving with Substitution

1. Solve for one variable in one equation:

From $a+o=50$, solve for $o$:

2. Substitute $o=50-a$ into the second equation:

3. Simplify:

4. Substitute $a=20$ back into $o=50-a$:

Answer:

The farmer sold $20$ pounds of apples and $30$ pounds of oranges.

🌟 Summary Tips for Solving Word Problems

• Define your variables clearly: highlight what each variable represents in the given text.

• Read the question carefully to ensure you understand the given relationships (totals, costs, etc.).

• Take your time setting up equations correctly. This is the most crucial step; if your equations are wrong, the solution will be wrong too.

• Practice both substitution and elimination methods so you can decide which is faster depending on the problem.

The SAT is designed to test your precision and problem-solving skills, so questions involving systems of two linear equations often include subtle traps. Falling into these traps can result in easily avoidable mistakes, but recognizing them ahead of time will set you up for success. Below are common pitfalls to watch out for, along with tips and examples to help you avoid them.

⚠️ Trap 1: The Question Asks for Another Expression, Not Just $x$ or $y$

One of the most common traps on the SAT is being asked for a combination of variables (e.g., $x+y$, $3x+3y$) instead of the individual values of a variable. If you forget to calculate the specific expression requested, you might stop too early or choose the wrong answer.

Example Trap:

The question asks: What is the value of $2x+y$?

Mistake: Many students stop after finding $x=5$ and $y=-2$ and think they're done. They choose the wrong answer because they don't calculate $2x+y$.

Tip: Always double-check what the question is asking for, and you can highlight it with the highlighter tool provided by Bluebook.

⚠️ Trap 2: Confusing "No Solution" with "Infinite Solutions"

The SAT frequently tests your understanding of when a system has no solution (parallel lines) versus infinite solutions (identical lines). The mistake many students make is failing to consider $y$-intercept after their same slope.

Example Trap:

Determine how many solutions does the given system have:

Mistake: Students recognize the slopes are the same but ignore or miscalculate the $y$-intercept, concluding there are no solution instead of infinite solutions. In this case:

• The first equation has slope $-\frac{3}{2}$ and intercept $3$.

• The second equation has slope $-\frac{3}{2}$ and the same intercept $3$, so the lines are identical, i.e. infinitely many solutions.

Tip:

• Write both equations in slope-intercept form ($y=mx+b$) when comparing slopes and $y$-intercepts.

• Parallel lines = Same slope + Different $y$-intercepts = No solutions.

• Identical/overlapping lines = Same slope + same $y$-intercepts = Infinitely many solutions.

⚠️ Trap 3: Misreading the Graph or Coordinate Axes

When students look at a line graph, they might overlook the scale intervals on the axes (in many cases, one unit on the graph represents 2 instead of 1). Additionally, they might mix up the values on the x-axis and y-axis, , which can lead to errors in identifying points of intersection.

Example Trap:

The graph of a system of equations has shown. We can see the coordinates of the intersection point are 3 grids on the $x$-axis and 4 grids on the $y$-axis. However, the scale of $x$-axis and $y$-axis is 2 units per tick mark.

Mistake: Students may misread the graph and assume the intersection is: $(3,4)$, $(7,8)$ or $(8,6)$ without checking carefully. The correct intersection is actually $(6,8)$.

Tip:

• Carefully check the scale of the axes on graphs, especially when the tick marks don't represent units of 1.

• For convenience, it is recommended to read the value of the $x$-axis first, followed by the value of the $y$-axis.

⚠️ Trap 4: Misinterpreting Variables in Word Problems

In word problems, students sometimes choose incorrect or unclear variables, leading to mistakes in writing the equations.

Example Trap:

A zoo ticket costs $$10$ for adults and $$6$ for children. On a certain day, the total revenue was $$2,680$, and the total number of tickets sold was $300$. How many adult tickets were sold?

Mistake:

After assigning the unknowns as the variables ($x$ and $y$), students may forget:

• $x=$ adult tickets or child tickets?

• $y=$ revenue or number of child tickets?

This confusion results in incorrect equations.

Tip:

• Define your variables in words before setting up equations. For example: $a=\text{number of adult tickets},c=\text{number of child tickets}$.

• Use units (e.g., dollars, tickets) to make equations clear.

• Use highlights of different color to mark if the given text has already defined the variables.

⚠️ Trap 5: Failing to Recognize Shortcut Opportunities

Some questions in real SAT are all well-structured in a way that allows for quick elimination, but students who try to solve everything "the long way" may waste valuable time.

Example Trap:

Given a system of linear equations:

Mistake: Students go through a full process of substitution or elimination, not realizing that the second equation is simply a multiple of the first. $6x+8y=60$ simplifies to $3x+4y=30$, meaning the lines are identical.

Tip: Simplify the equations first to save time and avoid unnecessary work.