The questions encountered on the SAT may not align with what you’ve encountered in your classroom or anticipate initially. They take familiar concepts and present them in an unfamiliar manner, aiming to increase the difficulty without necessarily raising the complexity beyond what can be expected from a typical high school student.
To tackle these questions effectively, it’s crucial to solidify your fundamental understanding of the concept. Once you have a firm grasp, you can then prepare for the common tricks and challenges that the SAT is known for.
In this article, we will delve into questions related to systems of equations. We’ll start by breaking down the core of this problem type, and then we’ll explore the unique twists and challenges that the SAT introduces in this context.
Systems of equations entail dealing with multiple variables spread across different equations. Typically, these equations are linear, avoiding the complexity of quadratics or exponentials. Solving these systems can be achieved through methods like substitution or elimination. If you were to graph these equations, solving the system boils down to identifying where the graphs intersect. Consequently, when you solve a system of equations, your answer often takes the form of one or more points.
The simplest way to solve a system of equations is by using a graph. If you have a straightforward graph of the equations, you can visually identify where the functions intersect. In the example below, it’s clear where the lines meet, and you can express your solution as the coordinate pair of (1, 3)
For the substitution method, you substitute one variable in one equation with its expression from another equation. This allows you to simplify the system to just one equation with one variable, making it easier to find the solution. In the example provided, you’ll replace one of the variables in a way that simplifies the equations, allowing you to easily determine the values that make the system work.
4x + 2y = 10
x - y = 13
We will use equation 2 since it is easier to isolate.
x = 13 + y
4(13 + y) + 2y = 10
52 + 4y + 2y = 10
52 + 6y = 10
6y = -42
y = -7
We will use equation 2 again for simplicity
x - (-7) = 13
x + 7 = 13
x = 6
(6, -7)
As for the elimination method, it involves getting rid of one variable by adding or subtracting the equations. This manipulation helps create a new equation with only one variable, making it simpler to find the solution. In the given example, you’ll strategically add or subtract the equations to eliminate one variable and make the system more manageable.
6x - 12y = 24
-x - 6y = 4
Here, we will manipulate the second equation by multiplying by -2. This will line up the y-coefficients to mirror each other. The resulting system is as follows:
6x - 12y = 24
2x + 12y = -8
Here, we can eliminate the y variable thanks to our efforts earlier. We will do this by adding the first and second equations vertically. The result is as follows:
8x + 0 = 16
8x = 16
x = 2
We will do this by plugging in our value to the second of the original equations.
-(2) - 6y = 4
-6y = 6
y = -1
(2, -1)
The SAT has many ways of making these straightforward problems difficult, we at Oahu Prep are here to support you in improving your skills and understanding how to solve these problems.
One common tactic is for the SAT to create a question which involves constants in place of numbers.
The SAT also loves to throw difficult to parse word problem in the mix, hoping you get tripped up on the language before ever getting to the math!
Instead of a system of equations, you may find a system of inequalities in your path!
With all your practice on solving systems of equations, you may see a system and run right through it. However the SAT can and will give you a system of equations and then not ask for the solution, but something related to the solution instead.
With that, you know what to expect on test day for systems of equations on the SAT. If you are having trouble with these or any other problem types, feel free to reach out to us for support at any time!