Mastering Systems of Equations for SAT Math
Introduction
Systems of equations are a frequent topic in the SAT Math section. Questions often require you to solve for variables using two or more equations. Understanding the substitution and elimination methods is key to solving these questions quickly and accurately.
This guide will cover:
- Substitution and elimination methods.
- Step-by-step examples for SAT questions.
- Common mistakes and tips for success.
What are Systems of Equations?
A system of equations involves two or more equations with two variables (x and y). The goal is to find the values of both variables that satisfy both equations.
Example:
\[
2x + y = 10 \quad \text{and} \quad x – y = 4
\]
Solving Using the Substitution Method
Step-by-Step Process:
- Isolate one variable in one of the equations.
- Substitute the isolated variable into the other equation.
- Solve for the remaining variable.
- Plug the value back to find the other variable.
Example:
Solve:
\[
x + y = 6 \quad \text{and} \quad 2x – y = 4
\]Solution:
- From the first equation: \( y = 6 – x \).
- Substitute \( y = 6 – x \) into the second equation: \( 2x – (6 – x) = 4 \). Simplify: \( 2x – 6 + x = 4 \implies 3x = 10 \implies x = \frac{10}{3} \).
- Plug \( x = \frac{10}{3} \) into \( y = 6 – x \): \( y = 6 – \frac{10}{3} = \frac{8}{3} \).
Solution: \( x = \frac{10}{3}, y = \frac{8}{3} \).
Solving Using the Elimination Method
Step-by-Step Process:
- Align the equations so they are in the same form.
- Add or subtract the equations to eliminate one variable.
- Solve for the remaining variable.
- Substitute back to find the other variable.
Example:
Solve:
\[
2x + y = 10 \quad \text{and} \quad x – y = 4
\]Solution:
- Add the equations to eliminate \( y \): \( (2x + y) + (x – y) = 10 + 4 \). Simplify: \( 3x = 14 \implies x = \frac{14}{3} \).
- Substitute \( x = \frac{14}{3} \) into \( x – y = 4 \): \( \frac{14}{3} – y = 4 \). Simplify: \( y = \frac{2}{3} \).
Solution: \( x = \frac{14}{3}, y = \frac{2}{3} \).
Practice Question
Question: Solve for \( x \) and \( y \):
\[
3x – 2y = 7 \quad \text{and} \quad x + y = 5
\]Solution:
- From the second equation: \( y = 5 – x \).
- Substitute into the first equation: \( 3x – 2(5 – x) = 7 \). Simplify: \( 3x – 10 + 2x = 7 \implies 5x = 17 \implies x = \frac{17}{5} \).
- Substitute \( x = \frac{17}{5} \): \( y = 5 – \frac{17}{5} = \frac{8}{5} \).
Answer: \( x = \frac{17}{5}, y = \frac{8}{5} \).
Summary
Systems of equations can be solved using either substitution or elimination methods. Practise both approaches to determine which works best for you on test day.
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