Solve The System Of Linear Equations. 4x + 6y = 12 X + 2y = 10 The Solution Is
Understanding how to solve systems of linear equations is a fundamental skill in algebra, mathematics, engineering, and various applied sciences. In this article, we will explore the methods to solve the specific system of equations:
\[
\begin{cases}
4x + 6y = 12 \\
x + 2y = 10
\end{cases}
\]
We will discuss various strategies, including substitution, elimination, and graphical methods, to find the solutions. This comprehensive guide aims to help students, educators, and professionals grasp the concepts involved in solving such systems efficiently and accurately.
What Is a System of Linear Equations?
A system of linear equations consists of two or more equations involving the same set of variables. The goal is to find the values of these variables that satisfy all equations simultaneously. The solutions to the system can be:
- A single unique solution (the equations intersect at one point)
- Infinitely many solutions (the equations are the same line)
- No solution (the equations are parallel and never intersect)
In our case, the system involves two equations with two variables, x and y, which typically has a unique solution unless the equations are dependent or inconsistent.
Methods for Solving the System
There are several methods to solve systems of linear equations:
- Substitution Method
- Elimination Method
- Graphical Method
- Matrix Method (using determinants or matrices)
Each method has its advantages and is suitable for different types of systems.
1. Substitution Method
This method involves solving one of the equations for one variable and substituting that expression into the other equation.
Step-by-step process for our system:
- Solve the second equation \( x + 2y = 10 \) for x:
\[
x = 10 - 2y
\]
- Substitute this expression into the first equation \( 4x + 6y = 12 \):
\[
4(10 - 2y) + 6y = 12
\]
- Simplify:
\[
40 - 8y + 6y = 12
\]
\[
40 - 2y = 12
\]
- Isolate y:
\[
-2y = 12 - 40
\]
\[
-2y = -28
\]
\[
y = \frac{-28}{-2} = 14
\]
- Find x by substituting y back into \( x = 10 - 2y \):
\[
x = 10 - 2(14) = 10 - 28 = -18
\]
Solution:
\[
x = -18, \quad y = 14
\]
This pair of values is the solution to the system.
2. Elimination Method
The elimination method involves manipulating the equations to eliminate one variable, making it easier to solve for the other.
Step-by-step process:
- The system:
\[
\begin{cases}
4x + 6y = 12 \quad (1) \\
x + 2y = 10 \quad (2)
\end{cases}
\]
- Multiply equation (2) by a factor that makes the coefficients of x in both equations equal (or opposites). Here, multiply (2) by 4:
\[
4x + 8y = 40 \quad (3)
\]
- Now, subtract equation (1) from equation (3):
\[
(4x + 8y) - (4x + 6y) = 40 - 12
\]
\[
(4x - 4x) + (8y - 6y) = 28
\]
\[
0 + 2y = 28
\]
\[
2y = 28
\]
\[
y = 14
\]
- Substitute y into equation (2):
\[
x + 2(14) = 10
\]
\[
x + 28 = 10
\]
\[
x = 10 - 28 = -18
\]
Solution:
\[
x = -18, \quad y = 14
\]
Both methods lead to the same solution, confirming its accuracy.
3. Graphical Method
Graphing involves plotting both equations on a coordinate plane and identifying the point where they intersect.
- Rewrite equations in slope-intercept form \( y = mx + b \):
From \( 4x + 6y = 12 \):
\[
6y = -4x + 12
\]
\[
y = -\frac{2}{3}x + 2
\]
From \( x + 2y = 10 \):
\[
2y = -x + 10
\]
\[
y = -\frac{1}{2}x + 5
\]
- Plot the two lines:
- Line 1: Slope \( -\frac{2}{3} \), y-intercept 2
- Line 2: Slope \( -\frac{1}{2} \), y-intercept 5
- Find the intersection point manually or using graphing tools:
Set the right sides equal to find the intersection:
\[
-\frac{2}{3}x + 2 = -\frac{1}{2}x + 5
\]
Multiply through by 6 (least common denominator) to clear fractions:
\[
6 \times \left( -\frac{2}{3}x + 2 \right) = 6 \times \left( -\frac{1}{2}x + 5 \right)
\]
\[
-4x + 12 = -3x + 30
\]
Bring variables to one side:
\[
-4x + 12 + 3x = 30
\]
\[
- x + 12 = 30
\[
- x = 18
\[
x = -18
\]
Substitute into one of the equations to find y:
\[
y = -\frac{1}{2}(-18) + 5 = 9 + 5 = 14
\]
Graphical solution:
Point \((-18, 14)\)
This confirms the solution found via algebraic methods.
Understanding the Solution
The solution to the system:
\[
\boxed{
x = -18, \quad y = 14
}
\]
represents the unique point where both lines intersect on the coordinate plane. This point satisfies both original equations, meaning substituting these values back into the equations confirms their correctness.
Verifying the Solution
Verification is an essential step to ensure the solution is accurate.
- Substitute \( x = -18 \) and \( y = 14 \) into the first equation:
\[
4(-18) + 6(14) = -72 + 84 = 12
\]
- Substitute into the second equation:
\[
(-18) + 2(14) = -18 + 28 = 10
\]
Both equations are satisfied, confirming that \((-18, 14)\) is the solution.
Additional Tips for Solving Systems of Equations
- Always check whether the system has a unique solution, infinitely many solutions, or no solution.
- When equations are in different forms, convert to the same form for easier comparison.
- Use graphing tools or software for complex systems or when visual understanding helps.
- Practice with different methods to become proficient and recognize which method is most efficient for a given system.
Applications of Solving Systems of Linear Equations
Understanding how to solve systems of equations is vital in various real-world applications:
- Engineering: Designing circuits, analyzing forces in structures.
- Economics: Equilibrium analysis, cost-profit optimization.
- Physics: Kinematic equations, motion analysis.
- Computer Science: Algorithm development, graphics programming.
- Business: Budget planning, resource allocation.
Conclusion
Solving systems of linear equations like:
\[
\begin{cases}
4x + 6y = 12 \\
x + 2y = 10
\end{cases}
\]
is a fundamental skill that can be approached through substitution, elimination, or graphical methods. In our case, all methods yield the solution:
\[
\boxed{
x = -18, \quad y = 14
}
\]
Mastering these techniques enables problem-solvers to analyze and interpret complex systems across various disciplines efficiently. Whether working with simple algebraic systems or advanced matrix methods, the core principles remain consistent and essential for mathematical literacy and practical problem-solving.
---
If you want to enhance your understanding further, consider practicing with similar systems, exploring matrix solutions, or applying these methods in real-world scenarios to see their effectiveness firsthand.
