Substituting Points into Both Equations

Oct 13, 2025 by orovajewels.com 140 views

Introduction

In mathematics, a system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables. These equations are in the form of ax + by = c, where a, b, and c are constants, and x and y are the variables. In this article, we will explore a system of two linear equations and find the point that lies on both lines.

The System of Equations

The given system of equations is:

\begin{cases} 6x + 4y - 2 = 0 \\ 6x + 5y - 4 = 0 \end{cases}

Substituting Points into Both Equations

To find the point that lies on both lines, we need to substitute each point into both equations and check if the result is true. Let's start with the first point, (-3, 5).

Substituting (-3, 5) into the First Equation

Substituting x = -3 and y = 5 into the first equation, we get:

6(-3) + 4(5) - 2 = 0

Expanding and simplifying the equation, we get:

-18 + 20 - 2 = 0

0 = 0

The result is true, which means that the point (-3, 5) lies on the first line.

Substituting (-3, 5) into the Second Equation

Substituting x = -3 and y = 5 into the second equation, we get:

6(-3) + 5(5) - 4 = 0

Expanding and simplifying the equation, we get:

-18 + 25 - 4 = 0

3 = 0

The result is false, which means that the point (-3, 5) does not lie on the second line.

Conclusion

Based on the results, we can conclude that the point (-3, 5) lies on the first line but not on the second line. Therefore, the point (-3, 5) is not the common point that lies on both lines.

What's Next?

In the next section, we will explore the process of finding the common point that lies on both lines.

Finding the Common Point

To find the common point that lies on both lines, we need to solve the system of equations. One way to do this is by using the method of substitution or elimination.

Method of Substitution

Let's use the method of substitution to solve the system of equations. We can solve the first equation for x and substitute it into the second equation.

From the first equation, we can solve for x:

6x + 4y - 2 = 0

6x = 2 - 4y

x = \frac{2 - 4y}{6}

Substituting x into the second equation, we get:

6\left(\frac{2 - 4y}{6}\right) + 5y - 4 = 0

Simplifying the equation, we get:

2 - 4y + 5y - 4 = 0

y = 2

Substituting y back into the first equation, we get:

6x + 4(2) - 2 = 0

Simplifying the equation, we get:

6x + 8 - 2 = 0

6x + 6 = 0

x = -1

Therefore, the common point that lies on both lines is (-1, 2).

Conclusion

Based on the results, we can conclude that the common point that lies on both lines is (-1, 2).

What's Next?

In the next section, we will explore the process of graphing the system of equations.

Graphing the System of Equations

To graph the system of equations, we need to plot the two lines on a coordinate plane.

Plotting the First Line

The first line is given by the equation:

6x + 4y - 2 = 0

We can plot the line by finding two points on the line and drawing a line through them.

Let's find two points on the line by substituting x = 0 and x = 1 into the equation.

Substituting x = 0 into the equation, we get:

6(0) + 4y - 2 = 0

Simplifying the equation, we get:

4y - 2 = 0

y = \frac{1}{2}

Therefore, the point (0, 1/2) lies on the first line.

Substituting x = 1 into the equation, we get:

6(1) + 4y - 2 = 0

Simplifying the equation, we get:

4 + 4y - 2 = 0

4y = -2

y = -\frac{1}{2}

Therefore, the point (1, -1/2) lies on the first line.

Plotting the two points (0, 1/2) and (1, -1/2) on a coordinate plane, we get a line with a slope of -1/3 and a y-intercept of 1/2.

Plotting the Second Line

The second line is given by the equation:

6x + 5y - 4 = 0

We can plot the line by finding two points on the line and drawing a line through them.

Let's find two points on the line by substituting x = 0 and x = 1 into the equation.

Substituting x = 0 into the equation, we get:

6(0) + 5y - 4 = 0

Simplifying the equation, we get:

5y - 4 = 0

y = \frac{4}{5}

Therefore, the point (0, 4/5) lies on the second line.

Substituting x = 1 into the equation, we get:

6(1) + 5y - 4 = 0

Simplifying the equation, we get:

6 + 5y - 4 = 0

5y = -2

y = -\frac{2}{5}

Therefore, the point (1, -2/5) lies on the second line.

Plotting the two points (0, 4/5) and (1, -2/5) on a coordinate plane, we get a line with a slope of -2/3 and a y-intercept of 4/5.

Conclusion

Based on the results, we can conclude that the two lines intersect at the point (-1, 2).

What's Next?

In the next section, we will explore the process of solving a system of linear equations using matrices.

Solving a System of Linear Equations using Matrices

To solve a system of linear equations using matrices, we need to represent the system of equations as a matrix equation.

Representing the System of Equations as a Matrix Equation

Let's represent the system of equations as a matrix equation:

\begin{bmatrix} 6 & 4 \\ 6 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix}

Finding the Inverse of the Coefficient Matrix

To solve the system of equations, we need to find the inverse of the coefficient matrix.

The coefficient matrix is:

\begin{bmatrix} 6 & 4 \\ 6 & 5 \end{bmatrix}

We can find the inverse of the coefficient matrix using the formula:

A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix}

where det(A) is the determinant of the coefficient matrix.

The determinant of the coefficient matrix is:

\det(A) = 6(5) - 4(6) = 30 - 24 = 6

Therefore, the inverse of the coefficient matrix is:

A^{-1} = \frac{1}{6} \begin{bmatrix} 5 & -4 \\ -6 & 6 \end{bmatrix}

Solving the System of Equations

To solve the system of equations, we can multiply both sides of the matrix equation by the inverse of the coefficient matrix:

\begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} \begin{bmatrix} 2 \\ 4 \end{bmatrix}

Substituting the inverse of the coefficient matrix, we get:

\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 5 & -4 \\ -6 & 6 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \end{bmatrix}

Simplifying the equation, we get:

\begin<br/> **Q&A: Solving a System of Linear Equations** =============================================

Q: What is a system of linear equations?

A: A system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables. These equations are in the form of ax + by = c, where a, b, and c are constants, and x and y are the variables.

Q: How do I solve a system of linear equations?

A: There are several methods to solve a system of linear equations, including the method of substitution, the method of elimination, and the use of matrices. The method of substitution involves solving one equation for one variable and substituting that expression into the other equation. The method of elimination involves adding or subtracting the equations to eliminate one variable. The use of matrices involves representing the system of equations as a matrix equation and solving for the variables using matrix operations.

Q: What is the method of substitution?

A: The method of substitution involves solving one equation for one variable and substituting that expression into the other equation. For example, if we have the system of equations: