Solve the following system using Cramer's rule. 2x + 3y = 12 %3D 6x - y = -4 Step 1 of 4 Identify the 2x2 coefficient matrix for the system of equations. Find the valu
Solve the following system using Cramer's rule. 2x + 3y = 12 %3D 6x - y = -4 Step 1 of 4 Identify the 2x2 coefficient matrix for the system of equations. Find the valu
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Solve the following system using Cramer's rule.
\[
\begin{align*}
2x + 3y &= 12 \\
6x - y &= -4
\end{align*}
\]
**Step 1 of 4**
Identify the 2x2 coefficient matrix for the system of equations. Find the value of the determinant.
When you have a system of two equations in the form:
\[
\begin{cases}
ax + cy = k_1 \\
bx + dy = k_2
\end{cases}
\]
the coefficient matrix determinant is written in the form \(\begin{vmatrix} a & c \\ b & d \end{vmatrix}\).
In this case, the determinant is given by the following:
\[
\begin{vmatrix}
\text{[Input box]} & \text{[Input box]} \\
\text{[Input box]} & \text{[Input box]}
\end{vmatrix}
\]
The numerical value of the determinant is found by calculating \(ad - bc\).
\[
2(\text{[Input box]}) - 6(\text{[Input box]})
\]
\[
= -2 - \text{[Input box]}
\]
\[
= \text{[Input box]}
\]
The determinant, \(D\), is not equal to zero, so there is one real solution to the system.
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Transcribed Image Text:Solve the following system using Cramer's rule.
\[
\begin{align*}
2x + 3y &= 12 \\
6x - y &= -4
\end{align*}
\]
**Step 1 of 4**
Identify the 2x2 coefficient matrix for the system of equations. Find the value of the determinant.
When you have a system of two equations in the form:
\[
\begin{cases}
ax + cy = k_1 \\
bx + dy = k_2
\end{cases}
\]
the coefficient matrix determinant is written in the form \(\begin{vmatrix} a & c \\ b & d \end{vmatrix}\).
In this case, the determinant is given by the following:
\[
\begin{vmatrix}
\text{[Input box]} & \text{[Input box]} \\
\text{[Input box]} & \text{[Input box]}
\end{vmatrix}
\]
The numerical value of the determinant is found by calculating \(ad - bc\).
\[
2(\text{[Input box]}) - 6(\text{[Input box]})
\]
\[
= -2 - \text{[Input box]}
\]
\[
= \text{[Input box]}
\]
The determinant, \(D\), is not equal to zero, so there is one real solution to the system.
[Submit] [Skip (you cannot come back)]
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