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Graph the system of linear equations on your calculator and select the correct solution.

[tex]\[
\begin{cases}
6.87x + 7.66y = -9.24 \\
-1.34x + 5.78y = 15.49
\end{cases}
\][/tex]

A. [tex]\((1.03, 0.59)\)[/tex]
B. [tex]\((-3.44, 1.89)\)[/tex]
C. [tex]\((-3.75, 1.56)\)[/tex]
D. [tex]\((2.20, -1.32)\)[/tex]


Sagot :

To solve this system of linear equations:
[tex]\[ \left\{ \begin{array}{l} 6.87x + 7.66y = -9.24 \\ -1.34x + 5.78y = 15.49 \end{array} \right. \][/tex]

we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously.

Start by placing the equations in a matrix form, where [tex]\( A \)[/tex] is the matrix of coefficients and [tex]\( B \)[/tex] is the constant vector:

Matrix A:
[tex]\[ A = \begin{bmatrix} 6.87 & 7.66 \\ -1.34 & 5.78 \end{bmatrix} \][/tex]

Vector B:
[tex]\[ B = \begin{bmatrix} -9.24 \\ 15.49 \end{bmatrix} \][/tex]

Using linear algebra techniques (e.g., Gaussian elimination or matrix inversion), solve the system [tex]\( AX = B \)[/tex] for the variable vector [tex]\( X \)[/tex]:

[tex]\[ X = \begin{bmatrix} x \\ y \end{bmatrix} \][/tex]

After solving the system, we find:
[tex]\[ x \approx -3.44, \quad y \approx 1.89 \][/tex]

Thus, the solution to the system of equations is [tex]\( x \approx -3.44 \)[/tex] and [tex]\( y \approx 1.89 \)[/tex].

Given multiple-choice options:
[tex]\[ (1.03, 0.59), \quad (-3.44, 1.89), \quad (-3.75, 1.56), \quad (2.20, -1.32) \][/tex]

The correct solution is:

[tex]\[ (-3.44, 1.89) \][/tex]