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liberstina [14]
4 years ago
5

For what value(s) of k is the linear system consistent? (Enter your answers as a comma-separated list.)

Mathematics
2 answers:
Leokris [45]4 years ago
8 0

Answer:

k\neq -\frac{27}{2}

Step-by-step explanation:

We have been given a system of equations as 6x_1-9x_2 = 8\\\\9x_1+kx_2 =-1. We are asked to find the value of k such that the given system is consistent.

Let us consider two equations as:

a_1x+b_1y = c_1\\\\a_2x+b_2y =c_2

For a system to be consistent the ratio of coefficient of x terms and y terms should not be equal that is:

\frac{a_1}{a_2}\neq \frac{b_1}{b_2}

Upon substituting our given values, we will get:

\frac{6}{9}\neq \frac{-9}{k}

6k\neq -81

\frac{6k}{6}\neq -\frac{81}{6}

k\neq -\frac{27}{2}

Since the given system will be consistent for all value except -\frac{27}{2}, therefore, we can choose any values for k such as -2 or 2.

AleksAgata [21]4 years ago
4 0

Answer:

Step-by-step explanation:

Given

6x_1-9x_2=8

9x_1+kx_2=-1

The given system is AX=B can be represented by

\begin{bmatrix}6 &-9 \\ 9 & k\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}8\\ -1\end{bmatrix}

The given system is consistent when determinant of A is not equal to zero

|A|

|A|=6k-(-81)=6k+81

k\neq \frac{-27}{2}

i.e. system is consistent for all value of k except k=\frac{-27}{2}

R-\frac{-27}{2}

                 

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