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rac{ $6$ }{ $5$ }p + kq = rac{ $4$ }{ $5$ } q = rac{ $3$ }{ $5$ }p - rac{ $2$ }{ $5$ } Consider the system of equations, where $k$ is a constant. For which value of $k$ is there no $(p,q)$ solutions?

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Full solution

Q. rac{

6

}{

5

}p + kq = rac{

4

}{

5

} q = rac{

3

}{

5

}p - rac{

2

}{

5

} Consider the system of equations, where

k

is a constant. For which value of

k

is there no

(p,q)

solutions?

Equations Given: We have the system of equations: $\newline$ $1$ ) $\frac{6}{5}p + kq = \frac{4}{5}$ $\newline$ $2$ ) $q = \frac{3}{5}p - \frac{2}{5}$ $\newline$ We need to find the value of $k$ for which there is no solution to this system.
Substitute $q$ into Eq $1$ : First, let's substitute the expression for $q$ from equation $2$ into equation $1$ . $\newline$ $\frac{6}{5}p + k\left(\frac{3}{5}p - \frac{2}{5}\right) = \frac{4}{5}$
Distribute $k$ : Distribute $k$ to the terms inside the parentheses. $\newline$ $\frac{6}{5}p + \left(\frac{3k}{5}\right)p - \left(\frac{2k}{5}\right) = \frac{4}{5}$
Combine like terms: Combine like terms. $\newline$ $(\frac{6}{5} + \frac{3k}{5})p - \frac{2k}{5} = \frac{4}{5}$
Conditions for no solution: To have no solution, the coefficients of $p$ on both sides of the equation must be equal, and the constants must be different. This would make the system inconsistent. $\newline$ So, we need to find $k$ such that: $\newline$ $\frac{6}{5} + \frac{3k}{5} = 0$
Clear denominator: Multiply both sides of the equation by $5$ to clear the denominator. $\newline$ $6 + 3k = 0$
Solve for $k$ : Subtract $6$ from both sides to solve for $k$ . $3k = -6$
Calculate $k$ : Divide both sides by $3$ to find the value of $k$ . $k = \frac{-6}{3}$
Calculate $k$ : Divide both sides by $3$ to find the value of $k$ . $k = \frac{-6}{3}$ Calculate the value of $k$ . $k = -2$

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