{:[(6)/(5)p+kq=(4)/(5)],[q=(3)/(5)p-(2)/(5)]:} Consider the system of equations, where k is a constant. For which value of k is there no (p,q) solutions? Choose 1 answer: (A) -2 (B) 0 (C) 2 (D) None of the above

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{:[(6)/(5)p+kq=(4)/(5)],[q=(3)/(5)p-(2)/(5)]:} Consider the system of equations, where  k is a constant. For which value of  k is there no  (p,q) solutions? Choose 1 answer: (A) -2 (B) 0 (C) 2 (D) None of the above

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Write Equations: Write down the system of equations. We have the following system of equations: $$ \begin{align*} \frac{6}{5}p + kq &= \frac{4}{5} \ q &= \frac{3}{5}p - \frac{2}{5} \end{align*} $$ We need to find the value of k for which there is no solution to this system.Substitute and Simplify: Substitute the second equation into the first equation. By substituting the value of q from the second equation into the first equation, we get: $$ \frac{6}{5}p + k\left(\frac{3}{5}p - \frac{2}{5}\right) = \frac{4}{5} $$Distribute and Combine: Distribute k and combine like terms. Distribute k across the terms in the parentheses and then combine the p terms: $$ \frac{6}{5}p + \frac{3k}{5}p - \frac{2k}{5} = \frac{4}{5} $$ Combine the p terms: $$ \left(\frac{6}{5} + \frac{3k}{5}\right)p - \frac{2k}{5} = \frac{4}{5} $$Identify No Solution Condition: Identify the condition for no solution. For there to be no solution, the p terms must cancel out, and we must be left with a false statement. This happens when the coefficient of p is zero, and the constant terms do not equal each other. So we need: $$ \frac{6}{5} + \frac{3k}{5} = 0 $$Solve for k: Solve for k. Solve the above equation for k: $$ \frac{6}{5} + \frac{3k}{5} = 0 $$ Multiply through by 5 to clear the denominators: $$ 6 + 3k = 0 $$ Subtract 6 from both sides: $$ 3k = -6 $$ Divide by 3: $$ k = -2 $$

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