{:[(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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Analyze Equations: Analyze the system of equations to understand when there would be no solutions. A system of linear equations has no solutions when the lines represented by the equations are parallel. This happens when the coefficients of the variables are proportional, but the constants are not.Standard Form: Write the system of equations in a standard form to compare the coefficients. The given system is: [(6/5)p + kq = (4/5)] [q = (3/5)p - (2/5)] We need to express q from the second equation in terms of p and substitute it into the first equation.Substitute q into First: Substitute the value of q from the second equation into the first equation. [(6/5)p + k((3/5)p - (2/5)) = (4/5)] Now distribute k into the terms inside the parentheses.Distribute and Simplify: Distribute k and simplify the equation. [(6/5)p + (3k/5)p - (2k/5) = (4/5)] Combine like terms. [(6/5 + 3k/5)p - (2k/5) = (4/5)]Combine Like Terms: Combine the coefficients of p. [(6/5 + 3k/5)p - (2k/5) = (4/5)] [(6 + 3k)/5)p - (2k/5) = (4/5)] Now we have the equation in terms of p with the coefficients clearly visible.Combine Coefficients: Determine the condition for no solutions. For there to be no solutions, the coefficients of p must be the same on both sides of the equation, but the constants must be different. Since there is no p term on the right side of the equation, the coefficient of p on the left side must be zero. So, we set the coefficient of p to zero: (6 + 3k)/5 = 0Condition for No Solutions: Solve for k. Multiply both sides by 5 to get rid of the denominator: 6 + 3k = 0 Now, solve for k: 3k = -6 k = -6 / 3 k = -2

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