Which value for the constant d makes y=7 an extraneous solution in the following equation? {:[sqrt(4y-3)=d-y],[d=]:}

Question

Which value for the constant  d makes  y=7 an extraneous solution in the following equation?  {:[sqrt(4y-3)=d-y],[d=]:}

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Understanding extraneous solutions: Understand what an extraneous solution is. An extraneous solution is a solution that emerges from the process of solving an equation but is not a valid solution to the original equation. In this case, we want to find a value for d such that y=7 appears to be a solution after solving the equation but does not satisfy the original equation.Substituting y=7 into the left side: Substitute y=7 into the left side of the equation. We substitute y=7 into the equation sqrt(4y-3) to see what value it produces. Calculation: sqrt(4*7-3) = sqrt(28-3) = sqrt(25) = 5Substituting y=7 into the right side: Substitute y=7 into the right side of the equation. We substitute y=7 into the equation d-y to see what expression we get. Expression: d-7Setting the expressions equal: Set the expressions from Step 2 and Step 3 equal to each other. Since we want y=7 to be an extraneous solution, the expressions from the left and right sides of the equation when y=7 is substituted should not be equal. Equation: 5 = d-7Solving for d: Solve for d. We solve the equation from Step 4 for d. Calculation: d = 5 + 7 = 12Verifying y=7 as an extraneous solution: Verify that y=7 is an extraneous solution for the found value of d. We substitute d=12 and y=7 back into the original equation to check if it holds true. Original equation: sqrt(4y-3) = d-y Substitution: sqrt(4*7-3) = 12-7 Verification: sqrt(25) = 5 ≠ 12-7 = 5 Since the equation does not hold true, y=7 is indeed an extraneous solution for d=12.

Which value for the constant $d$ makes $y=7$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{4 y-3}=d-y \\ d= \end{array}$

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Which value for the constant d d d makes y=7 y=7 y=7 an extraneous solution in the following equation?\newline4y−3=d−yd= \begin{array}{l} \sqrt{4 y-3}=d-y \\ d= \end{array} 4y−3​=d−yd=​

Which value for the constant $d$ makes $y=7$ an extraneous solution in the following equation? $\newline$ $\begin{array}{l} \sqrt{4 y-3}=d-y \\ d= \end{array}$