The main idea here is to "translate" the words into maths. First we need to identify the unknowns and label these.
We need to know the number if dimes and the number of quarters. So lets say x: number of dimes y: number of quarters
Now lets write equations from the written problem. We know that there are 36 coind total, thus: x + y = 36 We also know that the coins total 5.85 dollars, but it is better to count in cents, that is 585 cents. x are the number of dimes, their value is x*10 y are quarters with value of y*25 thus: 10x+25y=585
We have two equations and two unknowns now, that needs to be solved to get the answer.
Remember that <span>an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.
Lets solve our equation to find out what is the extraneous solution: </span> and and <span> So, the solutions of our equation are </span> and . Lets replace each solution in our original equation to check if they are valid solutions: - For We can conclude that 7 is a valid solution of the equation.
- For We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.
We can conclude that the correct answer is: <span>D. the extraneous solution is x = 4</span>