(a) Fred bought a car for $5,600.00 and later sold it at 90% of the cost price. He spent $1,310.00 out of the amount received and invested the rest at 6% per annum simple interest. Calculate the interest earned in 3 years.
(b) Solve the equations 2\(^x\)(4\(^{-7}\)) = 2 and 3\(^{-x}\)(9\(^{2y}\)) = 3 simultaneously.
a) First, let's calculate the amount Fred received after selling the car. He sold the car for 90% of its cost price, so he received $5,600 * 0.9 = $5,040.00. After spending $1,310.00, he was left with $5,040 - $1,310 = $3,730.00. This is the amount he invested at 6% per annum simple interest.
The formula for simple interest is I = P * R * T, where I is the interest, P is the principal (the amount invested), R is the rate of interest (as a decimal), and T is the time (in years). Plugging in the values, we have I = $3,730 * 0.06 * 3 = $676.20.
So Fred earned $676.20 in interest over a period of 3 years.
b) To solve the equations 2x(4-7) = 2 and 3-x(92y) = 3 simultaneously, we need to find the values of x and y that make both equations true.
Starting with the first equation, we can simplify the left-hand side to get 2x * 4-7 = 2x * (1/16) = 2. We can then divide both sides of the equation by 2x to get (1/16) = 1. Since this is not true, there are no values of x that make this equation true.
Next, let's simplify the second equation to get 3-x * 92y = 3-x * 81y = 3. Dividing both sides of the equation by 3-x gives us 81y = 3x. Taking the log base 3 of both sides gives us y = x.
So the only solution for x and y is x = y = 1.
In conclusion, there is only one solution for the system of equations, x = y = 1.