Select the set of equations that represents the following situation: Mary invested one amount at 7% simple interest, and a second amount at 5% interest, earning $29.80 in one year. If she had switched the amounts, she would have earned $35.00. What were the two amounts?
Invested $X @ 7%.
Invested $Y @ 5%.
Eq1: 0.07X + 0.05Y = $29.80.
Eq2: 0.05X + 0.07Y=35(Amountswitched).
Multiply both Eqs by 100:
7X + 5Y = 2980
5X + 7Y = 3500
Multiply 1st Eq by -5 and 2nd Eq by7:
-35X -25Y = -14900
35X + 49Y = 24500
Add the 2 Eqs and get:
24Y = 9600,
Y = $400.
Substitute 400 for Y in Eq2:
0.05X + 0.07*400 = 35,
0.05X + 28 = 35,
0.05X = 35 - 28 = 7,
X = 7 / 0.05 = $140.
To solve this problem, we can set up a system of equations based on the given information.
Let's assume Mary's first investment is denoted by "x" and the second investment is denoted by "y".
According to the problem, Mary earned $29.80 in one year with the first arrangement, so we can write the equation:
0.07x + 0.05y = 29.80 -- Equation (1)
If she had switched the amounts, she would have earned $35.00, so the new equation is:
0.07y + 0.05x = 35.00 -- Equation (2)
We now have a system of two equations with two variables.
To solve this system, we can apply the method of substitution or elimination. Let's use substitution to solve the system.
Solve Equation (1) for x:
0.07x = 29.80 - 0.05y
x = (29.80 - 0.05y) / 0.07
Now substitute the value of x into Equation (2):
0.07y + 0.05((29.80 - 0.05y) / 0.07) = 35.00
Simplify the equation:
0.07y + 0.05(29.80 - 0.05y) = 245.00
Distribute and solve for y:
0.07y + 1.49 - 0.0025y = 35.00
0.0675y = 33.51
y ≈ 496.89
Now substitute the value of y back into Equation (1) to find x:
0.07x + 0.05(496.89) = 29.80
0.07x + 24.84 = 29.80
0.07x = 4.96
x ≈ 70.86
Therefore, Mary invested approximately $70.86 at 7% interest and approximately $496.89 at 5% interest.