Doris invested some money at 7% and some money at 8%. She did at 7%. Her yearly total from the two investments was $7.80.
How much did Doris invest at each rate?
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To solve this problem, we can use a system of equations approach. Let's assume Doris invested x amount of money at 7% and y amount of money at 8%.
According to the given information, Doris earned a total of $7.80 in a year. So we can set up the following equation:
0.07x + 0.08y = 7.80
We also know that Doris invested a certain amount of money, so the second equation represents the total investment:
x + y = Total investment
To eliminate one of the variables, we can rearrange the second equation to solve for x:
x = Total investment - y
Now we can substitute this value of x into the first equation:
0.07(Total investment - y) + 0.08y = 7.80
Next, distribute and combine like terms:
0.07Total investment - 0.07y + 0.08y = 7.80
Combine like terms:
0.07Total investment + 0.01y = 7.80
Now we have an equation with only one variable, y. To isolate y, subtract 0.07Total investment from both sides:
0.01y = 7.80 - 0.07Total investment
Finally, divide both sides by 0.01:
y = (7.80 - 0.07Total investment) / 0.01
Now we can substitute this value of y back into the second equation to solve for x:
x + (7.80 - 0.07Total investment) / 0.01 = Total investment
Rearrange the equation to isolate x:
x = Total investment - (7.80 - 0.07Total investment) / 0.01
With these equations, we can find the values of x and y, which represent the amounts Doris invested at 7% and 8% respectively.