Final answer:
To solve the investment problem, we corrected a typo and used substitution and addition methods to find that $11,480 is invested at 3% and $10,280 is invested at 2%.
Step-by-step explanation:
The student is solving a system of linear equations related to a word problem involving investment at different rates of interest, specifically 3% and 2%. To find the amounts invested at each rate, let's denote x as the amount invested at 3% and y as the amount invested at 2%. According to the problem we are given the following system:
- x = y + 1200
- 0.03x + 0.02y = 550
To solve the system using the method of addition, we must first express one of the variables in terms of the other using the first equation, then substitute this expression into the second equation. However, there is a typo in the provided system of equations. Instead of 'x = y + 12000', it should read 'x = y + 1200' according to the original statement of the problem.
We correct the typo and proceed:
- Substitute x in the second equation with y + 1200: 0.03(y + 1200) + 0.02y = 550.
- Expand and simplify: 0.03y + 36 + 0.02y = 550.
- Combine like terms: 0.05y + 36 = 550.
- Subtract 36 from both sides: 0.05y = 514.
- Divide by 0.05: y = 10280.
- Substitute y back into the first equation to find x: x = 10280 + 1200, so x = 11480.
In conclusion, $11,480 is invested at 3%, and $10,280 is invested at 2%.