Betty invested some money at 15% interest.Betty also invested $237 more than 2 times than amount at 13%

From the given problem, it is evident that there are two different investments. One investment earns 15% interest and the other investment earns 13%.

Let us say that the amount X is invested at 15% interest, amount Y is invested at 13% interest.

Since there are two variables X & Y, We need to frame two equations to solve for the two given variables.

Amount of interest earned by the amount X at 15% interest = $x\cdot \dfrac {15}{100}=0\cdot 15x$

Amount of interest earned by the amount Y at 13% interest = $y\cdot \dfrac {13}{100}=0.13y$

Given $\$1668.03$ is the total interest earned from the two investments.

Therefore $0.15x+0.13y=1668.03$

Multiplying with 100 throughout the equation to avoid the decimals.

Given that the “Betty invested $237 more than 2 times that amount at 13%”

Therefore we can say

By using the substitution method, we can replace y as 2x +237 in Equation 1.

15x+13(2x+237)=168803

15x+26x+3081=168803

41x=168803-3081=165722

$x=\dfrac {165722}{41}$

X=4042

Therefore y= 2x+237 =2(4042)+237

Y=8321

Therefore,In conclusion,the amount invested at 15% is $\$ 4042$ and the amount invested at 13% interest is $\$ 8321$ .

We can cross check the answers we got for the X =4042 and Y =8321 values.

When the values of X and Y are plugged back into the original equations, they must hold good.

Let us take up the equation1

15x+13y=166803

15(4042)+13(8321)=168803

60630+108173=168803

168803=168803.Therefore, the solutions we arrived work for the equation1

Let us cross-check with the equation2.

y=2x+237

8321=2(4042)+237

8321=8321 Therefore, the solutions we arrived work for the equation2 as well.

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