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Problem 1263

# $$\ 10,000$$ is to be invested for one year. A part of the $$\ 10,000$$ is invested at $$6 \%$$ and the remainder at $$7 \%$$. The income from the $$\ 10,000$$ is $$\ 640$$ per year. Find the amount invested at each rate.

Expert verified
The amount invested at 6% is $$\ 6,000$$ and the amount invested at 7% is $$\ 4,000$$.
See the step by step solution

## Step 1: Assigning Variables

Let x be the amount invested at 6%, and y be the amount invested at 7%. Our goal is to find the values of x and y.

## Step 2: Set up the system of linear equations

First, since the total investment is $10,000, we can write that as the sum of the amount invested at 6% and the amount invested at 7%. That gives us the first equation: $$x + y = 10000$$ Now, we know that the total income is$640. The income from the amount invested at 6% is 6% of x, or $$\frac{6}{100}x$$. The income from the amount invested at 7% is 7% of y, or $$\frac{7}{100}y$$. So the total income can be represented as: $$\frac{6}{100}x + \frac{7}{100}y = 640$$

## Step 3: Solve the system of linear equations

Now we have two equations with two variables. Our system of linear equations looks like this: 1) $$x + y = 10000$$ 2) $$\frac{6}{100}x + \frac{7}{100}y = 640$$ To solve the system, we can first multiply equation (2) by 100 to get rid of the fractions: 2) $$6x + 7y = 64000$$ Now we can solve equation (1) for x: $$x = 10000 - y$$ And substitute this expression back into equation 2: $$6(10000 - y) + 7y = 64000$$

## Step 4: Solve for y

Now we have an equation with one variable: $$60000 - 6y + 7y = 64000$$ Solve for y: $$y = 4000$$

## Step 5: Find x

Now we know the value of y, we can plug it back into the expression for x: $$x = 10000 - 4000$$ $$x = 6000$$

## Step 6: Interpret the result

We have found that $$x = 6000$$ and $$y = 4000$$. This means that $$\ 6,000$$ was invested at 6% and $$\ 4,000$$ was invested at 7%.

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