# A person plans to invest no more than 20,000 in 2 different interest-bearing accounts. Each account is to contain at least 5,000. Moreover, one account should have at least twice the amount that is...

A person plans to invest no more than 20,000 in 2 different interest-bearing accounts. Each account is to contain at least 5,000. Moreover, one account should have at least twice the amount that is in the other account. Find a system of inequalities describing the various amounts that can be deposited in each account and sketch the graph of the system.

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### 1 Answer

You should use the following notations for the 2 accounts, such that:

`x` - the first account

`y` - the second account

The investment in the two accounts is no more than 20,000, hence, you need to set up the following inequality, such that:

`x + y <= 20,000`

One account should have, at least, twice the amount that is in the other account, hence, you may set up the following inequality, such that:

`x >= 2y => x - 2y >= 0`

Each account contains, at least 5,000, hence, you may set up the following inequality, such that:

`x >= 5000`

`y>=5000`

The system of inequalities that describes the situation of the two accounts is the following, such that:

`{(x + y <= 20,000),(x - 2y >= 0),(x >= 5000),(y>=5000):}`

You need to put each inequality in slope-intercept form, to evaluate the solution, such that:

`{(y <= -x + 20,000),(y <= x/2),(x >= 5000),(y>=5000):}`

The solution of the first inequality lies below the line -x + 20,000.

Replacing `x/2` for `y` in the first inequality yields:

`x/2 <= -x + 20,000 => 3x/2 <= 20,000 => 3x <= 40,000 => x <= 13,333 => y <= 6666`

**Hence, the solution to the system of inequalitites is `x in [5000;13333]` and y in [5000;6666]. You should notice that that point `(13,333; 6,666)` lies in the region that collects all solutions to the system of inequalities.**

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