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15.2: Examples

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Dec 21, 2020
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- 15.1: Youtube
- 15.3: Exercise 15

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Solving Linear Equations. Investment Problems

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Some of the problems here are simple. The solution can be worked out fast by quick reasoning. The benefit of these problems is not to find the solution by reasoning, but by learning algebraic steps applicable to more challenging situations.

The general method is not to read a problem first and understand it. Many of my colleagues will disagree with me. I propose creating a preamble of symbols for each element in the problem.

Look at the end of the problem which asks what to solve for. Let be the number we are looking for. Write this as the first step in the preamble. Start reading the problem and develop the preamble step by step. Write the symbol(s) for each step on a new line. Finish converting all the steps in the problem into symbols. Now look at your preamble and understand it.

It will be easier to obtain an equation using the symbols from the preamble. Then solve the equation by the method introduced this far for linear equations. (A linear equation has a variable to the first degree (exponent).)

Investment problems normally deal with compound interest. We need to understand simple interest first. We deal with simple interest in the following problems.

When you invest money you want to be rewarded. Your reward, called interest, depends on three quantities:

1) The more your investment (Principal ) the more your reward.

2) The more money you get per dollars invested (rate of interest, or just rate ), the higher your reward.

3) The longer you invest your money (time usually for a number of years or a fraction of a year). We assume here that the rate of interest is in years.

Thus $\parbox is only supported in math mode\fbox{\parbox[t]{2.0in}{\begin{center}\boldmath \Large I\ =\ P\ r\ t\end{center}}}$

Examples

Example 1:

Your son invests at % and your daughter invests at % simple interest for one year. What is the total interest earned?

Solution:

Preamble:

Let be the total interest earned (not too useful in our problem).

Equation:

Total Interest= Interest from son added to interest from daughter.

Total interest

Example 2:

Your grandmother invests part of at and the rest at . Her annual interest income from both accounts is . How much money did she invest at ?

Solution:

Preamble:
Avoid using two variables. You would need two equations. You will learn how to handle a system of two equations in two variables later.

Let be the amount invested at (like ).

Then is the amount invested at

(like )

You have the option to chose as the % account instead of the account.

Equation:

$\) is only supported in math mode\begin{array}{rcl lll lll} \hbox{Interest (\)8%$)}+\hbox{interest ($6%\()}&=&290\\[6pt] 0.08x+0.06(4,500-x)&=&290\\[6pt] 0.08x+270-0.06x&=&290\\[6pt] 0.02x+270&=&290\\[6pt] 0.02x+270-270&=&290-270\\[6pt] 0.02x&=&20\\[8pt] \displaystyle \frac{0.02}{0.02}x&=&\displaystyle \frac{20}{0.02}\\[8pt] x&=&\displaystyle \frac{2,000}{2}\\[8pt] x&=&1,000\\ \end{array}$

dollars was invested at .

Example 3:

An entrepreneur borrowed some money at and more than hat account at annually. His interest is for two years. How much did the entrepreneur borrow at ?

Solution:

Preamble:

Let be the amount borrowed at (like ).
Then is the amount borrowed at
(like )

Equation:

$\) is only supported in math mode\begin{array}{rcl lll lll} \hbox{Interest (\)4%$)}+\hbox{interest ($7%\()}&=&768\\[5pt] (0.04)(2)x+(0.07)(2)(x+1,400)&=&768\\[5pt] (0.08)x+(0.14)(x+1,400)&=&768\\[5pt] (0.08)x+(0.14)x+(0.14)(1,400)&=&768\\[5pt] 0.22x+196&=&768\\[5pt] 0.22x&=&572\\[5pt] \displaystyle \frac{0.22}{0.22}x&=&\displaystyle \frac{572}{0.22}\\[15pt] x&=&\displaystyle \frac{572}{0.22}\\[15pt] x&=&2,600\\ \end{array}$

dollars was invested at .
Example 4:

A banker invests of a client’s money at and the rest at annual interest rate. Both investments are for months. The interest earned is . How much money did the client make available to the banker?

Solution:

Preamble:

Let be the client’s money.
was invested at (like of ).
The rest of the money is .
Then is the amount invested at
(like of )

The time is months which means

year.

$Extra close brace or missing open brace\begin{array}{|c|c|c| c|c|c| c|c|c| r|r|r|}\hline &\!\hbox{Principal}\!&\cdot&\hbox{Rate}&\cdot&\!\hbox{Time}\!&=&\hbox{Interest}\\ \hline \quad}&0.3x&\cdot&0.05&\cdot&0.25&=&(0.05)(0.25)(0.3)x\\ \hline \quad&0.7x&\cdot&0.06&\cdot&0.25&=&\!\!(0.06)(0.25)(0.7x)\\ \hline \end{array}$

Equation:

$\) is only supported in math mode\begin{array}{rcl lll lll} \hbox{Interest (\)5%$)}+\hbox{interest ($6%\()}&=&712.5\\[3pt] (0.05)(0.25)(0.3)x+(0.06)(0.25)(0.7x)&=&712.5\\[3pt] (0.0125)(0.3)x+(0.015)(0.7x)&=&712.5\\[3pt] 0.00375x+0.01050x&=&712.5\\[3pt] 375x+1,050x&=&71,250,000\\[3pt] 1,425x&=&71,250,000\\[3pt] \displaystyle \frac{1,425}{1,425}x&=&\displaystyle \frac{71,250,000}{1,425}\\[10pt] x&=&\displaystyle \frac{71,250,000}{1,425}\\[10pt] x&=&50,000 \end{array}$

The client trusted the banker with .

Exercise 15

Armand invests at and Bonnie invests at simple interest for one year. What is the total interest earned?
Carina invests part of at and the rest at simple annual interest rate. Her annual interest income is . How much money did she invest at ?
An engineer borrowed some money at and more than that amount at simple annual interest rates. His (her) interest is for one year. How much did the engineer borrow at ?
A realtor invests of a client’s money at and the rest at simple annual interest, both for months. The interest earned is . How much money did the client make available to the realtor?

Solution:
Preamble:

Let be the total interest (not too useful in the present example).

Equation:

Total Interest= Interest from son added to interest from daughter.
Total interest
Solution:
Preamble:

Let be the amount invested at (for example ).
is the amount invested at

Note: could have been the account.

Equation:

$\) is only supported in math mode\begin{array}{rcl lll lll} \hbox{Interest (\)9%$)}+\hbox{interest ($7%\()}&=&475\\[5pt] 0.09x+0.07(6,000-x)&=&475\\[5pt] 0.09x+0.07(6,000)-(0.07)x&=&475\\[5pt] 0.09x+420-0.07x&=&475\\[5pt] 0.02x+420&=&475\\[5pt] 0.02x+420-420&=&475-420\\[5pt] -0.02x&=&-55\\[5pt] \displaystyle \frac{-0.02}{-0.02}x&=&\displaystyle \frac{-55}{-0.02}\\[15pt] x&=&\displaystyle \frac{5,500}{2}\\[15pt] x&=&2,750\\ \end{array}$

Carina invested at .
Solution:
Preamble:

: amount borrowed at (for example ).
: amount borrowed at

Equation:

$\) is only supported in math mode\begin{array}{rcl lll lll} \hbox{Interest (\)6%$)}+\hbox{interest ($10%\()}&=&1,450\\[5pt] 0.06x+0.1(x+2,500)&=&1,450\\[5pt] 0.06x+0.1x+250&=&1,450\\[5pt] 0.16x+250&=&1,450\\[5pt] 0.16x+250-250&=&1,450-250\\[5pt] 0.16x&=&1,200\\[5pt] \displaystyle \frac{0.16}{0.16}x&=&\displaystyle \frac{1,200}{0.16}\\[15pt] x&=&\displaystyle \frac{1,200}{0.16}\\[15pt] x&=&7,500\\ \end{array}$

The engineer borrowed dollars at .
Solution:
Preamble:

Let be the client’s money. (for example )
was invested at of .
Rest of money: .
is the amount invested at of

The time is months which means
year.

Equation:

$\) is only supported in math mode\begin{array}{rcl lll lll} \hbox{Interest (\)6%$)}+\hbox{interest ($11%\()}&=&1,068.75\\[5pt] (0.06)(0.5)(0.4)x+(0.11)(0.5)(0.6x)&=&1,068.75\\[5pt] (0.03)(0.4)x+(0.055)(0.6x)&=&1,068.5\\[5pt] 0.012x+0.033x&=&1,068.75\\[5pt] 0.045x&=&1,068.75\\[5pt] \displaystyle \frac{0.045}{0.045}x&=&\displaystyle \frac{1,068.75}{0.045}\\[15pt] x&=&\displaystyle \frac{1,068.75}{0.045}\\[15pt] x&=&23,750 \end{array}$

The client trusted the realtor with .

Solving Linear Equations. Investment Problems

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Examples

Exercise 15

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