15.2: Examples

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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 $x$ 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 $P$) the more your reward.

2) The more money you get per $\!100\!$ dollars invested (rate of interest, or just rate $r$), the higher your reward.

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

Thus \[\fbox{\parbox[t]{2.0in}{\begin{center}\boldmath \Large I\ =\ P\ r\ t\end{center}}}\]

Examples

Example 1:

Your son invests $\$20,000$ at $5.5$% and your daughter invests $\$15,000$ at $3.5$% simple interest for one year. What is the total interest earned?

Solution:

Preamble:

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

\[\begin{array}{|c|c|c| c|c|c| r|r|r| r|r|r|}\hline &\hbox{Principal}&\cdot&\hbox{Rate}&=&\hbox{Interest}\\ \hline \hbox{Son}&20,000&\cdot&0.055&=&1,100\\ \hline \hbox{Daughter}&15,000&\cdot&0.035&=&525\\ \hline \end{array}\]

Equation:

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

Total interest $=1,100+525=\$1,625$

Example 2:

Your grandmother invests part of $\$4,500$ at $8\%$ and the rest at $6\%$. Her annual interest income from both accounts is $\$290$. How much money did she invest at $8\%$?

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 $x$ be the amount invested at $8\%$ (like $\$1,000$).

Then $4,500-x$ is the amount invested at $6\%$

(like $4,500-1,000$)

You have the option to chose $x$ as the $8$% account instead of the $4\%$ account.

\[\begin{array}{|c|c|c| c|c|c| r|r|r| r|r|r|}\hline &\hbox{Principal}&\cdot&\hbox{Rate}&=&\hbox{Interest}\\ \hline \quad &x&\cdot&0.08&=&0.08x\\ \hline \quad &4,500-x&\cdot&0.06&=&0.06(4,500-x)\\ \hline \end{array}\]

Equation:

$\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}$

$1,000$ dollars was invested at $8\%$.

Example 3:

An entrepreneur borrowed some money at $4\%$ and $\$1,400$ more than hat account at $7\%$ annually. His interest is $\$768$ for two years. How much did the entrepreneur borrow at $7\%$?

Solution:

Preamble:

Let $x$ be the amount borrowed at $4\%$ (like $\$1,000$).
Then $x+1,400$ is the amount borrowed at $7\%$
(like $1,000+1,400$)

\[\begin{array}{|c|c|c| c|c|c| c|c|c| r|r|r|}\hline &\!\hbox{Principal}\!&\cdot&\hbox{Rate}&\cdot&\!\hbox{Time}\!&=&\hbox{Interest}\\ \hline \quad &x&\cdot&0.04&\cdot&2&=&(0.04)(2)x\\ \hline \quad &\!x\!+\!1,400\!&\cdot&0.07&\cdot&2&=&\!(0.07)(2)(x\!+\!1,400)\\ \hline \end{array}\]

Equation:

$\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}$

$2,600+1,400=4,000$ dollars was invested at $7\%$.
Example 4:

A banker invests $30\%$ of a client’s money at $5\%$ and the rest at $6\%$ annual interest rate. Both investments are for $3$ months. The interest earned is $\$712.50$. How much money did the client make available to the banker?

Solution:

Preamble:

Let $x$ be the client’s money.
$0.30x$ was invested at $5\%$ (like $30\%$ of $\$1,000=300$).
The rest of the money is $100\%x-30\%x=70\%x$.
Then $0.70x$ is the amount invested at $6\%$
(like $70\%$ of $1,000=700$)

The time is $3$ months which means

$\displaystyle \frac{3\hbox{ months}}{1}\cdot \frac{1 \hbox{ year}}{12\hbox{ months}}=\frac{1}{4}\hbox{ year}=0.25$ year.

\[\begin{array}{|c|c|c| c|c|c| c|c|c| r|r|r|}\hline &\!\hbox{Principal}\!&\cdot&\hbox{Rate}&\cdot&\!\hbox{Time}\!&=&\hbox{Interest}\\ \hline \quad}&0.3x&\cdot&0.05&\cdot&0.25&=&(0.05)(0.25)(0.3)x\\ \hline \quad&0.7x&\cdot&0.06&\cdot&0.25&=&\!\!(0.06)(0.25)(0.7x)\\ \hline \end{array}\]

Equation:

$\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 $\$50,000$.

Exercise 15

Armand invests $\$10,000$ at $2.5\%$ and Bonnie invests $\$5,000$ at $4\%$ simple interest for one year. What is the total interest earned?
Carina invests part of $\$6,000$ at $9\%$ and the rest at $7\%$ simple annual interest rate. Her annual interest income is $\$475$. How much money did she invest at $9\%$?
An engineer borrowed some money at $6\%$ and $\$2,500$ more than that amount at $10\%$ simple annual interest rates. His (her) interest is $\$1,450$ for one year. How much did the engineer borrow at $6\%$?
A realtor invests $40\%$ of a client’s money at $6\%$ and the rest at $11\%$ simple annual interest, both for $6$ months. The interest earned is $\$1,068.75$. How much money did the client make available to the realtor?

Solution:
Preamble:

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

\[\begin{array}{|c|c|c| c|c|c| r|r|r| r|r|r|}\hline &\hbox{Principal}&\cdot&\hbox{Rate}&=&\hbox{Interest}\\ \hline \hbox{Armand}&10,000&\cdot&0.025&=&250\\ \hline \hbox{Bonnie}&5,000&\cdot&0.04&=&200\\ \hline \end{array}\]

Equation:

Total Interest= Interest from son added to interest from daughter.
Total interest $=250+200=\$450$
Solution:
Preamble:

Let $x$ be the amount invested at $9\%$ (for example $\$1,000$).
$6,000-x$ is the amount invested at $7\%$ $6,000-1,000$

Note: $x$ could have been the $7\%$ account.

\[\begin{array}{|c|c|c| c|c|c| r|r|r| r|r|r|}\hline &\hbox{Principal}&\cdot&\hbox{Rate}&=&\hbox{Interest}\\ \hline \quad&x&\cdot&0.09&=&0.09x\\ \hline \quad&6,000-x&\cdot&0.07&=&0.07(6,000-x)\\ \hline \end{array}\]

Equation:

$\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 $2,750$ at $9\%$.
Solution:
Preamble:

$x$: amount borrowed at $6\%$ (for example $\$1,000$).
$x+2,500$: amount borrowed at $10\%$ $1,000+1,500$

\[\begin{array}{|c|c|c| c|c|c| c|c|c| r|r|r|}\hline &\!\hbox{Principal}\!&\cdot&\hbox{Rate}&\cdot&\!\hbox{Time}\!&=&\hbox{Interest}\\ \hline \quad&x&\cdot&0.06&\cdot&1&=&(0.06)x\\ \hline \quad&\!x\!+\!2,500\!&\cdot&0.1&\cdot&1&=&(0.1)(x\!+\!2,500)\\ \hline \end{array}\]

Equation:

$\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 $7,500$ dollars at $6\%$.
Solution:
Preamble:

Let $x$ be the client’s money. (for example $\$1,000$)
$0.40x$ was invested at $6\%$ $40\%$ of $\$1,000=400$.
Rest of money: $100\%x-40\%x=60\%x$.
$0.60x$ is the amount invested at $11\%$ $60\%$ of $1,000=600$

The time is $6$ months which means
$\displaystyle \frac{6\hbox{ months}}{1}\cdot \frac{1 \hbox{ year}}{12\hbox{ months}}=\frac{1}{2}\hbox{ year}=0.5$ year.

\[\begin{array}{|c|c|c| c|c|c| c|c|c| r|r|r|}\hline &\!\hbox{Principal}\!&\cdot&\hbox{Rate}&\cdot&\hbox{Time}&=&\hbox{Interest}\\[5pt] \hline \quad&0.4x&\cdot&0.06&\cdot&0.5&=&(0.06)(0.5)(0.4)x\\[5pt] \hline \quad&0.6x&\cdot&0.11&\cdot&0.5&=&\!\!(0.11)(0.5)(0.6x)\\ \hline \end{array}\]

Equation:

$\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 $\$23,750$.

Solving Linear Equations. Investment Problems

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Examples

Exercise 15

Supplemental homework