EDGE Math 8

Investment and Mixture Problems

Problems involving money and mixtures can also be solved using systems of two linear equations in two variables. The following are some examples:

Example 1

Mrs. Richy invested ₱100,000 in two separate banks. One bank gave an 8% annual simple interest while the other gave a 10% annual simple interest. If Mrs. Richy's total annual interest from these investments was ₱8700, how much was invested at each rate?

Solution

Represent each amount (in pesos) invested at each rate. Let $x = amount invested at 8 %$ and $y = amount invested at 10 % .$

Use the formula $I = P r t$ to solve the given problem. In the formula, I is the simple interest, P is the principal or the amount of investment, r is the simple interest rate, and t is the length of time in years.

Then set up two equations involving the variables x and y.

First equation: amount invested at 8% (x) + amount invested at 10% (y) = ₱100,000, or

x + y = 100 000.

Second equation (using the formula

I = P r t,

where

t = 1) :

annual simple interest at 8% (or 0.08) + annual simple interest at 10% (or 0.10) = ₱8700, or

0.08 x + 0.10 y = 8700.

The following is the resulting system of equations.

{\begin{array}{l} x + y = 100 000 \\ 0.08 x + 0.10 y = 8700 \end{array}

Using the elimination method, multiply both sides of the first equation by $- 0.08,$ and then add the result to the second equation.

\begin{array}{l} - 0.08 x - 0.08 y = - 8000 \\ \underline{+ 0.08 x + 0.10 y = 8700} \\ 0.02 y = 700 \end{array}

Solve for y in the resulting equation.

$0.02 y$	$= 700$
$y$	$= \frac{700}{0.02}$
$y$	$= 35 000$

Then substitute $35 000$ for y in either the first or second equation of the system. Using the first equation, you will get

$x + y$	$= 100 000$
$x + 35 000$	$= 100 000$
$x$	$= 100 000 - 35 000$
$x$	$= 65 000.$

The solution of the system is $(65 000, 35 000) .$ The amount invested at 8% was ₱65,000 and the amount invested at 10% was ₱35,000. As an exercise, check if the two numbers satisfy the conditions of the given problem.