Problem 102 - College Algebra

Question

The group should write four different word problems that can be solved using a system of linear equations in two variables. All of the problems should be on different topics. The group should turn in the four problems and their algebraic solutions.

Step-by-Step Solution

Answer

The problems created touched on diverse topics: family income, fruit prices, geometry and investment, all solvable using a system of two linear equations.

Step 1: Problem 1 - Family Income

Let \( X \) denote mother's salary and \( Y \) denotes father's salary. If both together bring home $5000 a month and father earns $700 more than the mother, the system of equations will be: \( X + Y = 5000 \) and \( Y = X + 700 \). By substituting the second equation in the first, the values of \( X \) (mother's salary) and \( Y \) (father's salary) will be found.

Step 2: Problem 2 - Fruit Prices

Suppose \( X \) denote the price of a kilo of apples and \( Y \) the price of a kilo of strawberries. If 3 kilos of apples and 2 kilos of strawberries cost $13 and 1 kilo of apples and 4 kilos of strawberries cost $14, the system of equations will look like this: \( 3X + 2Y = 13 \) and \( X + 4Y = 14 \). By solving this system, we will find the price per kilo for each type of fruit.

Step 3: Problem 3 - Geometry Problem

Let \( X \) denote the length of a rectangle and \( Y \) its width. If the perimeter of the rectangle is 24 units and the length is twice the width, the system of equations will be: \( 2X + 2Y = 24 \) and \( X = 2Y \). Solving this system will determine the dimensions of the rectangle.

Step 4: Problem 4 - Investment Problem

Suppose a person invests \( X \) dollars in a savings account and \( Y \) dollars in a bond. If the total amount invested is $10,000 and the bond yields three times more than the saving account, we can express this as: \( X + Y = 10000 \) and \( Y = 3X \). Solving this system will give the amount invested in each instrument.