Solving for Numbers Given a Difference and Product

Introduction

When dealing with equations, it's common to encounter scenarios where two conditions are given: a difference between two numbers and their product. This article will guide you through solving such a problem, specifically when a number is 3 more than another and their product is 40.

Problems and Solutions

The given conditions are that one number is 3 more than another (x y 3), and their product is 40 (xy 40). Let's break down the problem and find the possible pairs of numbers that satisfy these conditions.

Step-by-step Solution

Step 1: Set up the equations based on the given conditions.

We start with the two equations:

"x y 3 "xy 40

Substitute the expression for ( x ) from the first equation into the second:

"y(y 3) 40"

This simplifies to:

"y^2 3y - 40 0"

Solving the Quadratic Equation

To solve the quadratic equation "y^2 3y - 40 0", we can use the quadratic formula:

"y frac{-b pm sqrt{b^2 - 4ac}}{2a}"

Here, ( a 1 ), ( b 3 ), and ( c -40 ).

"y frac{-3 pm sqrt{3^2 - 4 cdot 1 cdot -40}}{2 cdot 1}"

Calculate the discriminant:

"3^2 - 4 cdot 1 cdot -40 9 160 169"

Now, substitute back into the quadratic formula:

"y frac{-3 pm 13}{2}"

This gives us two possible values for ( y ):

"y frac{-3 13}{2} 5" "y frac{-3 - 13}{2} -8"

Finding the Corresponding Values for ( x )

Now that we have the possible values for ( y ), we can find the corresponding values for ( x ):

For ( y 5 ):"x y 3 5 3 8". So one pair is ( 8, 5 ). For ( y -8 ):"x y 3 -8 3 -5". So the other pair is ( -5, -8 ).

Therefore, the possible pairs of numbers are ( 8, 5 ) and ( -5, -8 ).