Solving a System of Equations with Given Sum and Product

Given a system of equations where the sum and product of two numbers are known, we can find the individual numbers by solving the resulting quadratic equation. This article explains how to solve such a system and provides a detailed step-by-step guide.

Problem Statement

Let the two numbers be x and y. The problem states that the sum of the two numbers is 8, and their product is -33. We can write this as the following system of equations:

x y 8 xy -33

Our goal is to find the values of x and y.

Solving the System of Equations

First, we express y in terms of x using the first equation:

y 8 - x

Next, we substitute this expression for y into the second equation:

x(8 - x) -33

Expanding the left side gives:

8x - x2 -33

Rearranging terms to form a standard quadratic equation:

-x2 8x 33 0

Multiplying through by -1 to simplify:

x2 - 8x - 33 0

Now we solve this quadratic equation using the quadratic formula:

x (-b ± √(b2 - 4ac)) / 2a

Here a 1, b -8, and c -33. Substituting these values into the quadratic formula:

x (8 ± √((-8)2 - 4(1)(-33))) / 2(1)

Calculating the discriminant:

√((-8)2 - 4(1)(-33)) √(64 132) √196 14

Now substituting back into the quadratic formula:

x (8 ± 14) / 2

This gives us two possible solutions for x:

x (8 14) / 2 22 / 2 11 x (8 - 14) / 2 -6 / 2 -3

Now we substitute these values back into the expression for y:

When x 11, y 8 - 11 -3 When x -3, y 8 - (-3) 11

Thus, the two numbers are 11 and -3.

Alternative Methods

Here are a couple of alternative methods to solve the same system of equations:

Method 1

Let the two numbers be x and y.

Sum of the numbers: xy 8 Product of the numbers: xy -33

Express y in terms of x

y 8 - x

Substitute into the second equation:

(8 - x)(x) -33

Expand and rearrange:

8x - x2 -33

x2 - 8x - 33 0

Solve the quadratic equation:

x (-b ± √(b2 - 4ac)) / 2a

Here a 1, b -8, and c -33. Substituting:

x (8 ± 14) / 2

This gives us x 11 and x -3.

Method 2

Given: xy 8 and xy -33.

Let b -33/a and substitute into the first equation:

a × (-33/a) 8

Now solve:

-33 8a - a2

a2 - 8a - 33 0

Solve the quadratic equation:

a (-b ± √(b2 - 4ac)) / 2a

Here a 1, b -8, and c -33. Substituting:

a (8 ± 14) / 2

This gives us a 11 and a -3.

Thus, the numbers are 11 and -3.

Verification

To verify, we check that the solutions satisfy the original equations:

11 (-3) 8 11 × (-3) -33

The values are correct.

Conclusion

In conclusion, we have successfully solved the system of equations where the sum is 8 and the product is -33. The two numbers are 11 and -3. This problem involves quadratic equations and demonstrates the practical application of algebraic techniques in solving real-world problems.

Keywords

quadratic equation, sum and product of numbers, system of equations