# 3 variables 3 equations (2 linear 1 quadratic)

• Appleton
In summary, the book suggests that in order to solve a quadratic equation with three unknowns, express the equation in terms of the third unknown from the first two linear equations. This is done by solving the two linear equations and substituting the results into the last equation. If z is known to be equal to 2, then this equation can be solved easily.
Appleton

## Homework Statement

Find three numbers x,y,z satisfying the following equations:
9x-6y-10z=1
-6x+4y+7z=0
x2+y2+z2=9

## The Attempt at a Solution

The book suggests that the best approach is to express 2 unknowns in terms of the third unknown from the first 2 linear equations and, by substituting these expressions in the last equation, obtain a quadratic equation for the third unknown.
So i try this:
4(9x-6y-10z=1)= 36x-24y-40z=4
6(-6x+4y+7z=0)= -36x+24y+42z=0
(36x-24y-40z=4) + (-36x+24y+42z=0)= 2z=4
z=2
I figure I've gone wrong somewhere because this is not the strategy outlined in the book (it's also not the correct answer for z according to the book)

Appleton said:

## Homework Statement

Find three numbers x,y,z satisfying the following equations:
9x-6y-10z=1
-6x+4y+7z=0
x2+y2+z2=9

## The Attempt at a Solution

The book suggests that the best approach is to express 2 unknowns in terms of the third unknown from the first 2 linear equations and, by substituting these expressions in the last equation, obtain a quadratic equation for the third unknown.
So i try this:
4(9x-6y-10z=1)= 36x-24y-40z=4
6(-6x+4y+7z=0)= -36x+24y+42z=0
(36x-24y-40z=4) + (-36x+24y+42z=0)= 2z=4
z=2
I figure I've gone wrong somewhere because this is not the strategy outlined in the book (it's also not the correct answer for z according to the book)

If you have stated the problem correctly, you are right: z = 2, no matter what the book says. In fact, there are two solutions; they both have z = 2 but have different x,y values.

Thanks for your reply. On further inspection I realize that the book does in fact say that z is equal to 2. However this still leaves me a bit confused about the purpose of the quadratic equation since it seems it is superfluous to the requirements for providing a solution.

The problem is question 1A, page 44 (in the pdf digital copy. Or 22 in the hard copy)
of this book:
http://www.scribd.com/doc/39412845/Mathematics-and-Plausible-Reasoning-Vol1-Induction-and-Analogy-in-Mathematics-Polya-1954

Last edited by a moderator:
Appleton said:
Thanks for your reply. On further inspection I realize that the book does in fact say that z is equal to 2. However this still leaves me a bit confused about the purpose of the quadratic equation since it seems it is superfluous to the requirements for providing a solution.

The problem is question 1A, page 44 (in the pdf digital copy. Or 22 in the hard copy)
of this book:
http://www.scribd.com/doc/39412845/Mathematics-and-Plausible-Reasoning-Vol1-Induction-and-Analogy-in-Mathematics-Polya-1954

If you know z = 2 you can put this into the first and third equations (for example), to get two equations in the two unknowns x and y. One of the equations is linear, so that helps.

Last edited by a moderator:
Thanks for the help I'll try that.

## 1. What is a 3 variable 3 equation (2 linear 1 quadratic) problem?

A 3 variable 3 equation problem refers to a system of equations that has 3 variables and 3 equations, with two of the equations being linear and one being quadratic. This type of problem can be solved by using algebraic techniques such as elimination or substitution.

## 2. How do you solve a 3 variable 3 equation (2 linear 1 quadratic) problem?

To solve a 3 variable 3 equation problem, first eliminate one variable by using one of the linear equations. Then, substitute the resulting equation into the other linear equation to eliminate another variable. Finally, substitute the values for the two eliminated variables into the quadratic equation to solve for the remaining variable.

## 3. Can a 3 variable 3 equation (2 linear 1 quadratic) problem have multiple solutions?

Yes, a 3 variable 3 equation problem can have multiple solutions. This can occur when the three equations have three different planes of intersection, resulting in a unique point of intersection for each solution.

## 4. Is it possible for a 3 variable 3 equation (2 linear 1 quadratic) problem to have no solution?

Yes, it is possible for a 3 variable 3 equation problem to have no solution. This can happen when the three equations do not intersect at a common point, indicating that there is no solution that satisfies all three equations simultaneously.

## 5. What are some real-life applications of 3 variable 3 equation (2 linear 1 quadratic) problems?

3 variable 3 equation problems can be used to model and solve various real-life situations, such as optimizing production levels in a manufacturing plant, determining the break-even point for a business, or calculating the trajectory of a projectile. These problems can also be found in physics, engineering, and economics, among other fields.

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