Math Problem Statement

3-2zx-4zy=0 2-z4x-2zy=0 xx+4xy+yy=66

Solution

We have the following system of equations:

(1)32zx4zy=0\text{(1)} \quad 3 - 2zx - 4zy = 0 (2)24zx2zy=0\text{(2)} \quad 2 - 4zx - 2zy = 0 (3)x2+4xy+y2=66\text{(3)} \quad x^2 + 4xy + y^2 = 66

Let's break these down step by step.

Step 1: Solve for zz in terms of xx and yy from the first two equations.

From the first equation, we can isolate zz: 32zx4zy=03 - 2zx - 4zy = 0 2zx+4zy=32zx + 4zy = 3 z(2x+4y)=3z(2x + 4y) = 3 z=32x+4yz = \frac{3}{2x + 4y}

Now, from the second equation: 24zx2zy=02 - 4zx - 2zy = 0 4zx+2zy=24zx + 2zy = 2 z(4x+2y)=2z(4x + 2y) = 2 z=24x+2y=12x+yz = \frac{2}{4x + 2y} = \frac{1}{2x + y}

Step 2: Set the two expressions for zz equal.

Since both expressions represent zz, we can set them equal to each other: 32x+4y=12x+y\frac{3}{2x + 4y} = \frac{1}{2x + y}

Step 3: Solve for xx and yy.

Cross-multiply to solve: 3(2x+y)=1(2x+4y)3(2x + y) = 1(2x + 4y) 6x+3y=2x+4y6x + 3y = 2x + 4y Simplify: 6x2x=4y3y6x - 2x = 4y - 3y 4x=y4x = y

So, we now know that y=4xy = 4x.

Step 4: Substitute y=4xy = 4x into the third equation.

Substitute y=4xy = 4x into x2+4xy+y2=66x^2 + 4xy + y^2 = 66: x2+4x(4x)+(4x)2=66x^2 + 4x(4x) + (4x)^2 = 66 x2+16x2+16x2=66x^2 + 16x^2 + 16x^2 = 66 33x2=6633x^2 = 66 x2=2x^2 = 2 x=±2x = \pm \sqrt{2}

Step 5: Find the value of yy.

Since y=4xy = 4x, we substitute x=±2x = \pm \sqrt{2}: y=4(±2)=±42y = 4(\pm \sqrt{2}) = \pm 4\sqrt{2}

Step 6: Find zz.

Now, substitute x=±2x = \pm \sqrt{2} and y=±42y = \pm 4\sqrt{2} into one of the expressions for zz. Using z=32x+4yz = \frac{3}{2x + 4y}, we get: z=32(±2)+4(±42)z = \frac{3}{2(\pm \sqrt{2}) + 4(\pm 4\sqrt{2})} z=322±162=3(2±16)2z = \frac{3}{2\sqrt{2} \pm 16\sqrt{2}} = \frac{3}{(2 \pm 16)\sqrt{2}} Thus, z=3182z = \frac{3}{18\sqrt{2}} or z=3142z = \frac{3}{-14\sqrt{2}}.

Simplifying: z=162orz=1142z = \frac{1}{6\sqrt{2}} \quad \text{or} \quad z = \frac{-1}{14\sqrt{2}}

Final Solutions:

The possible solutions for (x,y,z)(x, y, z) are:

  1. x=2,y=42,z=162x = \sqrt{2}, y = 4\sqrt{2}, z = \frac{1}{6\sqrt{2}}
  2. x=2,y=42,z=1142x = -\sqrt{2}, y = -4\sqrt{2}, z = \frac{-1}{14\sqrt{2}}

Let me know if you need more clarification or a step-by-step breakdown of the solution!


Here are some related questions for you:

  1. How do you approach solving systems of nonlinear equations?
  2. How do you handle substitutions and eliminations in a system like this?
  3. Can you solve similar problems when the equations involve higher degrees (like cubic terms)?
  4. What’s the importance of symmetry in such systems of equations?
  5. How do you verify that your solutions are correct after solving?

Tip: When solving nonlinear systems like this, always check if substitution or elimination works best based on the structure of the equations.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Systems of Nonlinear Equations
Algebraic Manipulation
Substitution
Quadratic Equations

Formulas

z = 3/(2x + 4y)
z = 2/(4x + 2y)
x^2 + 4xy + y^2 = 66

Theorems

Solving systems of nonlinear equations
Substitution method for solving equations

Suitable Grade Level

Grades 11-12