Solving Homework Equations for x & y with 2x + 3y + 1 = 0

• Buffu
In summary, the equation attempts to solve for x when z=2x+3y, but fails because the following conditions are not met: dz/dx does not equal f(x) and 1-z/2z+1 does not equal 0.
Buffu

Homework Statement

##(2x + 3y + 1)dx + (4x + 6y + 1) dy = 0##

##y(-2) = 2##

The Attempt at a Solution

Let ##z = 2x + 3y##
then ##z^\prime = 2 + 3y^prime##

##\displaystyle \dfrac{(z + 1)}{2z + 1} + \dfrac13\left({dz \over dx} - 2\right) = 0##
##\dfrac{dz}{dx} = \dfrac{z- 1}{2z + 1}##

##x = 2z + 3\log(1 - z) + C = 2(2x + 3y) + 3\log|(1 - 2x - 3y)| + C##

Plugging the values, I get ##C = -6##

So, ##0 = x + 2y + log|3y + 2x - 1| - 2##

But the answer given is ##0 = x + 2y \color{red}{-} log|3y + 2x - 1| - 2##.

Is the answer in the book wrong ?

Buffu said:

Homework Statement

##(2x + 3y + 1)dx + (4x + 6y + 1) dy = 0##

##y(-2) = 2##

The Attempt at a Solution

Let ##z = 2x + 3y##
then ##z^\prime = 2 + 3y^prime##

##\displaystyle \dfrac{(z + 1)}{2z + 1} + \dfrac13\left({dz \over dx} - 2\right) = 0##
##\dfrac{dz}{dx} = \dfrac{z- 1}{2z + 1}##

##x = 2z + 3\log(1 - z) + C = 2(2x + 3y) + 3\log|(1 - 2x - 3y)| + C##

Plugging the values, I get ##C = -6##

So, ##0 = x + 2y + log|3y + 2x - 1| - 2##

But the answer given is ##0 = x + 2y \color{red}{-} log|3y + 2x - 1| - 2##.

Is the answer in the book wrong ?
@Buffu I looked over your solution=I don't see any mistakes.

Hi Buffu:

I think your equation where dz/dx = f(x)is wrong.
You want to do two things first.
1. Substitute z=2x+3y into the given equation.
2. Substitute an expression for dy=g(dz) into the result of step 1.

Hope that helps.

Regards,
Buzz

Buzz Bloom said:
Hi Buffu:

I think your equation where dz/dx = f(x)is wrong.
You want to do two things first.
1. Substitute z=2x+3y into the given equation.
2. Substitute an expression for dy=g(dz) into the result of step 1.

Hope that helps.

Regards,
Buzz
##\displaystyle \dfrac{(z + 1)}{2z + 1} + \dfrac13\left({dz \over dx} - 2\right) = 0 \iff \dfrac{(3z + 3) - 4z - 2}{2z + 1} + {dz \over dx} = 0\iff \dfrac{ 1 - z}{2z + 1} + {dz \over dx} = 0 ##

Buzz Bloom

1. How do I solve for x and y in the equation 2x + 3y + 1 = 0?

To solve for x and y in this equation, you can use the method of substitution or the method of elimination. In substitution, you solve one variable in terms of the other and then substitute that expression into the equation to solve for the remaining variable. In elimination, you manipulate the equation to eliminate one variable and then solve for the remaining variable.

2. Can I use any method to solve this equation?

Yes, you can use any method that is appropriate for solving systems of equations. The most commonly used methods are substitution, elimination, and graphing. It is important to choose the method that you are most comfortable with and that will give you the most accurate solution.

3. What is the importance of solving for x and y in this equation?

Solving for x and y in this equation allows you to find the specific values of x and y that satisfy the equation. This is important in order to understand the relationship between the variables and to make predictions or solve problems using the equation.

4. What are the steps for solving this equation?

The steps for solving this equation depend on the method you choose. However, the general steps include isolating one variable, substituting or eliminating to solve for the other variable, and then checking your solution by plugging in the values for x and y into the original equation.

5. Can I use this equation to solve real-world problems?

Yes, this equation can be used to solve real-world problems involving two variables. It can represent a variety of situations, such as finding the number of solutions for a given system, determining the intersection point of two lines, or calculating the relationship between two quantities.

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