**Assignment**

$\frac{25a+4b}{7y+4\cdot3+2}=9b$

What is the domain of the equation?

**Solution**

We must calculate for which values of $y$ it is forbidden to be equal to $\frac{25a+4b}{7y+4\cdot3+2}=9b$

For this equation, we can see that we have a rational function, so to calculate the domain, we have a restriction, which is that the denominator cannot be $0$. Therefore, we set the denominator to zero to determine which value $y$ cannot take:

$7y+12+2=0$

We proceed to solve the previous equation by isolating the variable $y$

$7y+14=0$

We move the $14$ to the right side and keep the corresponding sign

$7y=-14$

We divide by: $7$

$y=-2$

If $y$ is equal to: $-2$ then the denominator is equal to $0$ and the exercise has no solution

**Answer**

$y\ne-2$

**Assignment**

What is the domain of the equation?

$\frac{xyz}{2(3+y)+4}=8$

**Solution**

We need to calculate for which values of $y$ it is forbidden to be equal to zero

$2\left(3+y\right)+4=0$

Multiply by $2$ in both elements inside the parentheses

$6+2y+4=0$

Add them up

$10+2y=0$

Move the $10$ to the right side

$2y=-10$

Divide by $2$

$y=-5$

$y\ne-5$

If $y$ is equal to negative $5$, then the denominator equals $0$ and the exercise has no solution

**Answer**

$y\ne-5$

**Assignment**

$\frac{\sqrt{15}+34:z}{4y-12+8:2}=5$

What is the domain of the equation?

**Solution**

We must calculate for which Y it is forbidden to be equal to

$4y-12+8:2=0$

$4y-12+4=0$

We move the terms while keeping the corresponding signs

$4y=12-4$

$4y=8$

Divide by $4$

$y=2$

If $Y$ is equal to: $2$ then the denominator is equal to: $0$ and the exercise has no solution

$y\ne2$

**Answer**

$y\ne2$

**Assignment**

Solve the following equation:

$\frac{3}{(x+1)^2}+\frac{2x}{x+1}+x+1=3$

**Solution**

$\frac{3}{(x+1)^2}+\frac{2x}{x+1}+x+1=3$

Multiply by:

$\left(x+1\right)^2$

The domain of definition is

$x\ne-1$

$3+2x\left(x+1\right)+\left(x+1\right)^3=3$

Reduce by: $3$

$\left(x+1\right)\left\lbrack2x+\left(x+1\right)^2\right\rbrack=0$

$2x+\left(x+1\right)^2=0$

$2x+x^2+2x+1=0$

$x^2+4x+1=0$

$x_{1,2}=\frac{-4\pm\sqrt{4^2-4}}{2}$

$\frac{-4\pm\sqrt{12}}{2}=$

$-2\pm\frac{\sqrt{12}}{2}$

$-2\pm\frac{2\sqrt{3}}{2}=-2\pm\sqrt{3}$

**Answer**

$x=\sqrt{3}-2,-\sqrt{3}-2$

**Assignment**

Solve the following equation

$\frac{(2x+1)^2}{x+2}+\frac{(x+2)^2}{2x+1}=4.5x$

**Solution**

$\frac{(2x+1)^2}{x+2}+\frac{(x+2)^2}{2x+1}=4.5x$

Multiply by: $\left(x+2\right)\left(2x+1\right)$

The domain is $x\ne-2,-\frac{1}{2}$

$\left(2x+1\right)^3+\left(x+2\right)^3=4.5x\left(x+2\right)\left(2x+1\right)$

$\left(2x+1\right)\left(2x+1\right)^2+\left(x+2\right)\left(x+2\right)^2=4.5x(2x^2+5x+2)$

$\left(2x+1\right)\left(4x^2+4x+1\right)+\left(x+2\right)\left(x^2+4x+4\right)=9x^3+22.5x^2+9x$

Combine like terms

$9x^3+18x^2+18x+9=9x^3+22.5x^2+9x$

Divide by: $9$

$x^3+2x^2+2x+1=x^3+2.5x^2+x$

$0.5x^2-x-1=0$

Divide by: $0.5$

$x^2-2x-2=0$

$x_{1,2}=\frac{2\pm\sqrt{\left(-2\right)^2-4\cdot\left(-2\right)}}{2}$

$\frac{2\pm\sqrt{4+8}}{2}$

$\frac{2\pm\sqrt{12}}{2}$

$\frac{2\pm2\sqrt{3}}{2}=1\pm\sqrt{3}$

**Answer**

$x=1±\sqrt{3}$

**What does it mean for a function to be well-defined?**

A well-defined function means that it satisfies the definition of a function, which is:

To each element of a set $X$ (independent variable), which is called the domain of the function, corresponds a unique value from the set $Y$ (dependent variable), known as the codomain.

**What is the domain of a function?**

The domain of a function in mathematics is all the possible values that the independent variable $X$ can take, such that the function is well-defined when taking these values.

**What is the range of a function?**

The range, also called the image of a function, are those values that the dependent variable $Y$ takes, which depend on the set of numbers from the domain, hence the name dependent variable to the set $Y$.

**How is the domain of a function calculated?**

The domain of a function depends on the type of function you are working with, as some functions have certain restrictions or ambiguities for the function to exist, that is, to be well-defined.

For example: If we work with a rational function, our restriction for it to be a defined function is that the denominator cannot be equal to zero. Then we must check for which values of the independent variable this restriction is met.

If we work with a radical function in the real numbers, then the restriction is that we cannot have a negative number inside the radical. Similarly, we must observe for which values of the independent variable it is true that it is positive or equal to zero.

**How to calculate the domain of the following examples?**

**Example 1:**

**Assignment**

Determine the domain of the following equation:

$\frac{5x}{16-4x}=0$

We can see that it is a rational function, then we must determine for which values of $X$, our denominator is different from zero, for this we equal the denominator to zero

$16-4x=0$

And we proceed to solve this equation:

$-4x=-16$

We divide by $-4$ on the right side

$x=\frac{-16}{-4}$

$x=4$

Therefore, we conclude that when $x=4$ the denominator is equal to $0$, so the domain will be any number except $4$

**Answer:**

$x\ne4$

**Example 2:**

Determine the domain of the following equation:

$\sqrt{x+5}=0$

Here we can see that the equation is a radical function and our restriction is that what's inside the radical be positive or equal to zero, so let's see for which values of $X$ this is satisfied:

$x+5>0$

We solve

$x>-5$

We conclude that if the variable takes values equal to or greater than $-5$, it will give us the root of a positive number, that is, a defined function.

**Answer:**

$x>-5$