solve each equation.
1. $\frac{1}{x + 4} = 5$
2. $\frac{1}{x - 3} = 6$
3. $\frac{1}{2x - 3} + \frac{1}{5 - 2x}$
4. $2 - \frac{1}{x + 3} = \frac{1}{x + 3}$
5. $\frac{1}{x + 3} + \frac{1}{x - 3} = \frac{6}{x^2 - 9}$
6. $\frac{4}{x + 2} = \frac{x^2}{x + 2}$

Question

Accepted Answer

### Problem 1: Solve $\boldsymbol{\frac{1}{x + 4}=5}$
# Explanation:
## Step 1: Multiply both sides by $x + 4$
To eliminate the denominator, we multiply both sides of the equation $\frac{1}{x + 4}=5$ by $x + 4$ (assuming $x
eq - 4$ to avoid division by zero).
$$
\frac{1}{x + 4}\times(x + 4)=5\times(x + 4)
$$
This simplifies to:
$$
1 = 5(x + 4)
$$
## Step 2: Expand the right - hand side
Using the distributive property $a(b + c)=ab+ac$, where $a = 5$, $b=x$ and $c = 4$, we get:
$$
1=5x+20
$$
## Step 3: Solve for $x$
Subtract 20 from both sides of the equation:
$$
1-20=5x+20 - 20
$$
$$
- 19 = 5x
$$
Then divide both sides by 5:
$$
x=-\frac{19}{5}=-3.8
$$
We need to check if this solution makes the original denominator non - zero. When $x =-\frac{19}{5}$, $x + 4=-\frac{19}{5}+4=-\frac{19}{5}+\frac{20}{5}=\frac{1}{5}
eq0$. So the solution is valid.
# Answer:
$x =-\frac{19}{5}$

### Problem 2: Solve $\boldsymbol{\frac{1}{x - 3}=6}$
# Explanation:
## Step 1: Multiply both sides by $x - 3$
Multiply both sides of the equation $\frac{1}{x - 3}=6$ by $x - 3$ (assuming $x
eq3$ to avoid division by zero).
$$
\frac{1}{x - 3}\times(x - 3)=6\times(x - 3)
$$
This simplifies to:
$$
1 = 6(x - 3)
$$
## Step 2: Expand the right - hand side
Using the distributive property $a(b - c)=ab - ac$, where $a = 6$, $b=x$ and $c = 3$, we get:
$$
1=6x-18
$$
## Step 3: Solve for $x$
Add 18 to both sides of the equation:
$$
1 + 18=6x-18 + 18
$$
$$
19 = 6x
$$
Divide both sides by 6:
$$
x=\frac{19}{6}\approx3.17
$$
Check the denominator: When $x=\frac{19}{6}$, $x - 3=\frac{19}{6}-\frac{18}{6}=\frac{1}{6}
eq0$. So the solution is valid.
# Answer:
$x=\frac{19}{6}$

### Problem 3: Simplify $\boldsymbol{\frac{1}{2x - 3}+\frac{1}{5 - 2x}}$
# Explanation:
## Step 1: Notice the relationship between denominators
We can rewrite $5 - 2x$ as $-(2x - 5)=-(2x-3 + 2)=-(2x - 3)-2$, but a simpler way is to note that $5 - 2x=-(2x - 5)=-(2x-3 + 2)$ is not necessary. Instead, we can factor out a negative sign from the second denominator: $5 - 2x=-(2x - 5)=-(2x-3 + 2)$ is incorrect. The correct way is $5 - 2x=-(2x - 5)=-(2x-3 - 2)$ no, better: $5 - 2x=-(2x - 5)=-(2x-3 - 2)$ is wrong. Let's do it as $5 - 2x=-(2x - 5)=-(2x-3 - 2)$ no. The correct approach is:
$\frac{1}{2x - 3}+\frac{1}{5 - 2x}=\frac{1}{2x - 3}-\frac{1}{2x - 5}$ (since $\frac{1}{5 - 2x}=\frac{1}{-(2x - 5)}=-\frac{1}{2x - 5}$)
## Step 2: Find a common denominator
The common denominator of $2x - 3$ and $2x - 5$ is $(2x - 3)(2x - 5)$
$$
\frac{1}{2x - 3}-\frac{1}{2x - 5}=\frac{(2x - 5)-(2x - 3)}{(2x - 3)(2x - 5)}
$$
## Step 3: Simplify the numerator
Simplify the numerator: $(2x - 5)-(2x - 3)=2x-5 - 2x + 3=-2$
So the expression simplifies to $\frac{-2}{(2x - 3)(2x - 5)}=\frac{2}{(3 - 2x)(2x - 5)}$ (we can also multiply the denominator out: $(2x - 3)(2x - 5)=4x^{2}-10x-6x + 15=4x^{2}-16x + 15$, so $\frac{-2}{4x^{2}-16x + 15}$)
# Answer:
$\frac{-2}{(2x - 3)(2x - 5)}$ (or $\frac{2}{(3 - 2x)(2x - 5)}$ or $\frac{-2}{4x^{2}-16x + 15}$)

### Problem 4: Solve $\boldsymbol{2-\frac{1}{x + 3}=\frac{1}{x + 3}}$
# Explanation:
## Step 1: Add $\frac{1}{x + 3}$ to both sides
Add $\frac{1}{x + 3}$ to both sides of the equation $2-\frac{1}{x + 3}=\frac{1}{x + 3}$
$$
2-\frac{1}{x + 3}+\frac{1}{x + 3}=\frac{1}{x + 3}+\frac{1}{x + 3}
$$
Simplifying both sides, we get:
$$
2=\frac{2}{x + 3}
$$
## Step 2: Multiply both sides by $x + 3$
Multiply both sides by $x + 3$ (assuming $x
eq - 3$):
$$
2(x + 3)=\frac{2}{x + 3}\times(x + 3)
$$
$$
2x+6 = 2
$$
## Step 3: Solve for $x$
Subtract 6 from both sides:
$$
2x=2 - 6=-4
$$
Divide both sides by 2:
$$
x=-2
$$
Check the denominator: When $x=-2$, $x + 3=-2 + 3 = 1
eq0$. So the solution is valid.
# Answer:
$x=-2$

### Problem 5: Solve $\boldsymbol{\frac{1}{x + 3}+\frac{1}{x - 3}=\frac{6}{x^{2}-9}}$
# Explanation:
## Step 1: Factor the denominator on the right
We know that $x^{2}-9=(x + 3)(x - 3)$ by the difference of squares formula $a^{2}-b^{2}=(a + b)(a - b)$ where $a = x$ and $b = 3$
## Step 2: Find a common denominator
The common denominator of $x + 3$, $x - 3$ and $(x + 3)(x - 3)$ is $(x + 3)(x - 3)$
Multiply the first fraction $\frac{1}{x + 3}$ by $\frac{x - 3}{x - 3}$ and the second fraction $\frac{1}{x - 3}$ by $\frac{x + 3}{x + 3}$
$$
\frac{1\times(x - 3)}{(x + 3)(x - 3)}+\frac{1\times(x + 3)}{(x + 3)(x - 3)}=\frac{6}{(x + 3)(x - 3)}
$$
## Step 3: Simplify the left - hand side
Simplify the left - hand side numerator: $(x - 3)+(x + 3)=x-3+x + 3 = 2x$
So the equation becomes:
$$
\frac{2x}{(x + 3)(x - 3)}=\frac{6}{(x + 3)(x - 3)}
$$
## Step 4: Solve for $x$
Since the denominators are the same (and $x
eq\pm3$ to avoid division by zero), we can set the numerators equal:
$2x=6$
Divide both sides by 2:
$x = 3$
But we need to check if $x = 3$ is a valid solution. When $x = 3$, the denominators $x + 3=6$, $x - 3 = 0$ and $x^{2}-9 = 0$. So $x = 3$ makes the original equation undefined. Therefore, there is no solution.
# Answer:
No solution

### Problem 6: Solve $\boldsymbol{\frac{4}{x + 2}=\frac{x^{2}}{x + 2}}$
# Explanation:
## Step 1: Multiply both sides by $x + 2$
Multiply both sides of the equation $\frac{4}{x + 2}=\frac{x^{2}}{x + 2}$ by $x + 2$ (assuming $x
eq - 2$ to avoid division by zero)
$$
\frac{4}{x + 2}\times(x + 2)=\frac{x^{2}}{x + 2}\times(x + 2)
$$
This simplifies to:
$$
4=x^{2}
$$
## Step 2: Solve the quadratic equation
We can rewrite $x^{2}=4$ as $x^{2}-4 = 0$, and factor it using the difference of squares formula: $(x - 2)(x + 2)=0$
So $x - 2=0$ or $x + 2=0$
Which gives $x = 2$ or $x=-2$
## Step 3: Check for valid solutions
We need to check if these solutions make the original denominator non - zero. When $x=-2$, the denominator $x + 2 = 0$, so $x=-2$ is not a valid solution. When $x = 2$, the denominator $x + 2=4
eq0$. So the only valid solution is $x = 2$
# Answer:
$x = 2$

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QUESTION IMAGE

Question

Explanation:

Problem 1: Solve \(\boldsymbol{\frac{1}{x + 4}=5}\)

Step 1: Multiply both sides by \(x + 4\)

Step 2: Expand the right - hand side

Step 3: Solve for \(x\)

Step 1: Multiply both sides by \(x - 3\)

Step 2: Expand the right - hand side

Step 3: Solve for \(x\)

Step 1: Notice the relationship between denominators

Step 2: Find a common denominator

Step 3: Simplify the numerator

Answer:

Problem 2: Solve \(\boldsymbol{\frac{1}{x - 3}=6}\)