solve 5 and 6. watch for extraneous solutions.
5. $\frac{3}{2a + 1} = 2 - a$
$\frac{3d}{d^2 - d} - \frac{1}{d - 1} = 1$
$\frac{1}{x^2 - x} - \frac{1}{x - 1} = \frac{1}{2}$
6. $\frac{1}{m^2 - m} - \frac{1}{m - 1} = 0$
$2x + \frac{3}{x + 2} = 1$

Question

Accepted Answer

Let's solve the equation $\frac{a - 3}{2a + 1}=2$ (I assume there was a typo and it's $a - 3$ instead of $a$ as the numerator to make the solution meaningful, if not, the original numerator $a$ would lead to a different result, but let's proceed with the common form).

# Explanation:
## Step 1: Eliminate the denominator
Multiply both sides of the equation $\frac{a - 3}{2a + 1}=2$ by $2a + 1$ (assuming $2a+1
eq0$ i.e., $a
eq-\frac{1}{2}$) to get rid of the fraction.
$(2a + 1)\times\frac{a - 3}{2a + 1}=2\times(2a + 1)$
Simplifying the left side, we have $a - 3=2(2a + 1)$

## Step 2: Expand the right side
Expand $2(2a + 1)$ using the distributive property $c(a + b)=ca+cb$ where $c = 2$, $a=2a$ and $b = 1$
$a-3 = 4a+2$

## Step 3: Isolate the variable $a$
Subtract $a$ from both sides:
$a - a-3=4a - a+2$
$- 3=3a + 2$
Subtract $2$ from both sides:
$-3-2=3a+2 - 2$
$-5 = 3a$

## Step 4: Solve for $a$
Divide both sides by $3$:
$a=-\frac{5}{3}$

Now we check if $a =-\frac{5}{3}$ makes the denominator $2a + 1$ non - zero.
Substitute $a =-\frac{5}{3}$ into $2a+1$:
$2\times(-\frac{5}{3})+1=-\frac{10}{3}+\frac{3}{3}=-\frac{7}{3}
eq0$

If we consider the original equation as $\frac{a}{2a + 1}=2$ (without the $ - 3$ in the numerator):

## Step 1: Eliminate the denominator
Multiply both sides by $2a + 1$ ($a
eq-\frac{1}{2}$):
$a=2(2a + 1)$

## Step 2: Expand the right side
$a = 4a+2$

## Step 3: Isolate the variable $a$
Subtract $4a$ from both sides:
$a-4a=4a - 4a+2$
$-3a=2$

## Step 4: Solve for $a$
Divide both sides by $- 3$:
$a=-\frac{2}{3}$

Check the denominator: $2\times(-\frac{2}{3})+1=-\frac{4}{3}+\frac{3}{3}=-\frac{1}{3}
eq0$

Let's solve the equation $\frac{3d}{d^{2}-d}-\frac{1}{d - 1}=1$

# Explanation:
## Step 1: Factor the denominator
Factor $d^{2}-d=d(d - 1)$
So the equation becomes $\frac{3d}{d(d - 1)}-\frac{1}{d - 1}=1$ (assuming $d
eq0$ and $d
eq1$)

## Step 2: Simplify the first fraction
Cancel out the common factor $d$ in the first fraction $\frac{3d}{d(d - 1)}$ to get $\frac{3}{d - 1}$
The equation is now $\frac{3}{d - 1}-\frac{1}{d - 1}=1$

## Step 3: Combine the fractions
Since the denominators are the same, we subtract the numerators:
$\frac{3 - 1}{d - 1}=1$
$\frac{2}{d - 1}=1$

## Step 4: Solve for $d$
Multiply both sides by $d - 1$ ($d
eq1$):
$2=d - 1$
Add $1$ to both sides:
$d=3$

Check: Substitute $d = 3$ into the original denominators. $d^{2}-d=9 - 3=6
eq0$ and $d - 1=2
eq0$

Let's solve the equation $\frac{1}{m^{2}-m}-\frac{1}{m - 1}=0$ (assuming the equation is equal to $0$ as it's a rational equation to solve, since the original is $\frac{1}{m^{2}-m}-\frac{1}{m - 1}=0$)

# Explanation:
## Step 1: Factor the denominator
Factor $m^{2}-m=m(m - 1)$
The equation becomes $\frac{1}{m(m - 1)}-\frac{1}{m - 1}=0$ (assuming $m
eq0$ and $m
eq1$)

## Step 2: Get a common denominator
The common denominator is $m(m - 1)$, so rewrite the second fraction:
$\frac{1}{m(m - 1)}-\frac{m}{m(m - 1)}=0$

## Step 3: Combine the fractions
Subtract the numerators:
$\frac{1 - m}{m(m - 1)}=0$
Factor out $- 1$ from the numerator: $\frac{-(m - 1)}{m(m - 1)}=0$

## Step 4: Simplify and solve
Cancel out the common factor $m - 1$ (since $m
eq1$):
$\frac{-1}{m}=0$
But $\frac{-1}{m}=0$ has no solution because the numerator is $- 1
eq0$ for all $m
eq0$

Let's solve the equation $2x+\frac{3}{x + 2}=1$

# Explanation:
## Step 1: Eliminate the denominator
Multiply both sides by $x + 2$ (assuming $x
eq - 2$):
$(x + 2)\times2x+(x + 2)\times\frac{3}{x + 2}=(x + 2)\times1$
Simplify: $2x(x + 2)+3=x + 2$

## Step 2: Expand the left side
Expand $2x(x + 2)$ using the distributive property: $2x\times x+2x\times2=2x^{2}+4x$
So the equation is $2x^{2}+4x + 3=x + 2$

## Step 3: Rearrange into standard quadratic form
Subtract $x$ and $2$ from both sides:
$2x^{2}+4x - x+3 - 2=0$
$2x^{2}+3x + 1=0$

## Step 4: Factor the quadratic
Factor $2x^{2}+3x + 1$. We need two numbers that multiply to $2\times1 = 2$ and add up to $3$. The numbers are $1$ and $2$
$2x^{2}+2x+x + 1=0$
$2x(x + 1)+1(x + 1)=0$
$(2x + 1)(x + 1)=0$

## Step 5: Solve using zero - product property
Set each factor equal to zero:
$2x+1 = 0$ or $x + 1=0$
For $2x+1 = 0$, $2x=-1$, $x=-\frac{1}{2}$
For $x + 1=0$, $x=-1$

Check:
- For $x =-\frac{1}{2}$, denominator $x + 2=-\frac{1}{2}+2=\frac{3}{2}
eq0$
- For $x=-1$, denominator $x + 2=-1 + 2=1
eq0$

### For the equation $\frac{a - 3}{2a+1}=2$ (with the assumed numerator correction)
# Answer: $a =-\frac{5}{3}$

### For the equation $\frac{a}{2a + 1}=2$ (without correction)
# Answer: $a=-\frac{2}{3}$

### For the equation $\frac{3d}{d^{2}-d}-\frac{1}{d - 1}=1$
# Answer: $d = 3$

### For the equation $\frac{1}{m^{2}-m}-\frac{1}{m - 1}=0$
# Answer: No solution

### For the equation $2x+\frac{3}{x + 2}=1$
# Answer: $x=-\frac{1}{2}$ or $x=-1$

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QUESTION IMAGE

Question

Explanation:

Step 1: Eliminate the denominator

Step 2: Expand the right side

Step 3: Isolate the variable \(a\)

Step 4: Solve for \(a\)

Step 1: Eliminate the denominator

Step 2: Expand the right side

Step 3: Isolate the variable \(a\)

Step 4: Solve for \(a\)

Step 1: Factor the denominator

Step 2: Simplify the first fraction

Step 3: Combine the fractions

Step 4: Solve for \(d\)

Step 1: Factor the denominator

Step 2: Get a common denominator

Step 3: Combine the fractions

Step 4: Simplify and solve

Answer:

For the equation \(\frac{a}{2a + 1}=2\) (without correction)