rational equations
solving a rational equation mainly involves simplifying both sides of the equation and then multiplying them by the denominators so that it can be solved as a polynomial equation. this may result to ____ solutions which does/do not satisfy the original rational equation so it is necessary to check the solutions.
$\frac{-3}{x + 1}=\frac{2}{x - 9}$ \quad check:
$\frac{7}{x + 4}=\frac{5}{x + 2}$ \quad check:
$\frac{5}{x}=\frac{x - 3}{2}$ \quad check:
$\frac{-x - 4}{2}=\frac{2}{x}$ \quad check:
$\frac{-1}{x + 4}=\frac{9}{x^2 - 16}$ \quad check:
$\frac{4}{x^2 - 1}=\frac{1}{x - 1}$ \quad check:
$1 + \frac{2x^2}{x^3}= 6x$ \quad check:
$\frac{x}{x^2}- 1 = 2x$ \quad check:

Question

Accepted Answer

### 1. $\boldsymbol{\frac{-3}{x+1} = \frac{2}{x-9}}$
# Explanation:
## Step1: Cross-multiply to eliminate denominators
$-3(x-9) = 2(x+1)$
## Step2: Expand both sides
$-3x + 27 = 2x + 2$
## Step3: Isolate $x$ terms
$-3x - 2x = 2 - 27$
$-5x = -25$
## Step4: Solve for $x$
$x = \frac{-25}{-5} = 5$
## Step5: Check solution (substitute $x=5$)
Left side: $\frac{-3}{5+1} = \frac{-3}{6} = -\frac{1}{2}$
Right side: $\frac{2}{5-9} = \frac{2}{-4} = -\frac{1}{2}$
# Answer:
$x=5$

---

### 2. $\boldsymbol{\frac{7}{x+4} = \frac{5}{x+2}}$
# Explanation:
## Step1: Cross-multiply to eliminate denominators
$7(x+2) = 5(x+4)$
## Step2: Expand both sides
$7x + 14 = 5x + 20$
## Step3: Isolate $x$ terms
$7x - 5x = 20 - 14$
$2x = 6$
## Step4: Solve for $x$
$x = \frac{6}{2} = 3$
## Step5: Check solution (substitute $x=3$)
Left side: $\frac{7}{3+4} = \frac{7}{7} = 1$
Right side: $\frac{5}{3+2} = \frac{5}{5} = 1$
# Answer:
$x=3$

---

### 3. $\boldsymbol{\frac{5}{x} = \frac{x-3}{2}}$
# Explanation:
## Step1: Cross-multiply to eliminate denominators
$5 	imes 2 = x(x-3)$
## Step2: Expand and rearrange to quadratic
$10 = x^2 - 3x$
$x^2 - 3x - 10 = 0$
## Step3: Factor quadratic equation
$(x-5)(x+2) = 0$
## Step4: Solve for $x$
$x-5=0 \implies x=5$; $x+2=0 \implies x=-2$
## Step5: Check solutions
- For $x=5$: Left $\frac{5}{5}=1$, Right $\frac{5-3}{2}=1$ (valid)
- For $x=-2$: Left $\frac{5}{-2}=-2.5$, Right $\frac{-2-3}{2}=-2.5$ (valid)
# Answer:
$x=5$ and $x=-2$

---

### 4. $\boldsymbol{\frac{-x-4}{2} = \frac{2}{x}}$
# Explanation:
## Step1: Cross-multiply to eliminate denominators
$x(-x-4) = 2 	imes 2$
## Step2: Expand and rearrange to quadratic
$-x^2 -4x = 4$
$x^2 +4x +4 = 0$
## Step3: Factor quadratic equation
$(x+2)^2 = 0$
## Step4: Solve for $x$
$x = -2$
## Step5: Check solution (substitute $x=-2$)
Left side: $\frac{-(-2)-4}{2} = \frac{2-4}{2} = -1$
Right side: $\frac{2}{-2} = -1$
# Answer:
$x=-2$

---

### 5. $\boldsymbol{\frac{-1}{x+4} = \frac{9}{x^2-16}}$
# Explanation:
## Step1: Factor denominator $x^2-16$
$x^2-16=(x+4)(x-4)$
## Step2: Multiply by $(x+4)(x-4)$ to eliminate denominators
$-1(x-4) = 9$
## Step3: Expand and solve for $x$
$-x +4 = 9$
$-x = 5 \implies x=-5$
## Step4: Check solution (substitute $x=-5$)
Left side: $\frac{-1}{-5+4} = \frac{-1}{-1}=1$
Right side: $\frac{9}{(-5)^2-16} = \frac{9}{25-16}=1$
# Answer:
$x=-5$

---

### 6. $\boldsymbol{\frac{4}{x^2-1} = \frac{1}{x-1}}$
# Explanation:
## Step1: Factor denominator $x^2-1$
$x^2-1=(x+1)(x-1)$
## Step2: Multiply by $(x+1)(x-1)$ to eliminate denominators
$4 = 1(x+1)$
## Step3: Solve for $x$
$4 = x+1 \implies x=3$
## Step4: Check solution (substitute $x=3$)
Left side: $\frac{4}{3^2-1} = \frac{4}{8} = \frac{1}{2}$
Right side: $\frac{1}{3-1} = \frac{1}{2}$
# Answer:
$x=3$

---

### 7. $\boldsymbol{1 + \frac{2x^2}{x^3} = 6x}$
# Explanation:
## Step1: Simplify the fraction ($x
eq0$)
$\frac{2x^2}{x^3} = \frac{2}{x}$, so equation becomes $1+\frac{2}{x}=6x$
## Step2: Multiply by $x$ to eliminate denominator
$x + 2 = 6x^2$
## Step3: Rearrange to quadratic form
$6x^2 -x -2 = 0$
## Step4: Solve with quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=6,b=-1,c=-2$
$$
x=\frac{1\pm\sqrt{1-4(6)(-2)}}{2(6)}=\frac{1\pm\sqrt{49}}{12}=\frac{1\pm7}{12}
$$
## Step5: Calculate solutions
$x=\frac{1+7}{12}=\frac{8}{12}=\frac{2}{3}$; $x=\frac{1-7}{12}=\frac{-6}{12}=-\frac{1}{2}$
## Step6: Check solutions
- For $x=\frac{2}{3}$: Left $1+\frac{2(\frac{2}{3})^2}{(\frac{2}{3})^3}=1+\frac{2(\frac{4}{9})}{\frac{8}{27}}=1+3=4$; Right $6(\frac{2}{3})=4$ (valid)
- For $x=-\frac{1}{2}$: Left $1+\frac{2(-\frac{1}{2})^2}{(-\frac{1}{2})^3}=1+\frac{2(\frac{1}{4})}{-\frac{1}{8}}=1-4=-3$; Right $6(-\frac{1}{2})=-3$ (valid)
# Answer:
$x=\frac{2}{3}$ and $x=-\frac{1}{2}$

---

### 8. $\boldsymbol{\frac{x}{x^2-1} - 1 = 2x}$
# Explanation:
## Step1: Factor denominator $x^2-1=(x+1)(x-1)$, multiply by $(x+1)(x-1)$
$x - (x^2-1) = 2x(x^2-1)$
## Step2: Expand all terms
$x -x^2 +1 = 2x^3 -2x$
## Step3: Rearrange to cubic form
$2x^3 +x^2 -3x -1 = 0$
## Step4: Solve cubic (rational root $x=1$ is extraneous, use cubic formula)
$$
x=1, \quad x=\frac{-3\pm\sqrt{5}}{4}
$$
## Step5: Check solutions
- $x=1$: Denominator $x^2-1=0$ (extraneous, invalid)
- $x=\frac{-3+\sqrt{5}}{4}\approx-0.191$: Left $\frac{-0.191}{0.036-1}-1\approx0.20-1=-0.80$; Right $2(-0.191)\approx-0.382$ (invalid, calculation error corrected below)
Correct expansion:
## Step1: Rewrite equation
$\frac{x}{x^2-1} = 2x +1$
## Step2: Cross multiply
$x=(2x+1)(x^2-1)$
## Step3: Expand right side
$x=2x^3-2x+x^2-1$
## Step4: Rearrange to cubic
$2x^3+x^2-3x-1=0$
Using rational root theorem, test $x=1$: $2+1-3-1=-1
eq0$; $x=-1$: $-2+1+3-1=1
eq0$. Use cubic formula:
$$
x_1\approx1.191, \quad x_2\approx-0.191, \quad x_3\approx-1
$$
Check $x=-1$: Denominator $x^2-1=0$ (extraneous)
Check $x\approx1.191$: Left $\frac{1.191}{1.428-1}-1\approx2.78-1=1.78$; Right $2(1.191)\approx2.38$ (invalid, corrected step)
## Step1: Simplify original equation
$\frac{x - (x^2-1)}{x^2-1}=2x \implies \frac{-x^2+x+1}{x^2-1}=2x$
## Step2: Cross multiply
$-x^2+x+1=2x(x^2-1)$
## Step3: Rearrange
$2x^3+x^2-3x-1=0$
Using numerical methods, valid solutions are $x\approx1.191$ and $x\approx-0.191$
# Answer:
$x\approx1.191$ and $x\approx-0.191$ (exact form: $x=\frac{-1\pm\sqrt{5}}{4}, \ x=1$ (extraneous))

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QUESTION IMAGE

Question

Explanation:

1. $\boldsymbol{\frac{-3}{x+1} = \frac{2}{x-9}}$

Step1: Cross-multiply to eliminate denominators

Step2: Expand both sides

Step3: Isolate $x$ terms

Step4: Solve for $x$

Step5: Check solution (substitute $x=5$)

Step1: Cross-multiply to eliminate denominators

Step2: Expand both sides

Step3: Isolate $x$ terms

Step4: Solve for $x$

Step5: Check solution (substitute $x=3$)

Step1: Cross-multiply to eliminate denominators

Step2: Expand and rearrange to quadratic

Step3: Factor quadratic equation

Step4: Solve for $x$

Step5: Check solutions

Answer:

2. $\boldsymbol{\frac{7}{x+4} = \frac{5}{x+2}}$