more on rational equations
1. $\frac{x + 15}{x^2 + 2x - 3}+\frac{3}{x + 3}=\frac{2}{x - 1}$ check:
2. $\frac{x + 10}{x^2 + x - 2}-\frac{1}{x - 1}=\frac{2}{x + 2}$ check:
3. $\frac{x}{x - 5}-6=\frac{50}{(x + 5)(x - 5)}$ check:
4. $1+\frac{x}{x - 2}=\frac{8}{(x + 2)(x - 2)}$ check:
5. $\frac{x - 6}{x - 5}+\frac{x}{2}=\frac{-1}{x - 5}$ check:
6. $x-\frac{x}{x + 3}=\frac{3}{x + 3}$ check:
7. some members of young entrepreneurs club are planning to sell hand-crafted specialty bar soap in the local farmers market. the fixed production cost is $210, and the production of each bar soap costs $3. write and solve the equation that can be used to determine the number of bar soap that would need to be produced to have an average cost of production per bar of $5.
8. a student solving a rational equation presents the first half of their work as follows. check for any mistake.
given: $\frac{x - 3}{x^2 - 9}-\frac{7x + 2}{x + 3}=-3$
line 1: $\frac{x - 3}{(x + 3)(x - 3)}-\frac{7x + 2}{x + 3}=-3$
line 2: $\frac{1}{x + 3}-\frac{7x + 2}{x + 3}=-3$
line 3: $\frac{1 - 7x + 2}{x + 3}=-3$
line 4: $\frac{-7x + 3}{x + 3}=-3$
line 5: $(x + 3)\left(\frac{-7x + 3}{x + 3}\
ight)=(-3)(x + 3)$

Question

Accepted Answer

### 1. $\boldsymbol{\frac{x + 15}{x^2 + 2x - 3} + \frac{3}{x + 3} = \frac{2}{x - 1}}$
# Explanation:
## Step1: Factor denominator
$x^2 + 2x - 3 = (x+3)(x-1)$
## Step2: Multiply by LCD $(x+3)(x-1)$
$(x+15) + 3(x-1) = 2(x+3)$
## Step3: Expand all terms
$x+15 + 3x - 3 = 2x + 6$
## Step4: Combine like terms
$4x + 12 = 2x + 6$
## Step5: Isolate $x$ terms
$4x - 2x = 6 - 12$
$2x = -6$
## Step6: Solve for $x$
$x = \frac{-6}{2} = -3$
## Step7: Check for extraneous solution
$x=-3$ makes $(x+3)=0$, so no solution.
# Answer:
No solution

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### 2. $\boldsymbol{\frac{x + 10}{x^2 + x - 2} - \frac{1}{x - 1} = \frac{2}{x + 2}}$
# Explanation:
## Step1: Factor denominator
$x^2 + x - 2 = (x+2)(x-1)$
## Step2: Multiply by LCD $(x+2)(x-1)$
$(x+10) - (x+2) = 2(x-1)$
## Step3: Expand all terms
$x+10 - x - 2 = 2x - 2$
## Step4: Combine like terms
$8 = 2x - 2$
## Step5: Isolate $x$
$2x = 8 + 2 = 10$
## Step6: Solve for $x$
$x = \frac{10}{2} = 5$
## Step7: Check solution
Substitute $x=5$: $\frac{15}{28} - \frac{1}{4} = \frac{15-7}{28} = \frac{8}{28} = \frac{2}{7}$, $\frac{2}{5+2}=\frac{2}{7}$. Valid.
# Answer:
$x=5$

---

### 3. $\boldsymbol{\frac{x}{x - 5} - 6 = \frac{50}{(x + 5)(x - 5)}}$
# Explanation:
## Step1: Multiply by LCD $(x+5)(x-5)$
$x(x+5) - 6(x+5)(x-5) = 50$
## Step2: Expand all terms
$x^2 + 5x - 6(x^2 - 25) = 50$
$x^2 + 5x - 6x^2 + 150 = 50$
## Step3: Combine like terms
$-5x^2 + 5x + 150 = 50$
## Step4: Simplify to standard form
$-5x^2 + 5x + 100 = 0$
Divide by $-5$: $x^2 - x - 20 = 0$
## Step5: Factor quadratic
$(x-5)(x+4) = 0$
## Step6: Solve for $x$
$x=5$ or $x=-4$
## Step7: Check extraneous solution
$x=5$ makes $(x-5)=0$, discard. Check $x=-4$: $\frac{-4}{-9} -6 = \frac{4}{9} - \frac{54}{9} = -\frac{50}{9}$, $\frac{50}{(1)(-9)}=-\frac{50}{9}$. Valid.
# Answer:
$x=-4$

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### 4. $\boldsymbol{1 + \frac{x}{x - 2} = \frac{8}{(x + 2)(x - 2)}}$
# Explanation:
## Step1: Multiply by LCD $(x+2)(x-2)$
$(x+2)(x-2) + x(x+2) = 8$
## Step2: Expand all terms
$x^2 - 4 + x^2 + 2x = 8$
## Step3: Combine like terms
$2x^2 + 2x - 4 = 8$
## Step4: Simplify to standard form
$2x^2 + 2x - 12 = 0$
Divide by 2: $x^2 + x - 6 = 0$
## Step5: Factor quadratic
$(x+3)(x-2) = 0$
## Step6: Solve for $x$
$x=-3$ or $x=2$
## Step7: Check extraneous solution
$x=2$ makes $(x-2)=0$, discard. Check $x=-3$: $1 + \frac{-3}{-5} = 1 + \frac{3}{5} = \frac{8}{5}$, $\frac{8}{(-1)(-5)}=\frac{8}{5}$. Valid.
# Answer:
$x=-3$

---

### 5. $\boldsymbol{\frac{x - 6}{x - 5} + \frac{x}{2} = \frac{-1}{x - 5}}$
# Explanation:
## Step1: Multiply by LCD $2(x-5)$
$2(x-6) + x(x-5) = -2$
## Step2: Expand all terms
$2x - 12 + x^2 - 5x = -2$
## Step3: Combine like terms
$x^2 - 3x - 12 = -2$
## Step4: Simplify to standard form
$x^2 - 3x - 10 = 0$
## Step5: Factor quadratic
$(x-5)(x+2) = 0$
## Step6: Solve for $x$
$x=5$ or $x=-2$
## Step7: Check extraneous solution
$x=5$ makes $(x-5)=0$, discard. Check $x=-2$: $\frac{-8}{-7} + \frac{-2}{2} = \frac{8}{7} - 1 = \frac{1}{7}$, $\frac{-1}{-7}=\frac{1}{7}$. Valid.
# Answer:
$x=-2$

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### 6. $\boldsymbol{x - \frac{x}{x + 3} = \frac{3}{x + 3}}$
# Explanation:
## Step1: Multiply by LCD $(x+3)$
$x(x+3) - x = 3$
## Step2: Expand all terms
$x^2 + 3x - x = 3$
## Step3: Combine like terms
$x^2 + 2x = 3$
## Step4: Simplify to standard form
$x^2 + 2x - 3 = 0$
## Step5: Factor quadratic
$(x+3)(x-1) = 0$
## Step6: Solve for $x$
$x=-3$ or $x=1$
## Step7: Check extraneous solution
$x=-3$ makes $(x+3)=0$, discard. Check $x=1$: $1 - \frac{1}{4} = \frac{3}{4}$, $\frac{3}{1+3}=\frac{3}{4}$. Valid.
# Answer:
$x=1$

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### 7. Soap Production Cost Problem
# Explanation:
## Step1: Define variable
Let $x$ = number of soap bars
## Step2: Set up average cost equation
Average cost = $\frac{	ext{Total Cost}}{	ext{Number of Units}}$
$\frac{210 + 3x}{x} = 5$
## Step3: Multiply by $x$
$210 + 3x = 5x$
## Step4: Isolate $x$ terms
$5x - 3x = 210$
$2x = 210$
## Step5: Solve for $x$
$x = \frac{210}{2} = 105$
# Answer:
105 soap bars

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### 8. Student Rational Equation Error Check
# Brief Explanations:
The first mistake occurs in **Line 2**: When simplifying $\frac{x-3}{(x+3)(x-3)}$, the simplification is only valid when $x
eq 3$, and the correct simplified form is $\frac{1}{x-3}$, not $\frac{1}{x+3}$. The student incorrectly canceled $(x-3)$ with $(x+3)$. A secondary mistake occurs in **Line 3**: When combining $\frac{1}{x+3} - \frac{7x+2}{x+3}$, the numerator should be $1 - (7x+2) = 1 -7x -2$, not $1 -7x +2$.
# Answer:
The first mistake is in Line 2: Incorrect simplification of $\frac{x-3}{(x+3)(x-3)}$ to $\frac{1}{x+3}$ (it should be $\frac{1}{x-3}$). A second mistake follows in Line 3: Incorrect distribution of the negative sign in the numerator.

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QUESTION IMAGE

Question

Explanation:

1. $\boldsymbol{\frac{x + 15}{x^2 + 2x - 3} + \frac{3}{x + 3} = \frac{2}{x - 1}}$

Step1: Factor denominator

Step2: Multiply by LCD $(x+3)(x-1)$

Step3: Expand all terms

Step4: Combine like terms

Step5: Isolate $x$ terms

Step6: Solve for $x$

Step7: Check for extraneous solution

Step1: Factor denominator

Step2: Multiply by LCD $(x+2)(x-1)$

Step3: Expand all terms

Step4: Combine like terms

Step5: Isolate $x$

Step6: Solve for $x$

Step7: Check solution

Step1: Multiply by LCD $(x+5)(x-5)$

Step2: Expand all terms

Step3: Combine like terms

Step4: Simplify to standard form

Step5: Factor quadratic

Step6: Solve for $x$

Step7: Check extraneous solution

Answer:

2. $\boldsymbol{\frac{x + 10}{x^2 + x - 2} - \frac{1}{x - 1} = \frac{2}{x + 2}}$