(x-7)/(3x^(2)+x-2)+(x^(2)+3x-18)/(x^(2)-5x+6)×(x^(2)+3x-10)/(3x^(2)+13 x-10)

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Factor Denominators and Numerators: First, we need to factor the denominators and numerators where possible to see if there are any common factors that can be canceled out. Let's start with the denominator of the first fraction: $3x^2 + x - 2$. We look for two numbers that multiply to $3 \times -2 = -6$ and add to $1$. These numbers are $3$ and $-2$. So, $3x^2 + x - 2$ factors to $(3x - 2)(x + 1)$.Factor First Fraction Denominator: Next, we factor the denominator of the second fraction: $x^2 - 5x + 6$. We look for two numbers that multiply to $6$ and add to $-5$. These numbers are $-2$ and $-3$. So, $x^2 - 5x + 6$ factors to $(x - 2)(x - 3)$.Factor Second Fraction Denominator: Now, let's factor the numerator of the second fraction: $x^2 + 3x - 18$. We look for two numbers that multiply to $-18$ and add to $3$. These numbers are $6$ and $-3$. So, $x^2 + 3x - 18$ factors to $(x + 6)(x - 3)$.Factor Second Fraction Numerator: We also need to factor the numerator of the third fraction: $x^2 + 3x - 10$. We look for two numbers that multiply to $-10$ and add to $3$. These numbers are $5$ and $-2$. So, $x^2 + 3x - 10$ factors to $(x + 5)(x - 2)$.Factor Third Fraction Numerator: Lastly, we factor the denominator of the third fraction: $3x^2 + 13x - 10$. We look for two numbers that multiply to $3 \times -10 = -30$ and add to $13$. These numbers are $15$ and $-2$. So, $3x^2 + 13x - 10$ factors to $(3x - 2)(x + 5)$.Factor Third Fraction Denominator: Now we can rewrite the entire expression with the factored forms: $$\frac{x - 7}{(3x - 2)(x + 1)} + \frac{(x + 6)(x - 3)}{(x - 2)(x - 3)} \times \frac{(x + 5)(x - 2)}{(3x - 2)(x + 5)}$$Rewrite Expression with Factored Forms: We can cancel out the common factors in the second term of the expression: The $(x - 3)$ in the numerator cancels with the $(x - 3)$ in the denominator. The $(x - 2)$ in the numerator cancels with the $(x - 2)$ in the denominator. This simplifies the second term to just $x + 6$.Cancel Common Factors in Second Term: Similarly, in the third term of the expression, the $(3x - 2)$ in the denominator cancels with the $(3x - 2)$ in the numerator. The $(x + 5)$ in the numerator cancels with the $(x + 5)$ in the denominator. This simplifies the third term to just $1$.Simplify Third Term: Now we have: $$\frac{x - 7}{(3x - 2)(x + 1)} + (x + 6) \times 1$$Combine Terms with Common Denominator: We can now combine the two terms by finding a common denominator, which is $(3x - 2)(x + 1)$. The second term, $x + 6$, needs to be multiplied by the common denominator to combine the terms: $$\frac{x - 7}{(3x - 2)(x + 1)} + \frac{(x + 6)(3x - 2)(x + 1)}{(3x - 2)(x + 1)}$$Distribute Second Term in Numerator: Now we can add the two fractions together since they have the same denominator: $$\frac{x - 7 + (x + 6)(3x - 2)(x + 1)}{(3x - 2)(x + 1)}$$Substitute Back into Combined Fraction: Next, we need to distribute $(x + 6)$ across $(3x - 2)(x + 1)$ in the numerator: $$(x + 6)(3x - 2)(x + 1) = (3x^2 - 2x + 18x - 12)(x + 1)$$ $$= (3x^2 + 16x - 12)(x + 1)$$ $$= 3x^3 + 3x^2 + 16x^2 + 16x - 12x - 12$$ $$= 3x^3 + 19x^2 + 4x - 12$$State Restrictions on Variables: Now we substitute this back into the combined fraction: $$\frac{x - 7 + 3x^3 + 19x^2 + 4x - 12}{(3x - 2)(x + 1)}$$ $$= \frac{3x^3 + 19x^2 + 5x - 19}{(3x - 2)(x + 1)}$$Finally, we state the restrictions on the variables. The original denominators cannot be equal to zero, so we set them equal to zero and solve for $x$: $3x - 2 = 0$ gives $x = \frac{2}{3}$. $x + 1 = 0$ gives $x = -1$. $x - 2 = 0$ gives $x = 2$. $x - 3 = 0$ gives $x = 3$. $x + 5 = 0$ gives $x = -5$. So the restrictions on $x$ are that $x$ cannot be $\frac{2}{3}$, $-1$, $2$, $3$, or $-5$.

$\frac{x-7}{3 x^{2}+x-2}+\frac{x^{2}+3 x-18}{x^{2}-5 x+6} \times \frac{x^{2}+3 x-10}{3 x^{2}+13 x-10}$

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$\frac{x-7}{3 x^{2}+x-2}+\frac{x^{2}+3 x-18}{x^{2}-5 x+6} \times \frac{x^{2}+3 x-10}{3 x^{2}+13 x-10}$