Question

solve the equation. check each solution.

1. $\frac{10}{x+3}+\frac{10}{3}=6$
2. $\frac{2x}{5}=\frac{x^2-5x}{5x}$
3. $\frac{10}{x^2-2x}+\frac{4}{x}=\frac{5}{x-2}$
4. $\frac{-2}{x-1}=\frac{x-8}{x+1}$

Explanation:

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Problem 1: $\frac{10}{x+3}+\frac{10}{3}=6$

Step1: Isolate the rational term

Subtract $\frac{10}{3}$ from both sides.
$\frac{10}{x+3}=6-\frac{10}{3}$

Step2: Simplify the right-hand side

Calculate the difference of constants.
$\frac{10}{x+3}=\frac{18-10}{3}=\frac{8}{3}$

Step3: Cross-multiply to eliminate fractions

Set products of numerator/denominator equal.
$10\times3=8(x+3)$

Step4: Simplify and solve for $x$

Expand and isolate $x$.
$30=8x+24 \implies 8x=6 \implies x=\frac{6}{8}=\frac{3}{4}$

Step5: Check the solution

Substitute $x=\frac{3}{4}$ into the original equation.
$\frac{10}{\frac{3}{4}+3}+\frac{10}{3}=\frac{10}{\frac{15}{4}}+\frac{10}{3}=\frac{8}{3}+\frac{10}{3}=6$, which matches the right-hand side.

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Problem 2: $\frac{2x}{5}=\frac{x^2-5x}{5x}$

Step1: Note domain restriction

$x
eq0$ (denominator cannot be 0).

Step2: Simplify the right-hand side

Factor and cancel common terms.
$\frac{2x}{5}=\frac{x(x-5)}{5x}=\frac{x-5}{5}$

Step3: Multiply both sides by 5

Eliminate the shared denominator.
$2x=x-5$

Step4: Solve for $x$

Isolate $x$ on one side.
$2x-x=-5 \implies x=-5$

Step5: Check the solution

Substitute $x=-5$ into the original equation.
Left-hand side: $\frac{2(-5)}{5}=-2$; Right-hand side: $\frac{(-5)^2-5(-5)}{5(-5)}=\frac{25+25}{-25}=-2$, which matches.

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Problem 3: $\frac{10}{x^2-2x}+\frac{4}{x}=\frac{5}{x-2}$

Step1: Factor denominators, note restrictions

$x
eq0,2$ (denominators cannot be 0).
$\frac{10}{x(x-2)}+\frac{4}{x}=\frac{5}{x-2}$

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

Eliminate all denominators.
$10+4(x-2)=5x$

Step3: Expand and simplify

Distribute and combine like terms.
$10+4x-8=5x \implies 4x+2=5x$

Step4: Solve for $x$

Isolate $x$ on one side.
$x=2$

Step5: Check validity

$x=2$ is a restricted value (makes original denominators 0), so there is no solution.

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Problem 4: $\frac{-2}{x-1}=\frac{x-8}{x+1}$

Step1: Note domain restriction

$x
eq1,-1$ (denominators cannot be 0).

Step2: Cross-multiply to eliminate fractions

Set products of numerator/denominator equal.
$-2(x+1)=(x-8)(x-1)$

Step3: Expand both sides

Multiply out the terms.
$-2x-2=x^2-9x+8$

Step4: Rearrange into standard quadratic form

Bring all terms to one side.
$x^2-7x+10=0$

Step5: Factor and solve for $x$

Factor the quadratic equation.
$(x-2)(x-5)=0 \implies x=2$ or $x=5$

Step6: Check the solutions

• For $x=2$: $\frac{-2}{2-1}=\frac{2-8}{2+1} \implies -2=-2$, which matches.
• For $x=5$: $\frac{-2}{5-1}=\frac{5-8}{5+1} \implies \frac{-2}{4}=\frac{-3}{6} \implies -\frac{1}{2}=-\frac{1}{2}$, which matches.

Answer:

1. $x=\frac{3}{4}$
2. $x=-5$
3. No solution
4. $x=2$ and $x=5$