4. Solve the following equations: (a) $$ x-3=\frac{18}{x} $$ (c) $$ \frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2} $$ (e) $$ \frac{x}{x-2}=\frac{1}{x-3}-\frac{2}{2-x} $$ (f) $$ \frac{3 x+4}{x+6}=\frac{3 x-2}{x-3}-\frac{21 x}{x^{2}+3 x-18} $$

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**(a)**  
We start with  
$$
x-3=\frac{18}{x}.
$$  
Multiply both sides by $$x$$ (noting that $$x\neq 0$$):
$$
x(x-3)=18.
$$
This leads to
$$
x^2-3x-18=0.
$$
We factor the quadratic:
$$
(x-6)(x+3)=0.
$$
Thus, the solutions are
$$
x=6\quad \text{or}\quad x=-3.
$$

---

**(c)**  
The equation is
$$
\frac{x+6}{x^2-4}-\frac{2}{x-2}=\frac{-1}{x+2}.
$$
Notice that
$$
x^2-4=(x-2)(x+2).
$$
Multiply both sides by $$(x-2)(x+2)$$ (with $$x\neq 2$$ and $$x\neq -2$$):
$$
(x+6)-2(x+2)=-1\cdot (x-2).
$$
Simplify the left-hand side:
$$
x+6-2x-4=-x+2,
$$
and the right-hand side:
$$
- (x-2)=-x+2.
$$
Since both sides are identical, the equation is true for all $$x$$ except where the denominators vanish. Therefore, the solution is  
$$
\text{All } x \text{ such that } x\neq 2 \text{ and } x\neq -2.
$$

---

**(e)**  
We have
$$
\frac{x}{x-2}=\frac{1}{x-3}-\frac{2}{2-x}.
$$
Notice that
$$
2-x=-(x-2),
$$
so we rewrite the second term:
$$
-\frac{2}{2-x}=-\frac{2}{-(x-2)}=\frac{2}{x-2}.
$$
This converts the equation into
$$
\frac{x}{x-2}=\frac{1}{x-3}+\frac{2}{x-2}.
$$
Multiply both sides by $$(x-2)(x-3)$$ (with $$x\neq 2$$ and $$x\neq 3$$):
$$
x(x-3) = 1\cdot (x-2) + 2\cdot (x-3).
$$
Expanding both sides:
$$
x^2-3x = \,(x-2)+(2x-6)=3x-8.
$$
Bring all terms to one side:
$$
x^2-3x-3x+8=0 \quad\Longrightarrow\quad x^2-6x+8=0.
$$
Solve the quadratic by computing the discriminant:
$$
\Delta=36-32=4.
$$
Thus, 
$$
x=\frac{6\pm\sqrt{4}}{2}=\frac{6\pm2}{2}.
$$
This gives
$$
x=4\quad \text{or}\quad x=2.
$$
However, $$x=2$$ is not allowed because it makes a denominator zero. Therefore, the only valid solution is  
$$
x=4.
$$

---

**(f)**  
Consider the equation
$$
\frac{3x+4}{x+6}=\frac{3x-2}{x-3}-\frac{21x}{x^2+3x-18}.
$$
First, factor the quadratic in the last denominator:
$$
x^2+3x-18=(x+6)(x-3).
$$
Thus the equation becomes
$$
\frac{3x+4}{x+6}=\frac{3x-2}{x-3}-\frac{21x}{(x+6)(x-3)}.
$$
Multiply both sides by $$(x+6)(x-3)$$ (keeping in mind that $$x\neq -6$$ and $$x\neq 3$$):
$$
(3x+4)(x-3)=(3x-2)(x+6)-21x.
$$
Expand both sides:
$$
\text{Left side: } (3x+4)(x-3)=3x^2-9x+4x-12=3x^2-5x-12.
$$
$$
\text{Right side: } (3x-2)(x+6)=3x^2+18x-2x-12=3x^2+16x-12,
$$
so, after subtracting $$21x$$:
$$
3x^2+16x-12-21x=3x^2-5x-12.
$$
Since the two sides are identical, the equation holds for all values of $$x$$ except those that make a denominator zero. The restrictions from the denominators are:
$$
x+6\neq 0\quad \Longrightarrow\quad x\neq -6,
$$
$$
x-3\neq 0\quad \Longrightarrow\quad x\neq 3.
$$
Thus, the solution is  
$$
\text{All } x \text{ such that } x\neq -6 \text{ and } x\neq 3.
$$
**(a)** $$ x = 6 $$ or $$ x = -3 $$ **(c)** All $$ x $$ except $$ x = 2 $$ and $$ x = -2 $$ **(e)** $$ x = 4 $$ **(f)** All $$ x $$ except $$ x = -6 $$ and $$ x = 3 $$

4. Solve the following equations: (a) \( x-3=\frac{18}{x} \) (c) \( \frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2} \) (e) \( \frac{x}{x-2}=\frac{1}{x-3}-\frac{2}{2-x} \) (f) \( \frac{3 x+4}{x+6}=\frac{3 x-2}{x-3}-\frac{21 x}{x^{2}+3 x-18} \)

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