Hello studentUnion!
[tex] \huge \boxed{\mathfrak{Question} \downarrow}[/tex]
Solve the equation-
[tex] \frac{5}{x - 3} - \frac{4}{x + 4} = \frac{1}{x} \\ [/tex]
[tex] \large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}[/tex]
[tex] \frac{5}{x - 3} - \frac{4}{x + 4} = \frac{1}{x} \\ [/tex]
Let's multiply both the sides of the equation by x (x - 3) & (x + 4) which is the LCM of (x - 3), (x + 4) & x. We can't solve it directly without taking the LCM first as the denominators are not equal to each other. Then do cross multiplication. After all these steps, you'll get an equation as..
[tex]x\left(x+4\right)\times 5-x\left(x-3\right)\times 4=\left(x-3\right)\left(x+4\right) \\ [/tex]
Now, use the distributive property & simplify it.
[tex]x\left(x+4\right)\times 5-x\left(x-3\right)\times 4=\left(x-3\right)\left(x+4\right) \\ \left(x^{2}+4x\right)\times 5-x\left(x-3\right)\times 4=\left(x-3\right)\left(x+4\right) \\ 5x^{2}+20x-x\left(x-3\right)\times 4=\left(x-3\right)\left(x+4\right) \\ 5x^{2}+20x-\left(x^{2}-3x\right)\times 4=\left(x-3\right)\left(x+4\right) \\ 5x^{2}+20x-\left(4x^{2}-12x\right)=\left(x-3\right)\left(x+4\right) \\ 5x^{2}+20x-4x^{2}+12x=\left(x-3\right)\left(x+4\right) \\ x^{2}+20x+12x=\left(x-3\right)\left(x+4\right) \\ x^{2}+32x=\left(x-3\right)\left(x+4\right) \\ x^{2}+32x=x^{2}+x-12 \\ x^{2}+32x-x^{2}=x-12 \\ 32x=x-12 \\ 32x-x=-12 \\ 31x=-12 \\ \boxed{ \boxed{ \bf \: x=-\frac{12}{31} }}[/tex]
- The value of x is -12/31.
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Hope it'll help you!
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