studentUnion
Answered

Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Solve the equation-
[tex] \frac{5}{x - 3} - \frac{4}{x + 4} = \frac{1}{x} [/tex]
​

Sagot :

ᴜᴋɪʏᴏ

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.

__________________

Hope it'll help you!

ℓu¢αzz ッ

Hi1315

Answer:

[tex]\frac{5}{x-3} -\frac{4}{x+4} =\frac{1}{x}[/tex]

[tex]\frac{5(x+4)-4(x-3)}{(x-3)(x+4)} =\frac{1}{x} \\[/tex]

[tex]\frac{5x+20-4x+12}{(x-3)(x+4)} =\frac{1}{x}[/tex]

[tex]\frac{x+32}{(x-3)(x+4)} =\frac{1}{x}[/tex]

Use Cross multiplication

[tex]x(x+32)=(x-3)(x+4)[/tex]

[tex]x^{2} +32x = x^{2} +4x-3x-12[/tex]

[tex]x^{2} +32x=x^{2} +x-12[/tex]

[tex]x^{2} +32x-x^{2} -x=-12[/tex]

[tex]31x = -12[/tex]

[tex]\frac{31x}{31} =\frac{-12}{31} \\x = \frac{-12}{31}[/tex]

Hope this helps you.

Let me know if you have any other questions :-)

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.