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Solve the Following Equation and Verify Your Answer: 9 X − 7 3 X + 5 = 3 X − 4 X + 6 - Mathematics

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Question

Solve the following equation and verify your answer:

\[\frac{9x - 7}{3x + 5} = \frac{3x - 4}{x + 6}\]
Sum

Solution

\[\frac{9x - 7}{3x + 5} = \frac{3x - 4}{x + 6}\]

\[\text{ or }9 x^2 - 7x + 54x - 42 = 9 x^2 - 12x + 15x - 20 [\text{ After cross multiplication }]\]

\[\text{ or }9 x^2 - 9 x^2 + 47x - 3x = - 20 + 42\]

\[\text{ or }44x = 22\]

\[\text{ or }x = \frac{22}{44}\]

\[\text{ or }x = \frac{1}{2}\]

\[\text{ Thus, }x = \frac{1}{2}\text{ is the solution of the given equation .} \]

\[\text{ Check: }\]

\[\text{ Substituting }x = \frac{1}{2}\text{ in the given equation, we get: }\]

\[\text{ L . H . S . }= \frac{9(\frac{1}{2}) - 7}{3(\frac{1}{2}) + 5} = \frac{9 - 14}{3 + 10} = \frac{- 5}{13}\]

\[\text{ R . H . S }. = \frac{3(\frac{1}{2}) - 4}{\frac{1}{2} + 6} = \frac{3 - 8}{1 + 12} = \frac{- 5}{13}\]

\[ \therefore\text{ L . H . S . = R . H . S . for }x = \frac{1}{2}\]

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Chapter 9: Linear Equation in One Variable - Exercise 9.3 [Page 17]

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RD Sharma Mathematics [English] Class 8
Chapter 9 Linear Equation in One Variable
Exercise 9.3 | Q 16 | Page 17

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