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Solve the Following Equation and Verify Your Answer: 7 X − 2 5 X − 1 = 7 X + 3 5 X + 4 - Mathematics

Sum

Solve the following equation and verify your answer:

\[\frac{7x - 2}{5x - 1} = \frac{7x + 3}{5x + 4}\]
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Solution

\[\frac{7x - 2}{5x - 1} = \frac{7x + 3}{5x + 4}\]

\[\text{ or }35 x^2 + 28x - 10x - 8 = 35 x^2 + 15x - 7x - 3 [\text{ After cross multiplication }]\]

\[\text{ or }35 x^2 - 35 x^2 + 18x - 8x = - 3 + 8\]

\[\text{ or }10x = 5\]

\[\text{ or }x = \frac{5}{10} \text{ or } x = \frac{1}{2}\]

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

\[\text{ Check: }\]

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

\[\text{ L . H . S . }= \frac{7(\frac{1}{2}) - 2}{5(\frac{1}{2}) - 1} = \frac{7 - 4}{5 - 2} = \frac{3}{3} = 1\]

\[\text{ R . H . S .} = \frac{7(\frac{1}{2}) + 3}{5(\frac{1}{2}) + 4} = \frac{7 + 6}{5 + 8} = \frac{13}{13} = 1\]

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

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APPEARS IN

RD Sharma Class 8 Maths
Chapter 9 Linear Equation in One Variable
Exercise 9.3 | Q 13 | Page 17
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