If 2x-7-5x-8 - 2x-6-5x-4 - then x -

The correct option is A x= 62/3 Given, 2x+7 5x+8 = 2x+6 5x+4 ⟹ (2x+7)(5x+4)=(2x+6)(5x+8) ....[By cross multiplication] ⟹ 10 x ...

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Byju's Answer

Standard X

Mathematics

Sum of Infinite Terms of a GP

If 2x+7/5x+...

Question

If $\frac{2 x + 7}{5 x + 8} = \frac{2 x + 6}{5 x + 4}$ , then $x =$

x = - 6 \frac{2}{3}

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x = - 3 \frac{1}{2}

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x = - 2 \frac{1}{2}

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x = - 3 \frac{6}{7}

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Solution

The correct option is A $x = - 6 \frac{2}{3}$
$G i v e n, \frac{2 x + 7}{5 x + 8} = \frac{2 x + 6}{5 x + 4} ⟹ (2 x + 7) (5 x + 4) = (2 x + 6) (5 x + 8) . . . . [B y c r o s s - m u l t i p l i c a t i o n] ⟹ 10 x^{2} + 35 x + 8 x + 28 = 10 x^{2} + 30 x + 16 x + 48 ⟹ 43 x + 28 = 46 x + 48 . . . . . [C a n c e l l i n g 10 x^{2} o n b o t h s i d e s] ⟹ 43 x - 46 x = 48 - 28 . . . . [T r a n s p o s i n g 48 x t o L H S a n d 28 t o R H S] ⟹ - 3 x = 20 ⟹ x = \frac{- 20}{3} o r - 6 \frac{2}{3}$

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Similar questions

Solve :- $(3 x^{2} + 5 x - 7) (x - 1) - (x^{2} - 2 x + 3) (x + 4)$

\frac{x^{2} - 7 x + 10}{x^{2} - 5 x - 14} \times \frac{x^{3} + 8}{x^{3} - 8} \times \frac{x^{2} + 2 x + 4}{x^{2} - 2 x + 4} \times \frac{x - 7}{x - 5} = ?

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$