Solve for x- 1-45x-4-1-32x-1

The correct option is C 16/7 Given, 14(5x+4)=13(2x 1) ⇒ 3(5x+4)=4(2x 1) #160; #160; #160; #160; #160; #160; #160; #160; #160 ...

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Byju's Answer

Standard VIII

Mathematics

Solving Equations with Variables on Both Sides

Solve for x...

Question

Solve for $x$ : $\frac{1}{4} (5 x + 4) = \frac{1}{3} (2 x - 1)$

\frac{- 16}{7}

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\frac{10}{7}

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\frac{8}{7}

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\frac{- 15}{7}

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Solution

The correct option is C $\frac{- 16}{7}$
Given, $\frac{1}{4} (5 x + 4) = \frac{1}{3} (2 x - 1)$
$\Rightarrow 3 (5 x + 4) = 4 (2 x - 1)$ ( $cross multiply$ )
$\Rightarrow 15 x + 12 = 8 x - 4$
$\Rightarrow 15 x - 8 x = - 4 - 12$
$\Rightarrow 7 x = - 16$
$\Rightarrow x = - \frac{16}{7}$

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Similar questions

Q. Solve for

x : \frac{(3 x + 1)}{16} + \frac{(2 x - 3)}{7} = \frac{(x + 3)}{8} + \frac{(3 x - 1)}{14}

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$