Solve for x- 3x-1 6 - 2x-3 7 - x-3 8 - 3x-1 14

The correct option is A 5 3x+1 6 + 2x 3 7 = x+3 8 + 3x 1 14 ( 21x+7+12x 18 ) #160; 42 = 14x+42+24x 8 112 8( 33x 11 ) =3( 38x+34 ...

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Byju's Answer

Standard VIII

Mathematics

Applications (Word Problem)

Solve for x...

Question

Solve for $x$ :
$\frac{(3 x + 1)}{6} + \frac{(2 x - 3)}{7} = \frac{(x + 3)}{8} + \frac{(3 x - 1)}{14}$

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Solution

The correct option is A $5$
$\frac{3 x + 1}{6} + \frac{2 x - 3}{7} = \frac{x + 3}{8} + \frac{3 x - 1}{14}$
$\frac{(21 x + 7 + 12 x - 18)}{42} = \frac{14 x + 42 + 24 x - 8}{112}$
$8 (33 x - 11) = 3 (38 x + 34)$
$264 x - 88 = 114 x + 102$
$150 x = 190$
So, $x = \frac{190}{150} = \frac{19}{15}$

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Similar questions

Solve the equations:

(i) 5x = 3x + 24;

(ii) 8t + 5 = 2t − 31;

(iii) 7x − 10 = 4x + 11;

(iv) 4z + 3 = 6 + 2z;

(v) 2x − 1 = 14 − x;

(vi) 6x + 1 = 3(x − 1) + 7;

(vii) ;

(viii) ;

(ix) 3(x + 1) = 12 + 4 (x − 1);

(x) 2x − 5 = 3(x − 5);

(xi) 6(1 − 4x) + 7(2 + 5x) = 53;

(xii) 3(x + 6) + 2 (x + 3) = 64;

(xiii) ;

(xiv) .

Solve each of the following equations and also verify your solution :

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

Solve :

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