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Addition and Subtraction of Algebraic Expressions

Find the valu...

Question

Find the value of $x$ if $25 x - 19 - {3 - (4 x - 5)} = 3 x - (6 x - 5)$

1

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Solution

The correct option is A $1$
Given, $25 x - 19 - {3 - (4 x - 5)} = 3 x - (6 x - 5)$
$\Rightarrow 25 x - 19 - {3 - 4 x + 5} = 3 x - 6 x + 5$
$\Rightarrow 25 x - 19 - 8 + 4 x = - 3 x + 5$
$\Rightarrow 29 x - 27 = - 3 x + 5$
$\Rightarrow 32 x = 32$
$\Rightarrow x = 1$

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Similar questions

\frac{17 - 3 x}{5} - \frac{4 x + 2}{3} = 5 - 6 x + \frac{7 x + 14}{3}

x

f (x) = 2 x^{3} - 6 x^{2} + 4 x - 5

g (x) = 3 x^{2} - 9

f (1) + g (- 2)

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) .

3 x (4 x - 5) + 3

x = 3

x = \frac{1}{2}

{(3 x^{2} - \frac{x}{2})}^{5}

= {\frac{x}{2}}^{5} [(6 x)^{5} - 5 (6 x)^{4} + A (6 x)^{3} - 10 (6 x)^{2} + 5 (6 x) - 1]

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