We have, #160; 3 ×25 = 1x5 ⇒65 = 5 + x5 ⇒ 6 = 5 + x [ cancle 5 both sides] ⇒ x = 6 5 ∴ x = 1 Hence, the value x=1 ...

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Byju's Answer

Standard IX

Mathematics

Linear Equation in One Variable

3×25 = 1x5....

Question

$3 \times \frac{2}{5} =$ $1 \frac{x}{5}$ . Find x.

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Solution

We have,
$3 \times \frac{2}{5} = 1 \frac{x}{5}$

$\Rightarrow \frac{6}{5} = \frac{5 + x}{5}$

$\Rightarrow 6 = 5 + x [c a n c l e 5 b o t h s i d e s]$

$\Rightarrow x = 6 - 5$

$∴ x = 1$
Hence, the value $x = 1$

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Q. Solve for x:

(i)

\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}; x \neq 2, 4

(ii)

\frac{1}{x} - \frac{1}{x - 2} = 3, x \neq 0, 2

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\frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6}, x \neq 3, - 5

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3×25= 1x5. Find x.

We have, 3×25=1x5⇒65=5+x5⇒6=5+x[cancle5bothsides]⇒x=6−5∴x=1Hence, the value x=1

$3 \times \frac{2}{5} =$ $1 \frac{x}{5}$ . Find x.

We have,
$3 \times \frac{2}{5} = 1 \frac{x}{5}$

$\Rightarrow \frac{6}{5} = \frac{5 + x}{5}$

$\Rightarrow 6 = 5 + x [c a n c l e 5 b o t h s i d e s]$

$\Rightarrow x = 6 - 5$

$∴ x = 1$
Hence, the value $x = 1$