$4 (2 x - 1) - 2 (x - 5) = 5 (x + 1) + 3$

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Solution

$4 (2 x - 1) - 2 (x - 5) = 5 (x + 1) + 3$
$8 x - 4 - 2 x + 10 = 5 x + 5 + 3$
$6 x + 6 = 5 x + 8$
$x = 8 - 6$
$x = 2$ .

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Similar questions

4 (2 x - 1) - 2 (x - 5) = 5 (x + 1) + 3

Find $x$ , if:

(i) $4^{2 x} = \frac{1}{32}$

(ii) $\sqrt{2^{x + 3}} = 16$

(iii) ${(\sqrt{\frac{3}{5}})}^{x + 1} = \frac{125}{27}$

(iv) ${(\sqrt[3]{\frac{2}{3}})}^{x - 1} = \frac{27}{8}$

Find the values of x in each of the following:

$(i) 2^{5 x} \div 2^{x} = \sqrt[5]{2^{20}}$

$(i i) (2^{3})^{4} = (2^{2})^{x}$

$(i i i) {(\frac{3}{5})}^{x} {(\frac{5}{3})}^{2 x} = \frac{125}{27}$

$(i v) 5^{x - 2} \times 3^{2 x - 3} = 135$

$(v) 2^{x - 7} \times 5^{x - 4} = 1250$

$(v i) (\sqrt[3]{4})^{2 x + \frac{1}{2}} = \frac{1}{32}$

$(v i i) 5^{2 x + 3} = 1$

$(v i i i) (13)^{\sqrt{x}} = 4^{4} - 3^{4} - 6$

$(i x) {(\sqrt{\frac{3}{5}})}^{x + 1} = \frac{125}{27}$

5 (x + 3) = 4 (2 x + 1)

\frac{5 x + 4}{2 x + 1} = 2

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