2x-(2x-3x-1/8)=2/3(x+2/6)-1/4-এর সমাধান করুন

$$2 x - ( 2 x - \frac { 3 x - 1 } { 8 } ) = \frac { 2 } { 3 } ( \frac { x + 2 } { 6 } ) - \frac { 1 } { 4 }$$

Solve for x

$$x=\frac{7}{19}\approx 0.368421053$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 24, the least common multiple of 8,3,6,4.

$$48x-24\left(2x-\frac{3x-1}{8}\right)=\frac{8}{3}\left(x+2\right)-6$$

Use the distributive property to multiply \frac{8}{3} by x+2.

$$48x-24\left(2x-\frac{3x-1}{8}\right)=\frac{8}{3}x+\frac{8}{3}\times 2-6$$

Express \frac{8}{3}\times 2 as a single fraction.

$$48x-24\left(2x-\frac{3x-1}{8}\right)=\frac{8}{3}x+\frac{8\times 2}{3}-6$$

Multiply 8 and 2 to get 16.

$$48x-24\left(2x-\frac{3x-1}{8}\right)=\frac{8}{3}x+\frac{16}{3}-6$$

Convert 6 to fraction \frac{18}{3}.

$$48x-24\left(2x-\frac{3x-1}{8}\right)=\frac{8}{3}x+\frac{16}{3}-\frac{18}{3}$$

Since \frac{16}{3} and \frac{18}{3} have the same denominator, subtract them by subtracting their numerators.

$$48x-24\left(2x-\frac{3x-1}{8}\right)=\frac{8}{3}x+\frac{16-18}{3}$$

Subtract 18 from 16 to get -2.

$$48x-24\left(2x-\frac{3x-1}{8}\right)=\frac{8}{3}x-\frac{2}{3}$$

Divide each term of 3x-1 by 8 to get \frac{3}{8}x-\frac{1}{8}.

$$48x-24\left(2x-\left(\frac{3}{8}x-\frac{1}{8}\right)\right)=\frac{8}{3}x-\frac{2}{3}$$

To find the opposite of \frac{3}{8}x-\frac{1}{8}, find the opposite of each term.

$$48x-24\left(2x-\frac{3}{8}x-\left(-\frac{1}{8}\right)\right)=\frac{8}{3}x-\frac{2}{3}$$

The opposite of -\frac{1}{8} is \frac{1}{8}.

$$48x-24\left(2x-\frac{3}{8}x+\frac{1}{8}\right)=\frac{8}{3}x-\frac{2}{3}$$

Combine 2x and -\frac{3}{8}x to get \frac{13}{8}x.

$$48x-24\left(\frac{13}{8}x+\frac{1}{8}\right)=\frac{8}{3}x-\frac{2}{3}$$

Use the distributive property to multiply -24 by \frac{13}{8}x+\frac{1}{8}.

$$48x-24\times \frac{13}{8}x-24\times \frac{1}{8}=\frac{8}{3}x-\frac{2}{3}$$

Express -24\times \frac{13}{8} as a single fraction.

$$48x+\frac{-24\times 13}{8}x-24\times \frac{1}{8}=\frac{8}{3}x-\frac{2}{3}$$

Multiply -24 and 13 to get -312.

$$48x+\frac{-312}{8}x-24\times \frac{1}{8}=\frac{8}{3}x-\frac{2}{3}$$

Divide -312 by 8 to get -39.

$$48x-39x-24\times \frac{1}{8}=\frac{8}{3}x-\frac{2}{3}$$

Multiply -24 and \frac{1}{8} to get \frac{-24}{8}.

$$48x-39x+\frac{-24}{8}=\frac{8}{3}x-\frac{2}{3}$$

Divide -24 by 8 to get -3.

$$48x-39x-3=\frac{8}{3}x-\frac{2}{3}$$

Combine 48x and -39x to get 9x.

$$9x-3=\frac{8}{3}x-\frac{2}{3}$$

Subtract \frac{8}{3}x from both sides.

$$9x-3-\frac{8}{3}x=-\frac{2}{3}$$

Combine 9x and -\frac{8}{3}x to get \frac{19}{3}x.

$$\frac{19}{3}x-3=-\frac{2}{3}$$

Add 3 to both sides.

$$\frac{19}{3}x=-\frac{2}{3}+3$$

Convert 3 to fraction \frac{9}{3}.

$$\frac{19}{3}x=-\frac{2}{3}+\frac{9}{3}$$

Since -\frac{2}{3} and \frac{9}{3} have the same denominator, add them by adding their numerators.

$$\frac{19}{3}x=\frac{-2+9}{3}$$

Add -2 and 9 to get 7.

$$\frac{19}{3}x=\frac{7}{3}$$

Multiply both sides by \frac{3}{19}, the reciprocal of \frac{19}{3}.

$$x=\frac{7}{3}\times \frac{3}{19}$$

Multiply \frac{7}{3} times \frac{3}{19} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{7\times 3}{3\times 19}$$

Cancel out 3 in both numerator and denominator.

$$x=\frac{7}{19}$$