What is the answer to solvefor x,((2x+3))/((4y+1))=3+2x-12y ?

The answer to solvefor x,((2x+3))/((4y+1))=3+2x-12y is x=6y;\quad y\ne-1/4 ,y\ne 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,((2x+3))/((4y+1))=3+2x-12y

solve for

x, \frac{( 2 x + 3 )}{( 4 y + 1 )} = 3 + 2 x - 12 y

x = 6 y; y \neq = - \frac{1}{4}, y \neq = 0

Solution steps

\frac{( 2 x + 3 )}{( 4 y + 1 )} = 3 + 2 x - 12 y

Multiply both sides by

4 y + 1

2 x + 3 = 3 (4 y + 1) + 2 x (4 y + 1) - 12 y (4 y + 1); y \neq = - \frac{1}{4}

Expand

3 (4 y + 1) + 2 x (4 y + 1) - 12 y (4 y + 1) : - 48 y^{2} + 8 x y + 2 x + 3

2 x + 3 = - 48 y^{2} + 8 x y + 2 x + 3; y \neq = - \frac{1}{4}

Move

3

to the right side

2 x = - 48 y^{2} + 8 x y + 2 x; y \neq = - \frac{1}{4}

Subtract

8 x y + 2 x

from both sides

2 x - (8 x y + 2 x) = - 48 y^{2} + 8 x y + 2 x - (8 x y + 2 x); y \neq = - \frac{1}{4}

Simplify

- 8 x y = - 48 y^{2}; y \neq = - \frac{1}{4}

Divide both sides by

- 8 y; y \neq = - \frac{1}{4}, y \neq = 0

x = 6 y; y \neq = - \frac{1}{4}, y \neq = 0

u, 2 u + 7 v = 52

y, \frac{( d ^{2} \cdot y )}{( d \cdot x ^{2} )} - y = x - 1

f, x (x^{3} - 1) \cdot f = 1

\frac{( 5 x - 1 )}{4} - \frac{( 3 x - 1 )}{6} = \frac{32}{3}

x, \frac{x}{9} = \frac{3}{y} + \frac{1}{18}

What is the answer to solvefor x,((2x+3))/((4y+1))=3+2x-12y ?
The answer to solvefor x,((2x+3))/((4y+1))=3+2x-12y is x=6y;\quad y\ne-1/4 ,y\ne 0