Solve for x

Solve for x

(1/2)(-4+6x)=(1/3)x+(2/3)(x+9)

\(\begin{array}{rcll} \frac12 \cdot (-4+6x) &=& \frac13 \cdot x+ \frac23 \cdot (x+9)\\ \frac12 \cdot (-4+6x) &=& \frac13 \cdot x+ \frac23\cdot x + \frac23\cdot 9 \qquad & | \qquad \frac13 \cdot x+ \frac23\cdot x = \frac33 \cdot x = x\\ \frac12 \cdot (-4+6x) &=& x + \frac23\cdot 9 \qquad & | \qquad \frac23\cdot 9 = 2 \cdot 3 = 6\\ \frac12 \cdot (-4+6x) &=& x + 6\\ \frac12 \cdot (-4) + \frac12 \cdot 6x &=& x + 6 \qquad & | \qquad \frac12 \cdot (-4) = -2\\ -2 + \frac12 \cdot 6x &=& x + 6 \qquad & | \qquad \frac12 \cdot 6x = 3x\\ -2 + 3x &=& x + 6 \qquad & | \qquad -x\\ -2 + 2x &=& 6 \qquad & | \qquad +2\\ 2x &=& 8 \qquad & | \qquad :2\\ x &=& \frac82 \qquad & | \qquad \frac82 = 4\\ x &=& 4\\ \end{array}\)

Solve for x: (6 x-4)/2 = x/3+(2 (x+9))/3 x/3+(2 (x+9))/3 = (x+2 (x+9))/3: (6 x-4)/2 = (x+2 (x+9))/3 2 (x+9) = 2 x+18: (6 x-4)/2 = (2 x+18+x)/3 Grouping like terms, 2 x+x+18 = (x+2 x)+18: (6 x-4)/2 = ((x+2 x)+18)/3 x+2 x = 3 x: (6 x-4)/2 = (3 x+18)/3 Multiply both sides by 6: (6 (6 x-4))/2 = (6 (3 x+18))/3 6/2 = (2×3)/2 = 3: 3 (6 x-4) = (6 (3 x+18))/3 6/3 = (3×2)/3 = 2: 3 (6 x-4) = 2 (3 x+18) Expand out terms of the left hand side: 18 x-12 = 2 (3 x+18) Expand out terms of the right hand side: 18 x-12 = 6 x+36 Subtract 6 x from both sides: (18 x-6 x)-12 = (6 x-6 x)+36 18 x-6 x = 12 x: 12 x-12 = (6 x-6 x)+36 6 x-6 x = 0: 12 x-12 = 36 Add 12 to both sides: 12 x+(12-12) = 36+12 12-12 = 0: 12 x = 36+12 36+12 = 48: 12 x = 48 Divide both sides of 12 x = 48 by 12: (12 x)/12 = 48/12 12/12 = 1: x = 48/12 The gcd of 48 and 12 is 12, so 48/12 = (12×4)/(12×1) = 12/12×4 = 4: Answer: | | x = 4

1/2(-4+6x)=1/3x+2/3(x+9)     multiply through by the common denominator of 2 and 3  = 6

3(-4 + 6x)  = 2x  + 4(x + 9)      simplify

-12 + 18x  = 2x  + 4x  + 36

-12 + 18x  = 6x  + 36       add12 to both sides, subtract 6x from both sides

12x  = 48      divide both sides by 12

x = 4

u dumb