Question

Solution of equation 2(x - 1)- 3(4x + 1)=2(x -1 )-3(5x - 4) + 6​ Find the x​

Answer 1

Answer:

2(x-1) - 3(4x+1) = 2(x-1) - 3(5x - 4) + 6x

→ \red{\sf{ [ 2(x) - 2(1) ] }}[2(x)−2(1)] - 3(4x+1) = 2(x-1) - 3(5x - 4) + 6x

→ \red{ \sf (2x - 2) }(2x−2) - 3(4x+1) = 2(x-1) - 3(5x - 4) + 6x

→ (2x - 2) - \red{\sf[ 3(4x)+3(1)] }[3(4x)+3(1)] = 2(x-1) - 3(5x - 4) + 6x

→ (2x - 2) - \red{\sf (12x+3) }(12x+3) = 2(x-1) - 3(5x - 4) + 6x

→ (2x - 2) \red{\sf -12x-3 }−12x−3 = 2(x-1) - 3(5x - 4) + 6x

→ 2x - 2 - 12x - 3 = \pink{\sf [2(x) - 2(1)] }[2(x)−2(1)] - 3(5x - 4) + 6x

→ 2x - 2 - 12x - 3 = \pink{\sf (2x - 2) }(2x−2) - 3(5x - 4) + 6x

→ 2x - 2 - 12x - 3 = (2x - 2) - \pink{\sf [3(5x) - 3(4) ]}[3(5x)−3(4)] + 6x

→ 2x - 2 - 12x - 3 = (2x - 2) - \pink{\sf (15x - 12) ]}(15x−12)] + 6x

→ 2x - 2 - 12x - 3 = (2x - 2) - 15x + 12 + 6x

→ (2x - 12x) - 3 - 2 = 2x + 6x - 15x - 2 + 12

→ (- 10x) - 5 = 8x - 15x + 10

→ (-10x) - 5 = -7x + 10

→ (-10x) + 7x = 10 + 5

[ ∵ taking (-5) to RHS and (-7x) to LHS ]

→ (-3x) = 15

→ x = 15 ÷ (-3)

→ x = -5

Answer 2

Answer:

=  -5

Step-by-step explanation:

2(x-1) - 3(4x+1) = 2(x-1) - 3(5x - 4) + 6x

→ \red{\sf{ [ 2(x) - 2(1) ] }}[2(x)−2(1)] - 3(4x+1) = 2(x-1) - 3(5x - 4) + 6x

→ \red{ \sf (2x - 2) }(2x−2) - 3(4x+1) = 2(x-1) - 3(5x - 4) + 6x

→ (2x - 2) - \red{\sf[ 3(4x)+3(1)] }[3(4x)+3(1)] = 2(x-1) - 3(5x - 4) + 6x

→ (2x - 2) - \red{\sf (12x+3) }(12x+3) = 2(x-1) - 3(5x - 4) + 6x

→ (2x - 2) \red{\sf -12x-3 }−12x−3 = 2(x-1) - 3(5x - 4) + 6x

→ 2x - 2 - 12x - 3 = \pink{\sf [2(x) - 2(1)] }[2(x)−2(1)] - 3(5x - 4) + 6x

→ 2x - 2 - 12x - 3 = \pink{\sf (2x - 2) }(2x−2) - 3(5x - 4) + 6x

→ 2x - 2 - 12x - 3 = (2x - 2) - \pink{\sf [3(5x) - 3(4) ]}[3(5x)−3(4)] + 6x

→ 2x - 2 - 12x - 3 = (2x - 2) - \pink{\sf (15x - 12) ]}(15x−12)] + 6x

→ 2x - 2 - 12x - 3 = (2x - 2) - 15x + 12 + 6x

→ (2x - 12x) - 3 - 2 = 2x + 6x - 15x - 2 + 12

→ (- 10x) - 5 = 8x - 15x + 10

→ (-10x) - 5 = -7x + 10

→ (-10x) + 7x = 10 + 5

[ ∵ taking (-5) to RHS and (-7x) to LHS ]

→ (-3x) = 15

→ x = 15 ÷ (-3)

→ x = -5