How do you solve $3 (x - 1) - 4 (x - 3) = 0$ $3(x-1)-4(x-3)=0$ ?

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Answer 1 · 2018-04-16T01:52:09

$x = 9$ $x=9$

$3 (x - 1) - 4 (x - 3) = 0$ $3(x-1)-4(x-3)=0$

Start with the distributive property

$(3) (x) + (3) (- 1) + (- 4) (x) + (- 4) (- 3) = 0$ $(3)(x)+(3)(-1)+(-4)(x)+(-4)(-3)=0$

$3 x - 3 - 4 x + 12 = 0$ $3x - 3 - 4x + 12 =0$

$- x + 9 = 0$ $-x + 9 = 0$

Subtract $9$ $9$ on both sides

$- x + 9 - 9 = 0 - 9$ $-x + 9 - 9 = 0 - 9$

$- x = - 9$ $-x = -9$

Multiply both sides by $- 1$ $-1$

$- x \times - 1 = - 9 \times - 1$ $-x xx-1 = -9 xx -1$

$x = 9$ $x = 9$

How do you solve 3(x−1)−4(x−3)=03(x-1)-4(x-3)=0?