How do you check that the solution is 9 for the equation $3 (x - 1) - 2 (x + 3) = 0$ $3(x - 1) - 2(x + 3) = 0$ ?

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Answer 1 · 2014-09-28T17:48:35

$3 (x - 1) - 2 (x + 3) = 0$ $3( x-1) -2 (x+3) =0$

Distribute the $3$ $3$ and the $- 2$ $-2$

$3 \cdot x - 3 \cdot 1 - 2 \cdot x - 2 \cdot 3 = 0$ $3*x -3*1 -2*x -2*3 = 0$

$3 x - 3 - 2 x - 6 = 0$ $3x -3 -2x -6 = 0$

Group the like terms

$3 x - 2 x - 3 - 6 = 0$ $3x -2x -3 -6 =0$

Combine the like terms

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

adding $9$ $9$ on both sides to get value of $x$ $x$ ;

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

$x = 9$ $x =9$

In order to check if $9$ $9$ is the correct answer , substitute $9$ $9$ as value of $x$ $x$ in the above equation;

$3 (x - 1) - 2 (x + 3) = 0$ $3( x-1) -2 (x+3) =0$

$3 (9 - 1) - 2 (9 + 3) = 0$ $3 ( 9-1) -2 (9 +3) =0$

$3 (8) - 2 (12) = 0$ $3(8) -2 (12) =0$

$24 - 24 = 0$ $24 -24 =0$ ( Proved)

How do you check that the solution is 9 for the equation 3(x - 1) - 2(x + 3) = 03(x−1)−2(x+3)=03(x - 1) - 2(x + 3) = 0?