How do you solve $3 x + 4 (x - 2) = 5 x + 6$ $3x + 4( x - 2) = 5x + 6$ ?

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Answer 1 · 2017-07-01T15:50:16

$x = 7$ $x=7$

Use the distributive property on the $4 (x - 2)$ $4(x-2)$ term:

$3 x + 4 x - 8 = 5 x + 6$ $3x+4x-8=5x+6$

$7 x - 8 = 5 x + 6$ $7x-8=5x+6$

Let's get the $x$ $x$ terms on one side of the equation and the constant terms on the other.

Let's first subtract $5 x$ $5x$ from both sides of the equation:

$2 x - 8 = 6$ $2x-8=6$

Now, let's add 8 to both sides of the equation:

$2 x = 14$ $2x=14$

Finally, let's dive both sides of the equation by $2$ $2$ to find $x$ $x$ :

$x = 7$ $x=7$

How do you solve 3x + 4( x - 2) = 5x + 63x+4(x−2)=5x+63x + 4( x - 2) = 5x + 6?