How do you solve $3 (2 x + 3 x + 2) = - 4 x + 124$ $3( 2x + 3x + 2) = - 4x + 124$ ?

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How do you solve $3 (2 x + 3 x + 2) = - 4 x + 124$ $3( 2x + 3x + 2) = - 4x + 124$ ?

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Answer 1 · 2018-05-28T21:48:54

$x = \frac{118}{19}$ $x = 118/19$

Explanation:

First, expand by multiplying $3$ $3$ by every term in the brackets. Then group together the terms with $x$ $x$ in them and group the terms without $x$ $x$ s in them together.

Finally, divide the number in front of the $x$ $x$ by the number on the other side to solve for $x$ $x$ .

$3 (2 x + 3 x + 2) = - 4 x + 124$ $3 (2x + 3x + 2) = -4x + 124$

$3 \cdot 2 x + 3 \cdot 3 x + 3 \cdot 2 = - 4 x + 124$ $3*2x + 3*3x + 3*2 = -4x + 124$

$6 x + 9 x + 6 = - 4 x + 124$ $6x + 9x + 6 = -4x + 124$

$15 x + 6 = - 4 x + 124$ $15x + 6 = -4x + 124$

$15 x + 4 x = 124 - 6$ $15x + 4x = 124 - 6$

$19 x = 118$ $19x = 118$

$x = \frac{118}{19} = 6.21$ $x = 118/19 = 6.21$

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How do you solve $3 (2 x + 3 x + 2) = - 4 x + 124$ $3( 2x + 3x + 2) = - 4x + 124$ ?

1 Answer

Explanation:

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How do you solve 3(2x+3x+2)=−4x+1243( 2x + 3x + 2) = - 4x + 124?

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How do you solve $3 (2 x + 3 x + 2) = - 4 x + 124$ $3( 2x + 3x + 2) = - 4x + 124$ ?