How do you solve $3 (x - 2) + 2 (x + 5) = 5 (x + 1) + 1$ $3(x-2)+2(x+5)=5(x+1)+1$ ?

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How do you solve $3 (x - 2) + 2 (x + 5) = 5 (x + 1) + 1$ $3(x-2)+2(x+5)=5(x+1)+1$ ?

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Answer 1 · 2017-01-09T19:43:40

See below

There are no values of $x$ $x$ which are a solution to the problem therefore $x$ $x$ is the null set or $x = {\emptyset}$ $x = {O/}$

Explanation:

Expand the terms in parenthesis:

$3 x - 6 + 2 x + 10 = 5 x + 5 + 1$ $3x - 6 + 2x + 10 = 5x + 5 + 1$

$5 x + 4 = 5 x + 6$ $5x + 4 = 5x + 6$

Subtract $5 x$ $5x$ from each side of the equation:

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

$0 + 4 = 0 + 6$ $0 + 4 = 0 + 6$

$4 = 6$ $4 = 6$

Because $4 \neq 6$ $4 != 6$ there are no values of $x$ $x$ which are a solution to the problem therefore $x$ $x$ is the null set or $x = {\emptyset}$ $x = {O/}$

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How do you solve $3 (x - 2) + 2 (x + 5) = 5 (x + 1) + 1$ $3(x-2)+2(x+5)=5(x+1)+1$ ?

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