How do you solve $7(w+1)=5(2w-3)+1$ ?

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Answer 1 · 2016-07-14T07:38:35

$w = 7$

Start by expanding both sides of the equation:

$7(w + 1) = 5(2w - 3) + 1$

$= 7w + 7 = 10w - 15 + 1$

Then we simplify by getting the terms with $w$ on one side and the integers on the other:

$7w - 7w + 7 + 14 = 10w - 7w - 14 + 14$

Therefore,

$21 = 3w$

By dividing both sides by $3$ we see that:

$21/3 = 3w/3 = 7 = w$

You can verify this by putting $7$ back into the original equations:

$7(7 + 1) = 5(14 - 3) + 1$

$= 7(8) = 5(11) + 1$

$= 56 = 55 + 1$

How do you solve 7(w+1)=5(2w-3)+17(w+1)=5(2w-3)+1?