How do you solve $2 (y + 2) + y = 19 - (2 y + 3)$ $2(y+2)+y=19-(2y+3)$ ?

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How do you solve $2 (y + 2) + y = 19 - (2 y + 3)$ $2(y+2)+y=19-(2y+3)$ ?

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Answer 1 · 2015-03-25T03:32:25

First get rid of the brackets multiplying:
$2 y + 4 + y = 19 - 2 y - 3$ $2y+4+y=19-2y-3$
Isolate $y$ $y$ on the left and numbers on the right:
$2 y + y + 2 y = - 4 + 19 - 3$ $2y+y+2y=-4+19-3$
Add and subtract;
$5 y = 12$ $5y=12$
Divide by $5$ $5$ both sides to get:
$y = \frac{12}{5}$ $y=12/5$

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How do you solve $2 (y + 2) + y = 19 - (2 y + 3)$ $2(y+2)+y=19-(2y+3)$ ?

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How do you solve 2(y+2)+y=19−(2y+3)2(y+2)+y=19-(2y+3)?

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How do you solve $2 (y + 2) + y = 19 - (2 y + 3)$ $2(y+2)+y=19-(2y+3)$ ?