How do you solve $7 y + 2 (y - 2) = 6 (y + 1) - 3$ $7y + 2( y - 2) = 6( y + 1) - 3$ ?

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How do you solve $7 y + 2 (y - 2) = 6 (y + 1) - 3$ $7y + 2( y - 2) = 6( y + 1) - 3$ ?

2 Answers

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Answer 1 · 2018-04-04T11:45:32

$y = 3$ $y=3$

Explanation:

$7 y + 2 \cdot (y - 2) = 6 \cdot (y + 1) - 3$ $7y+2*(y-2)=6*(y+1)-3$

$7 y + 2 y - 6 = 6 y + 6 - 3$ $7y+2y-6=6y+6-3$

$9 y - 6 = 6 y + 3$ $9y-6=6y+3$

$3 y = 9$ $3y=9$

$y = 3$ $y=3$

Answer 2 · 2018-04-04T11:46:38

Put all the y on one side of the equation, all numbers on the other.

Explanation:

$7 \cdot y + 2 \cdot (y - 2) = 6 \cdot (y + 1) - 3$ $7*y+2*(y−2)=6*(y+1)−3$

Multiply the parenthesis.

$7 \cdot y + 2 \cdot y + 2 \cdot (- 2) = 6 \cdot y + 6 - 3$ $7*y+2*y+2*(-2)=6*y+6-3$
$7 \cdot y + 2 \cdot y - 4 = 6 \cdot y + 6 - 3$ $7*y+2*y-4=6*y+6-3$
$9 \cdot y - 4 = 6 \cdot y + 3$ $9*y-4=6*y+3$

Subtract 6y, and add 4.

$9 \cdot y - 6 \cdot y - 4 + 4 = 6 \cdot y - 6 \cdot y + 3 + 4$ $9*y-6*y-4+4=6*y-6*y+3+4$
$9 \cdot y - 6 \cdot y = 3 + 4$ $9*y-6*y=3+4$
$3 \cdot y = 7$ $3*y=7$

Divide with 3.

$y = \frac{7}{3}$ $y=7/3$

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How do you solve $7 y + 2 (y - 2) = 6 (y + 1) - 3$ $7y + 2( y - 2) = 6( y + 1) - 3$ ?

2 Answers

Explanation:

Explanation:

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How do you solve $7 y + 2 (y - 2) = 6 (y + 1) - 3$ $7y + 2( y - 2) = 6( y + 1) - 3$ ?