What is the answer to 2k-3(k+5)=2k-(3k-2)-(3k-7) ?

The answer to 2k-3(k+5)=2k-(3k-2)-(3k-7) is k=8

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2k-3(k+5)=2k-(3k-2)-(3k-7)

2 k - 3 (k + 5) = 2 k - (3 k - 2) - (3 k - 7)

k = 8

Solution steps

2 k - 3 (k + 5) = 2 k - (3 k - 2) - (3 k - 7)

Subtract

2 k

from both sides

2 k - 3 (k + 5) - 2 k = 2 k - (3 k - 2) - (3 k - 7) - 2 k

Simplify

- 3 (k + 5) = - (3 k - 2) - (3 k - 7)

Expand

- 3 (k + 5) : - 3 k - 15

Expand

- (3 k - 2) - (3 k - 7) : - 6 k + 9

- 3 k - 15 = - 6 k + 9

Move

15

to the right side

- 3 k = - 6 k + 24

Move

6 k

to the left side

3 k = 24

Divide both sides by

3

k = 8

7 x + 15 a x - 110 + \frac{185}{9} a x = 18 x - 5 a + \frac{12}{57} a

\frac{1 + r}{- 1.5} = 9

x, - 3 x - 6 y = 0

y = lo g_{4} (64)

5 x + 12 = - 13

What is the answer to 2k-3(k+5)=2k-(3k-2)-(3k-7) ?
The answer to 2k-3(k+5)=2k-(3k-2)-(3k-7) is k=8
Solve for k: 2k-3(k+5)=2k-(3k-2)-(3k-7)
The solution is k=8