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Solve for z. -6z+2=3z+4z+28

0 votes
How would you solve this equation?
asked Mar 6, 2013 in BASIC MATH by skylar Apprentice

2 Answers

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–6z+2 =3z +4z+28

Add 6z to each side.

2 =3z+4z+6z+28

2 =13z+28                              (Since 3z + 6z + 4z = 13z)

Subtract 2 from each side.

2– 2=13z+28– 2

13z+28– 2= 0                         (Since 2 - 2 = 0)

13z+26 =0                              (since 28–2 =26)

Subtract 26 from each side.

13z = –26

Divide each side by 13.

13z/13 = –26/13

z = –2.

answered Mar 6, 2013 by hussy Rookie
0 votes
-6z+2=3z+4z+28

combine like terms using distributive property a(b + c) = ab + ac
-6z+2=7z+28

addition property of equality: if a = b then a + c = b + c
add 6z to each side
2 = 13z + 28

subtraction property of equality: if a = b then a - c = b - c
subtract 28 from each side
-26 = 13z

division property of equality: if a = b then a / c = b / c (c not equal to zero)
divide each side by 13
z = -2
answered Mar 6, 2013 by steve Scholar

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