SOLUTION: Retaken Geometry This Semester And Proofs were my biggest error last year, need help! 1. 1/2(4m+3)+1/3 2. 4(3r+1)-1/2(10r+6) 3. 1-3(2a+3b)-7(4-2a)+2 SOLVE 4. 2m+6(5-3

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Question 910273: Retaken Geometry This Semester And Proofs were my biggest error last year, need help!
1. 1/2(4m+3)+1/3
2. 4(3r+1)-1/2(10r+6)
3. 1-3(2a+3b)-7(4-2a)+2
SOLVE
4. 2m+6(5-3m)+22=0
5. 1-3(2m+2)+2m=5

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi
The following are completed by first using the Distributive Principle:
1. 1/2(4m+3)+1/3
2m + 3/2 + 1/3 Distributive Principle
0r 2m + 11/3 simplification
2. 4(3r+1)-1/2(10r+6)
12r + 4 - 5r - 3 Distributive Principle
0r 7r+1 simplification
3. 1-3(2a+3b)-7(4-2a)+2
1 - 6a - 6b -28 + 14a + 2 Distributive Principle
0r 8a - 6b -25 simplification
4. 2m+6(5-3m)+22=0
2m + 30 - 18m + 22 = 0 Distributive Principle
52 = 16m Addition Property of Equality
52/16 = m Multiplication Property of Equality
m = 13/4 0r 3 1/4
5. 1-3(2m+2)+2m=5
1-6m - 6 + 2m = 5 Distributive Principle
-10 = 4m Addition Property of Equality
-10/4 = m Multiplication Property of Equality
m = = -5/2 0r -2 1/2
In Solving: GOAL is to isolate the variable on one side of the EQ
so as You can read x = its value

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