1)2x+3x=10+5 put the variable on one side
5x=15 addition
x=3 divide by 5
2)4x+10=2*2x+2*5 distribute the right side
4x+10=4x+10 solve the distribution
4x=4x subtract 10, then divide by 4
x=x so A
2006-10-20 06:02:35
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answer #1
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answered by suprasteve 3
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1) 2x - 5 = -3x +10
2x - 5 + 5 = -3x + 10 +5
2x = -3x +15
2x +3x = -3x +15 +3x
5x = 15
5x/5 = 15/5
x = 3
2) 4x +10 = 2(2x +5)
4x +10 = 4x +10
4x +10 -10 = 4x +10 -10
4x = 4x +0
4x -4x =4x +0 -4x
0=0
2006-10-20 20:55:45
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answer #2
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answered by Anonymous
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1) 2x - 5 = -3x + 10
2x = -3x +15
5x = 15
x = 3
2) 4x + 10 = 2(2x+5)
4x + 10 = 4x + 10
4x = 4x
x = x
*There are no real solutions. This formula just tells you that x equals x. For a real solution x has to equal a number, not an indefinite value like x or y.
2006-10-20 16:05:07
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answer #3
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answered by gremlin_lemon 2
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1) 2x-5= -3x+10
+5 +5(add 5 to both sides to cancel the 5 on the left)
_____________
2x= -3x+15
+3x +3x(add 3 to both sides to cancel the 3 on the right)
_________
5x=15(divide both sides by 5)
x=3(check by plugging 3 back into equation above
2)4x+10=2(2x+5)
4x+10=4x+10(distribute the 2 into the right side equation)
answer: A
2006-10-20 13:14:24
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answer #4
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answered by Caduceus89 4
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2x-5=-3x+10, so 5x=15 means x=3
4x+10=2(2x+5) i.e 4x+10 so x=x
taking both of the above a) is the corret choice
2006-10-21 01:55:49
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answer #5
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answered by grandpa 4
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first, do your own work. in first problem..add 3x to each side and add 5 to each side so you get 5x=5 which means x=1
on second problem, 4x+10=4x+10....they are the same so any real number you use for x will do the same thing for both equations so all real numbers has to be the answer...try it..put in x=3 and then x=5...see...it stays equal...
2006-10-20 13:18:20
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answer #6
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answered by h b 1
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1)1
2) x=0
2006-10-20 13:09:01
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answer #7
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answered by Tats 1
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well , if these two parts are related then a is correct . because the first state lead you to X=3 and the second one leads to X=X and for the answer you should look for something which is correct in both situation and that is A .
2006-10-20 13:17:46
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answer #8
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answered by arash 3
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