Hi again dear
Well Darling;
• -2x + 5 = 4x - 7
• -2x -4x = - 7 - 5
• -6x = -12
• x = -12 / -6
{ Here cancel ' - ' }
• x = 12 / 6
• x = + 2
Good Luck sweetheart ♣
2006-10-26 08:53:16
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answer #1
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answered by sweetie 5
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-2x + 5 = 4x - 7 place numbers on one side and xs' on the other
7+5=4x+2x note: signs changed when moved to the otherside
12=6x just been added together
12/6=x 12 is divided by 6
2=x always follow the same steps
2006-10-26 11:38:47
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answer #2
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answered by krash 2
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- 2x + 5 = 4x - 7
- 2x + 5 - 4x = 4x - 7 - 4x
- 6x + 5 = - 7
- 6x + 5 - 5 = - 7 - 5
- 6x = - 12
- 6x/- = - 12/- 6
x = 2
The answer is x = 2
Insert the x value into the equation
- - - - - - - - - - - - - -
- 2x + 5 = 4x - 7
- 2(2) + 5 = 4(2) - 7
- 4+ 5 = 8 - 7
1 = 1
- - - - - - s-
2006-10-26 12:15:35
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answer #3
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answered by SAMUEL D 7
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-2x + 5 = 4x - 7
+2x +7 = +2x +7
12 = 6x
then
12/6 = 6x/6
2=x
2006-10-26 11:37:22
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answer #4
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answered by jasonheavilin 3
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-2x + 5 = 4x - 7
-2x + 2x + 5 = 4x + 2x - 7 (add 2x on both sides)
5 = 6x - 7
7 + 5 = 6x -7 +7 (add 7 on both sides)
12 = 6x
12/6 = 6x/6 (divide both sides by 6)
2 = x
Ans: x = 2
2006-10-26 11:36:25
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answer #5
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answered by ludacrusher 4
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-2x +5 = 4x - 7
-2x + 2x +5 = 4x + 2x -7
5 +7 = 6x -7 +7
12 = 6x
2 = x
2006-10-26 11:36:21
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answer #6
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answered by YellaMelaDude 3
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-2x + 5 = 4x - 7
first move all the like on one side
5=2x + 4x - 7
then add them together:
5 + 7 = 6x
then solve:
12/6 = x
3 = x
2006-10-26 11:34:20
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answer #7
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answered by southernparadise27 2
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-2x + 12 = 4x (add 7 to both sides)
12 = 6x (add 2x to both sides)
2=x (divide by 6)
2006-10-26 11:33:31
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answer #8
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answered by MJFrog 2
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-2x + 5 = 4x - 7
add 2x to both sides
5=6x-7
add 7 to both sides: now you have all x on one side and numbers on other:
12=6x
divide both sides by 6 and you have the answer
x=2
Hope that helps!
2006-10-26 11:31:03
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answer #9
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answered by D 3
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x=2
Here's how: add 2x to both sides and you will have 6x on the right, add 7 to both sides and you have 12 on the left, divide both sides by six and you get x=2
2006-10-26 11:39:02
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answer #10
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answered by devin_mcabee 1
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