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solve this:

2x+7=3x-4

2006-10-29 16:44:03 · 15 answers · asked by strawberrydaiquiri 3 in Science & Mathematics Mathematics

15 answers

2x +7= 3x- 4
hence 3x-2x= 7+4
X= 11

2006-10-29 16:48:14 · answer #1 · answered by GREY MATTER 2 · 0 0

x=11
2x+7=3x-4 so 3x-2x=7+4 x=11

2006-10-30 02:43:10 · answer #2 · answered by peterwan1982 2 · 0 0

2x + 7 = 3x - 4
-2x -2x

7 = 1x - 4
+4
11 = x

check:
2(11) + 7 = 3(11) - 4
22 + 7 = 33 - 4
29 = 29

2006-10-30 00:55:55 · answer #3 · answered by cassie86 3 · 0 0

You are supposed to find x.

To do that you need to put all the numbers on one side, and all the X's on the other. Whatever you do to one side, you have to do to the other so both sides of the + sign are balanced.

2x+7=3x-4

First add 4, (or subtract 7)

now you have 2x+11=3x

Now subtract 2x from both sides

11=3x-2x

3x-2x=1x or just x

So 11=x is all you are left with.

To check it, substitute 11 for x in the original equation.

2*11+7=3*11-4 or

22+7=33-4

Both equal 29, so x is 11

2006-10-30 00:49:14 · answer #4 · answered by Coop 3 · 0 0

2x+7=3x-4 Subtract 2x from both sides

7=x-4 Add 4 to both sides

x=11

2006-10-30 00:48:28 · answer #5 · answered by futureastronaut1 3 · 0 0

2x + 7=3x - 4

Firstly,arrange the equations so that they are both in the form of x

3x - 2x=7 + 4

Then,solve the equations.Submit 3x with 2x and add 7 with 4.

x = 11

Therefore, x is 11.

2006-10-30 01:51:47 · answer #6 · answered by toshib 1 · 0 0

2x+7=3x-4

2x - 3x = -4 - 7
-x = -11
divide both sides by -1

x = 11

2x+7=3x-4
2(11) + 7 = 3*11 - 4
22 +7 = 33 -4
29 = 29

2006-10-30 00:51:57 · answer #7 · answered by lazareh 2 · 0 0

x=11

2006-10-30 00:57:04 · answer #8 · answered by mustin j 2 · 0 0

2x + 7 = 3x - 4
2x + 7 -3x + 4 = 0
2x - 3x + (7+4) = 0
2x - 3x = -(7+4)
2x-3x = -4 - 7
-1x = -13
x = 13

2006-10-30 01:00:27 · answer #9 · answered by honeymay 2 · 0 0

2x + 7 = 3x - 4
2x + 7 - 7 = 3x -7 - 4
2x - 3x =3x - 3x - 11
- x = - 11
cutting minus we have

x = 11

Ans = 11

2006-10-30 03:01:42 · answer #10 · answered by Anonymous · 0 0

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