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Where do I start, please help me understand..thank you.

2007-06-29 13:09:20 · 6 answers · asked by Christ' Soldier 2 in Science & Mathematics Mathematics

6 answers

1) 12x -6 = 11x (put 11x and 12x together)

12x - 11x = 6 (find x => the difference of 12x -11x)

x = 6

2) 5x + 8 +3x -x +5 = 6x - 3 (put variables togegher and
constants together)

5x + 3x -x -6x = -3 -8 -5 ( find the sum both sides of the
equation)

8x -7x = - 16

x = -16

2007-06-29 13:20:43 · answer #1 · answered by frank 7 · 0 0

what are you trying to do.. solve for x???

if that's the case..

1) 12x - 6 = 11x
1.a) 12x - 11x = 6
1.b) x = 6
check:
(12 * 6) - 6 =? (11 * 6)
72 - 6 =? 66
66 = 66

2) 5x + 8 + 3x - x + 5 = 6x - 3
2.a) 5x + 3x - x - 6x = -3 - 8 - 5
2.b) x = - 16
check:
(5 *( -16)) + 8 + (3 *( -16)) - (-16) + 5 =? (6 * (-16)) - 3
(-80) + 8 + (-48) + 16 + 5 =? -96 - 3
-99 = -99

hope this helps.

2007-06-29 20:25:14 · answer #2 · answered by Anonymous · 0 0

Remember that you can do the same thing to each side of any equation and both sides will still be equal.
In this equation , 12x-6=11x, just take away 11x from each side
Then, x-6 = 0. Now add +6 to both sides, so x = 6

For the second equation, same idea but first combine all those x's and numbers: 5x+8+3x-x+5=6x-3
= 7x + 13 = 6x -3.
Now subtract 6x from both sides x + 13 = -3
then subtract 13 from each side x= -16

:)

2007-06-29 20:11:38 · answer #3 · answered by ignoramus 7 · 0 0

12x-6=11x; 12x-11x=6. 1st group all like terms.In so doing the signs change. x=6. 5x+8+3x-x+5=6x-3= 5x+8+2x+5=6x-3=5x+2x+8+5=6x-3 =7x+13=6x-3 =7x-6x=-3-13 ; x=-16. I take it that this was two equations

2007-06-29 20:21:14 · answer #4 · answered by elva f 2 · 0 0

group similar terms together... all x terms on one side and all the whole numbers on the other
while taking thr number from one side to the other make sure to invert the signs
12x - 11x = 6
so,
x = 6
and for...
5x + 3x - x - 6x = -3 -8 -5
x = -16

2007-06-29 20:25:16 · answer #5 · answered by Sindhoor 2 · 0 0

gather like terms and solve from there

2007-06-29 20:12:52 · answer #6 · answered by Sid 4 · 0 0

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