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(I) stands for bars

2007-01-21 10:22:05 · 16 answers · asked by rh 2 in Science & Mathematics Mathematics

16 answers

|2x+7|=1

We know that the expression within the absolute value signs can be positive or negative, so let's set up two equations.

2x+7=1 and -2x-7=1

Solve each

2x+7=1
2x= -6
x= -3

-2x-7=1
-2x=8
x= -4

The two answers are x= -3, -4

I hope that this helps.

2007-01-21 10:27:03 · answer #1 · answered by Anonymous · 2 0

The absolute value of something is defined as the magnitude of the distance of that thing from zero without regard to direction (i.e., in a positive or negative direction). If the absolute value of 2x + 7 is 1, then 2x + 7 is 1 unit from zero, so it is either 1 or -1.

Set 2x+7 = 1 and solve for x; and then set 2x+7 = -1 and solve for x. You will have two possible answers.

Hope that helps.

2007-01-21 18:25:05 · answer #2 · answered by Tim P. 5 · 0 1

2x + 7 = 1 and 2x + 7 = -1
or
2x= -6 and 2x = -8
x = -3 and x = -4

-3, -4

2007-01-21 19:15:42 · answer #3 · answered by xoxsoccerloverxox 1 · 0 0

as in "absolute value"?

If so the two roots are:

2x + 7 = 1 and 2x + 7 = -1
or
2x= -6 and 2x = -8
x = -3 and x = -4

-3, -4

2007-01-21 18:26:31 · answer #4 · answered by Anonymous · 1 0

2x+7=1 2x+7=-1
2x=-6 2x=-8
x=-3 and x=-4

2007-01-21 18:26:41 · answer #5 · answered by Scarecrow 1 · 1 0

2x+7=1
2x=-6
x=-3

2x+7=-1
2x=-8
x=-4

x=-3 or -4

Check:
I 2(-3)+7 I=1
I -6+7 I=1
I 1 I=1
1=1

I 2(-4)+7 I=1
I -8+7 I=1
I -1 I=1
1=1

I hope this helps!

2007-01-21 18:28:15 · answer #6 · answered by Anonymous · 0 0

|a| = b

+/-a = b so:

2x+7 = 1
-2x-7 = 1

2x = -6
-2x = 8

x=-3
x=-4

2007-01-21 18:27:19 · answer #7 · answered by Dashes 6 · 1 0

place -3 into "X" so, it would be [ 2(-3)+7]=1
[-6+7]=1
[1]=1
1=1 <--- your answer.

2007-01-21 18:29:00 · answer #8 · answered by Alex K 1 · 0 0

2x+7=1
-7 -7
2x= -6
x= negative 3

2007-01-21 18:25:06 · answer #9 · answered by penpal_247 2 · 0 1

the absolute value of 2x+7 is 1
2x+7=1
2x=-6
x=-3
then 2x+7=-1
2x=-8
x=-4

2007-01-21 18:37:41 · answer #10 · answered by Matt 1 · 0 0

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