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# Math problem help? rational abs. value inequality?

|x+7/x-3| > or equal to 2

so that's the absolute value of (x+7 divided by x-3) is greater than or equal to 2

jesus christ your wrong too.

### 5 Answers

- Demiurge42Lv 710 years ago
Use the identity |a / b| = |a| / |b|

|x+7/x-3| ≥ 2

|x+7| / |x-3| ≥ 2

|x+7| ≥ 2 |x-3|

x+7 ≥ ±2 (x-3)

x+7 ≥ 2(x-3) = 2x - 6

-x ≥ -13

x ≤ 13

and

x+7 ≥ -2(x-3) = -2x + 6

3x ≥ -1

x ≥ -1/3

x doesn't equal 3 because this would give division by zero in the original inequality.

x = [-1/3, 3) U (3, 13]

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- 10 years ago
let me denote greater than equal to by >=

now when x+7 / x-3 >0

then | x+7 / x-3| = x+7 / x-3

so x+7 / x-3>=2

=>x+7>=2x-6

=>7+6>=2x-x=x

=>x<=13 [ <= means less than or equal to]

when x+7 / x-3 <0

| x+7 / x-3| = -( x+7 / x-3)

so -( x+7 / x-3)>=2

=> -x-7.>=2x-6

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

=>-1>=3x

=>3x<=-1

=>x<= -1/3

thus we get x<= -1/3

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- FazLv 710 years ago
Case 1:

(x+7) / (x-3) >= 2

Multiply both sides by (x-3)² to preserve sign of inequality, so

(x+7)(x-3) >= 2(x-3)²

-(x-13)(x-3) >= 0

x-intercepts at 3 and 13. Parabola opens down, so

3 < x <= 13

Case 2:

(x+7) / (x-3) <= -2

(x+7)(x-3) <= -2(x-3)²

(x-3) (3x+1) <= 0

x-intercepts at x=3, x=-1/3. Parabola opens up, so:

-1/3 <= x < 3

So our two conditions are

-1/3 <= x < 3

3 < x <= 13

Note that x=3, is a vertical asymptote.

Combining these conditions, inequality holds when:

[-1/3, 3) U (3, 13]

Here's a plot to confirm:

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- koscinskiLv 43 years ago
|(x+7)/(x-3)| ? 2 this suggests: (x+7)/(x-3) ? 2, or -((x+7)/(x-3)) ? 2. case a million: (x+7)/(x-3) ? 2 Multiply the two factors by technique of x-3, giving x+7 ? 2(x-3), or x+7 ? 2x - 6. Subtract x from the two factors and upload 6 to the two factors, giving 13 ? x, or x ? 13. case 2: -((x+7)/(x-3)) ? 2 Multiply the two factors by technique of -a million, giving (x+7)/(x-3) ? -2. Multiply the two factors by technique of x-3, giving x+7 ? -2(x-3), or x+7 ? -2x + 6. upload 2x to the two factors and subtract 7 from the two factors, giving 3x ? -a million. sparkling up for x: x ? -a million/3.

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- 10 years ago
I'm gonna use >_ as greater than or equal to.

(x+7) / (x+3) >_ 2

x + 7 >_ 2x +6

7>_x+6

1>_x

x_< 1

Soooo I got x is smaller than or equal to 1...

Hope I helped.

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