How do you solve the following math equation?

Question

(x-2)/(x+4) < 7

Is the answer: any number except -5?

Answer 1

G&#039;day,

Inequalites can be a royal pain the **** from time to time, but to tackle this question I would do the following,

(x-2)/(x+4) &lt; 7

(x-2) &lt; 7(x+4)

x - 2 &lt; 7x + 28

x &lt; 7x + 30

x-7x &lt; 30

-6x &lt; 30

-x &lt; 5

or

x &gt; 5

Hope this helps,

David

Answer 2

The above users are all totally wrong! You cannot multiply the two sides by (x+4) because you don&#039;t know if it is positive or negative. When it is positive, the sign won&#039;t change but when you multiply both sides of inequality by a negative number, &lt; becomes &gt; and vice versa! See here:

http://en.wikipedia.org/wiki/Inequality#...

First, subtract 7 from both sides in order to get 0 on the right side of the inequality:

(x-2) / (x+4) - 7 &lt; 0

Now, the least common denominator is (x+4), so you can write this as:

(x-2)/ (x+4) - 7(x+4)/(x+4) &lt; 0

(x-2 - 7x - 28) / (x+4) &lt; 0

(-6x -30) / x+4 &lt; 0

this means that you have two cases:

the numerator is positive and the denominator is negative, or

the numerator is negative and the denominator is positive.

1st case:

-6x-30&gt;0 and x+4 &lt;0

-6x&gt;30 and x&lt;-4

x&lt;-5 (note the change in the sign after dividing by -6) and x&lt;-4

the common solution of the two inequalities is (-infinity; -5)

2nd case:

-6x-30&lt;0 and x+4&gt;0

-6x&lt;30 and x&gt; -4

x&gt;-5 (again, &lt; became &gt;) and x&gt;-4

Here, the common solution is (-4; +infinity)

Finally, all the solutions from both cases are ( -infinity; -5) and (-4; +infinity)

Hope this helps you :)

Answer 3

x-2&lt;7(x+4), x must not be equal with -4

x-2&lt;7x+28

6x&gt;-30

x&gt;-5

x belongs to the interval from -5 to infinity, without -4 :)

Answer 4

(x - 2)/(x + 4) &lt; 7

x - 2 &lt; 7x + 28

- 6x &lt; 30

- x &lt; 5

x &gt; - 5

Answer: x &gt; - 5

Answer 5

(x-2)/(x+4) &lt; 7, or

x-2 &lt;7x + 28

6x&gt;-30

x &gt; -5

Answer 6

(x-2)/(x+4) &lt; 7

(x-2)/(x+4) -7&lt;0

(x-2)/(x+4) -7(x+4)/(x+4)&lt;0

(-6x-30)/(x+4)&lt;0

(x+5)/(x+4)&gt;0

x&gt;-4 or x&lt;-5

Answer 7

yes your answer is correct

x-2 / x+4 &lt; 7

x-2 &lt; 7(x+4)

x-2 &lt; 7x + 28

x &lt; 7x + 30

x - 7x &lt; 30

-6x / -6 &lt; 30 / -6

x = -5

Answer 8

No, the ans. is the no&#039;s greater than -5 i.e. {-5 , infinity} Look:-

(x-2) / (x+4) &lt; 7

x-2 &lt; 7x+28

-30 &lt; 6x

-5 &lt; x

Answer 9

(x - 2)/(x + 4) &lt; 7

(x - 2) &lt; 7(x + 4)

x - 2 &lt; 7x + 28

x - 7x &lt; 28 + 2

-6x &lt; 30

6x &gt; -30

x &gt; -30/6

x &gt; -5

Answer 10

x - 2 &lt; 7x + 28

-6x &lt; 30

x &gt; - (30/6)

The answer is -(30/6). Not -5

Answer 11

(x-2)/(x+4) -7 &lt; 0

(x-2-7x-28)/(x+4) &lt; 0

(-6x - 30)/(x+4) &lt; 0

- (6x + 30) / (x + 4)  &lt;  0

(6x + 30) / (x + 4)  &gt; 0

6x + 30 &gt; 0    ---&gt;    x &gt; -5

x + 4 &gt; 0       ---&gt;    x &gt; -4  

So (x-2)/(x+4)  is  &lt; 7    if x &lt; -5   or  x &gt;  -4

Infact if you give x these numbers then .......