Solve for x in this inequality problem

[chwala]

Homework Statement

Solve the inequality;

$$\dfrac {7x+12} {x}≥ 3$$

Relevant Equations

inequalities

$7x+12≥3x$
$7x−3x≥−12$
$4x≥−12$

→ $x≥−3$
or $12/x≥-4$

but by substituting say$x=−4$ we see that it also satisfies the equation implying that my solution may not be correct.

 

[Delta2]

Homework Statement:

``Solve the inequality (7x+12)/x is greater than or equal to 3
Homework Equations: inequalities

$12/x≥ -4$

Up to this point you are correct. But now we have to take cases whether x>0 or x<0, because when we multiply an inequality by a negative number the inequality changes direction(from $\geq$ becomes $\leq$ and vice versa).

So we have to take two cases
1) x>0, what happens in this case ?
2) x<0, now if we multiply by $x$ both sides of the inequality it will become $12\leq-4x$ and if we multiply again by -1/4 it will change direction again

 

[chwala]

Up to this point you are correct. But now we have to take cases whether x>0 or x<0, because when we multiply an inequality by a negative number the inequality changes direction(from $\geq$ becomes $\leq$ and vice versa).

So we have to take two cases
1) x>0, what happens in this case ?
2) x<0, now if we multiply by $x$ both sides of the inequality it will become $12\leq-4x$ and if we multiply again by -1/4 it will change direction again

when $x≥0$ in the inequality $12/x≥ -4$, then the $x$ values will satisfy the inequality...unless i am missing something, i still do not understand you. what step am i missing? i know at $x=0$ we have a discontinuity.

 

[Delta2]

when $x≥0$ in the inequality $12/x≥ -4$, then the $x$ values will satisfy the inequality...unless i am missing something, i still do not understand you. what step am i missing? i know at $x=0$ we have a discontinuity.

That is correct so for x>0 the inequality is satisifed.

Now check what happens for x<0. Have in mind that when you multiply the inequality by x now, because x<0 the inequality will change direction. In general everytime you multiply both sides of inequality with a negative number the inequality changes direction.

 

[Mark44]

This is very important to understand when you're working with inequalities.

In general everytime you multiply both sides of inequality with a negative number the inequality changes direction.

@chwala, when you mulitply both sides of an inequality by a variable (x in this case), you don't know its value, so it could be either negative or positive. That's why @Delta2 says you need to look at the two cases.

 

[chwala]

i think i got it! there's a colleague who shared with me an approach that is easier for anyone to understand. Its as follows:
let $7x+12=3x$
$7x-3x=12$
$4x=-12$
$x=-3$ is the critical value on the Real number line system, now we need to examine values between $(-∞,-3), (-3,0),(0,+∞)$
on picking the value $-4$ in the first interval, the inequality is satisfied.
$-2$in the second interval, the inequality is not satisfied.
$1$ in the third interval, the inequality is satisfied.
Therefore the solution to the inequality is
$(-∞<x≤-3)$ and $(0<x<+∞)$

 

[Delta2]

i think i got it! there's a colleague who shared with me an approach that is easier for anyone to understand. Its as follows:
let $7x+12=3x$
$7x-3x=12$
$4x=-12$
$x=-3$ is the critical value on the Real number line system, now we need to examine values between $(-∞,-3), (-3,0),(0,+∞)$
on picking the value $-4$ in the first interval, the inequality is satisfied.
$-2$in the second interval, the inequality is not satisfied.
$1$ in the third interval, the inequality is satisfied.
Therefore the solution to the inequality is
$(-∞<x≤-3)$ and $(0<x<+∞)$

This method is correct but it doesn't reveal to you the deeper why. (why we have to choose the intervals the way we do)

While my method tells you that we have to take cases x>0 and x<0 and the why is that because inequalities change direction when multiplied by a negative number, or they keep their direction when multiplied by a positive number.
The case $x>0$ was treated successfully by you.
The case $x<0$, when we multiply $\frac{12}{x}\geq -4$ by x that is now negative the inequality changes direction so it ll become $12\leq -4x$ and if we now multiply again by $-\frac{1}{4}$ the inequality changes direction again and becomes $-\frac {12}{4} \geq x \iff -3\geq x$.

 

[Mark44]

i think i got it! there's a colleague who shared with me an approach that is easier for anyone to understand. Its as follows:
let $7x+12=3x$
$7x-3x=12$
$4x=-12$
$x=-3$ is the critical value on the Real number line system, now we need to examine values between $(-∞,-3), (-3,0),(0,+∞)$
on picking the value $-4$ in the first interval, the inequality is satisfied.
$-2$in the second interval, the inequality is not satisfied.
$1$ in the third interval, the inequality is satisfied.
Therefore the solution to the inequality is
$(-∞<x≤-3)$ and $(0<x<+∞)$

Not quite. The "and" you wrote indicates that a number x has to be in the first interval, and it has to also be in the second interval, which is impossible. The correct conjunction is or.

Taking the approach suggested by @Delta2, with the two cases, we have:
Case 1: x < 0
$\frac{7x + 12 }x \ge 3 \Rightarrow 7x + 12 \le 3x$ Notice that the direction of the inequality changed, because we're multiplying by a negative quantity.
$4x \le -12$ so $x \le -3$, which is consistent with assuming that x < 0.

Case 2: x > 0
$\frac{7x + 12 }x \ge 3 \Rightarrow 7x + 12 \ge 3x$ No change in the direction of the inequality -- we're multiplying by a positive quantity.
$4x \ge -12$ so $x \ge -3$
But since it is assumed that x > 0, we can't have any values in the interval [-3, 0].

From the two cases, the solutions are $-\infty < x \le -3$ or $0 < x < \infty$.
As set, this would be $\{(-\infty, -3] \cup (0, \infty)\}$

 

[chwala]

Not quite. The "and" you wrote indicates that a number x has to be in the first interval, and it has to also be in the second interval, which is impossible. The correct conjunction is or.

Taking the approach suggested by @Delta2, with the two cases, we have:
Case 1: x < 0
$\frac{7x + 12 }x \ge 3 \Rightarrow 7x + 12 \le 3x$ Notice that the direction of the inequality changed, because we're multiplying by a negative quantity.
$4x \le -12$ so $x \le -3$, which is consistent with assuming that x < 0.

Case 2: x > 0
$\frac{7x + 12 }x \ge 3 \Rightarrow 7x + 12 \ge 3x$ No change in the direction of the inequality -- we're multiplying by a positive quantity.
$4x \ge -12$ so $x \ge -3$
But since it is assumed that x > 0, we can't have any values in the interval [-3, 0].

From the two cases, the solutions are $-\infty < x \le -3$ or $0 < x < \infty$.
As set, this would be $\{(-\infty, -3] \cup (0, \infty)\}$

Thanks Mark, ...grammatical error...thanks for the correction.

 

[neilparker62]

As a general rule , you don't want to be multiplying or dividing both sides by x (the variable) in an inequality because you don't know whether it is positive or negative and hence whether or not the inequality sign will change direction.

$$\frac{7x+12}{x}≥3⇒\frac{7x+12-3x}{x}≥0⇒\frac{4x+12}{x}≥0$$

Essentially this is now the same as solving quadratic inequality x(4x+12)≥ 0 except you need to ensure x=0 is not part of your solution. Hence: x>0 or x≤-3.

 

[chwala]

Thanks for all the input guys!

 

[chwala]

As a general rule , you don't want to be multiplying or dividing both sides by x (the variable) in an inequality because you don't know whether it is positive or negative and hence whether or not the inequality sign will change direction.

$$\frac{7x+12}{x}≥3⇒\frac{7x+12-3x}{x}≥0⇒\frac{4x+12}{x}≥0$$

Essentially this is now the same as solving quadratic inequality x(4x+12)≥ 0 except you need to ensure x=0 is not part of your solution. Hence: x>0 or x≤-3.

I realize that with the inequalities, particularly the quadratic ones...one can also check using the Critical values in determining the region satisfying the inequality.
In this case we can have two equations; namely
$4x+12≥0$ from numerator and $x≥0$ from denominator giving us the critical values $-3$ and $0$, on checking these values on the given inequality, we find our solution to be $x≤-3$ or $x≥0$

 

[neilparker62]

I realize that with the inequalities, particularly the quadratic ones...one can also check using the Critical values in determining the region satisfying the inequality.
In this case we can have two equations; namely
$4x+12≥0$ from numerator and $x≥0$ from denominator giving us the critical values $-3$ and $0$, on checking these values on the given inequality, we find our solution to be $x≥-3$ or $x≥0$

0 is excluded - cannot have zero denominator. So x > 0 and x<= -3.

 

[chwala]

0 is excluded - cannot have zero denominator. So x > 0 and x<= -3.

True...

 

[neilparker62]

https://www.desmos.com/calculator/csrfwxq5y2

The solution x>0 or x<=3 is illustrated in the above graph.

 

[Mark44]

0 is excluded - cannot have zero denominator. So x > 0 and x<= -3.

The connector in the inequalities has to be "or", not "and". It's impossible for any number to be both positive and less than or equal to any negative number.

The solution x>0 or x<=3 is illustrated in the above graph.

The graph shows x > 0 or x <= -3; i.e., not + 3.

The most straightforward way to solve the inequality, IMO, is to make two cases.
Case 1: x > 0
$\frac{7x + 12} x \ge 3 \Rightarrow 7x + 12 \ge 3x \Rightarrow 4x \ge -12 \Rightarrow x \ge -3$
Because it was stipulated that x must be positive, then we are left with x > 0.

Case 2: x < 0
$\frac{7x + 12} x \ge 3 \Rightarrow 7x + 12 \le 3x \Rightarrow 4x \le -12 \Rightarrow x \le -3$
Multiplying by x, which is assumed to be negative for this case, changes the direction of the inequality.
The solution in this case is $x \le -3$

The solution of the original inequality is $x \le -3$ OR $x > 0$.

 

[neilparker62]

The connector in the inequalities has to be "or", not "and". It's impossible for any number to be both positive and less than or equal to any negative number.The graph shows x > 0 or x <= -3; i.e., not + 3.

Thanks for the correction - not thinking. Had it right in post #10.