Math Inequality: $rac{x+7}{x-5}-rac{x-7}{x+5} eq 0$

Oct 25, 2025 by SLV Team 55 views

$Math Inequality: $ rac{x+7}{x-5}- rac{x-7}{x+5}$

eq 0$

Hey everyone, let's dive into a cool math problem today that involves inequalities. We're going to tackle $rac{x+7}{x-5}- rac{x-7}{x+5} eq 0$ . This might look a bit intimidating at first, but trust me, guys, it's all about breaking it down step-by-step. We want to find all the values of 'x' that make this statement true. Ready to get your math hats on?

Understanding the Inequality

So, what are we really trying to do here with $rac{x+7}{x-5}- rac{x-7}{x+5} eq 0$ ? Essentially, we're looking for the 'x' values that don't make the expression equal to zero. This means we need to find when the expression is equal to zero, and then exclude those 'x' values from our final answer. It's a bit like finding all the possible solutions and then crossing out the ones that don't fit the bill. This technique is super common when dealing with inequalities, especially when you have fractions involved. The first hurdle we need to overcome is combining those two fractions into a single one. To do this, we need a common denominator, which in this case is $(x-5)(x+5)$ . This common denominator is also known as the difference of squares, which simplifies nicely to $x^2 - 25$ . So, let's get to work combining these fractions. It's going to involve a bit of algebraic manipulation, but that's where the fun is, right? We'll multiply the numerator and denominator of the first fraction by $(x+5)$ and the second fraction by $(x-5)$ . Remember to be super careful with your signs, especially when distributing that negative sign in front of the second fraction. A small mistake here can lead to a totally different answer, and nobody wants that! Once we have a common denominator, subtracting the numerators becomes a straightforward process. We'll be left with a single fraction that we can then analyze further to solve our inequality. This is where the real magic happens, and we start to see the shape of our solution emerge.

Combining the Fractions

To combine $rac{x+7}{x-5}- rac{x-7}{x+5}$ , we need a common denominator. As we mentioned, the least common denominator is $(x-5)(x+5)$ . So, let's rewrite each fraction with this denominator:

$rac{(x+7)(x+5)}{(x-5)(x+5)} - rac{(x-7)(x-5)}{(x+5)(x-5)}$

Now, let's expand the numerators:

$(x+7)(x+5) = x^2 + 5x + 7x + 35 = x^2 + 12x + 35$

$(x-7)(x-5) = x^2 - 5x - 7x + 35 = x^2 - 12x + 35$

Substitute these back into our expression:

$rac{(x^2 + 12x + 35)}{(x-5)(x+5)} - rac{(x^2 - 12x + 35)}{(x-5)(x+5)}$

Now we can subtract the numerators:

$rac{(x^2 + 12x + 35) - (x^2 - 12x + 35)}{(x-5)(x+5)}$

Be really careful with the subtraction. The negative sign distributes to both terms in the second numerator:

$rac{x^2 + 12x + 35 - x^2 + 12x - 35}{(x-5)(x+5)}$

Simplify the numerator by combining like terms:

$rac{(x^2 - x^2) + (12x + 12x) + (35 - 35)}{(x-5)(x+5)}$

This gives us:

$rac{24x}{(x-5)(x+5)}$

So, our original inequality $rac{x+7}{x-5}- rac{x-7}{x+5} eq 0$ is equivalent to $rac{24x}{(x-5)(x+5)} eq 0$ . This looks way simpler, right? We've successfully combined the fractions into a single, manageable expression. This is a huge step, and it really shows how simplifying complex expressions can make them much easier to understand and solve. Remember, when combining fractions, the common denominator is your best friend, and paying close attention to signs during subtraction is absolutely crucial. A small slip-up there can send you down a rabbit hole of incorrect calculations.

Finding Critical Points

Now that we have our simplified inequality $rac{24x}{(x-5)(x+5)} eq 0$ , we need to find the critical points. These are the points where the numerator or the denominator equals zero, because these are the points where the expression could potentially change its sign or become undefined. These points will divide our number line into intervals, and we'll test each interval.

Numerator equals zero: $24x = 0$ $x = 0$
Denominator equals zero: $(x-5)(x+5) = 0$ This means either $x-5 = 0$ or $x+5 = 0$ . So, $x = 5$ and $x = -5$ .

Our critical points are $x = -5$ , $x = 0$ , and $x = 5$ . These points are extremely important because they mark the boundaries where the expression $rac{24x}{(x-5)(x+5)}$ can change its value from positive to negative, or vice versa. They also tell us where the expression is undefined (at $x=5$ and $x=-5$ ). It's crucial to remember that the values that make the denominator zero are never included in the solution set because division by zero is undefined. The value that makes the numerator zero might be included depending on the inequality sign, but since our original inequality is 'not equal to zero', we'll exclude $x=0$ as well. So, all our critical points are potential exclusion points from our solution set. Visualizing these on a number line is a great strategy. We draw a line, mark these points on it, and then we have distinct regions to test.

Testing Intervals

Our critical points divide the number line into four intervals:

$(-oldsymbol{\infty}, -5)$
$(-5, 0)$
$(0, 5)$
$(5, oldsymbol{\infty})$

We need to pick a test value within each interval and plug it into our simplified expression $rac{24x}{(x-5)(x+5)}$ to see if the result is non-zero. Since our original inequality is $eq 0$ , we are looking for intervals where the expression is not equal to 0. In other words, we're looking for intervals where the expression is either positive or negative.

Interval 1: $(-oldsymbol{\infty}, -5)$

Let's choose $x = -6$ . $rac{24(-6)}{(-6-5)(-6+5)} = rac{-144}{(-11)(-1)} = rac{-144}{11}$ . This is not zero.

Interval 2: $(-5, 0)$

Let's choose $x = -1$ . $rac{24(-1)}{(-1-5)(-1+5)} = rac{-24}{(-6)(4)} = rac{-24}{-24} = 1$ . This is not zero.

Interval 3: $(0, 5)$

Let's choose $x = 1$ . $rac{24(1)}{(1-5)(1+5)} = rac{24}{(-4)(6)} = rac{24}{-24} = -1$ . This is not zero.

Interval 4: $(5, oldsymbol{\infty})$

Let's choose $x = 6$ . $rac{24(6)}{(6-5)(6+5)} = rac{144}{(1)(11)} = rac{144}{11}$ . This is not zero.

It looks like in all of these intervals, the expression $rac{24x}{(x-5)(x+5)}$ is not equal to zero. This is because the only way for the fraction to be zero is if the numerator is zero, which happens only at $x=0$ . Our inequality is specifically asking for when the expression is not equal to zero. This means we need to consider all values of 'x' except for those that make the expression equal to zero or undefined.

Determining the Solution Set

We found that the expression $rac{24x}{(x-5)(x+5)}$ equals zero only when $x=0$ . The expression is undefined when $x=5$ or $x=-5$ . Our original inequality is $rac{x+7}{x-5}- rac{x-7}{x+5} eq 0$ , which simplifies to $rac{24x}{(x-5)(x+5)} eq 0$ .

This means we want all values of 'x' for which the expression is not zero and not undefined.

The expression is undefined at $x = -5$ and $x = 5$ . These values must be excluded.
The expression is equal to zero at $x = 0$ . Since our inequality is 'not equal to zero', this value must also be excluded.

Therefore, the solution includes all real numbers except for $-5$ , $0$ , and $5$ . We can write this in interval notation as:

$(-oldsymbol{\infty}, -5) oldsymbol{\cup} (-5, 0) oldsymbol{\cup} (0, 5) oldsymbol{\cup} (5, oldsymbol{\infty})$

This means that any number you pick from these intervals will satisfy the original inequality. For example, if you pick $x = -10$ , it will work. If you pick $x = -2$ , it will work. If you pick $x = 2$ , it will work. And if you pick $x = 100$ , it will work. The only numbers that won't work are $-5$ , $0$ , and $5$ . It's really satisfying to see how these inequalities play out, guys. We started with a complex fraction, simplified it, found our crucial boundary points, and then tested each region. The final solution set shows us all the numbers that make our original statement true. Keep practicing these, and you'll become inequality pros in no time!