How to divide rational expressions - PSAT Math
Card 0 of 7
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
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We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Tap to see back →
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Tap to see back →
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Tap to see back →
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Tap to see back →
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Tap to see back →
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Which of the following is equivalent to $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} ? Assume that denominators are always nonzero.
Tap to see back →
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.
We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of x-t.
In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.
In order to convert the fraction $\frac{1}{t}$ to a denominator with xt, we will need to multiply the top and bottom by x.
$\frac{1}{t}$=\frac{1cdot x}{tcdot x}$=\frac{x}{xt}$
Similarly, we will multiply the top and bottom of -$\frac{1}{x}$ by t.
$\frac{1}{x}$=\frac{1cdot t}{xcdot t}$=\frac{t}{xt}$
We can now rewrite $\frac{1}{t}$-$\frac{1}{x}$ as follows:
$\frac{1}{t}$-$\frac{1}{x}$ = $\frac{x}{xt}$-$\frac{t}{xt}$=\frac{x-t}{xt}$
Let's go back to the original fraction $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We will now rewrite the numerator:
$\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t} = $\frac{frac{x-t}{xt}$}{x-t}
To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$div (x-t) . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, adiv b=acdot $\frac{1}{b}$.
$\frac{x-t}{xt}$div (x-t) = $\frac{x-t}{xt}$cdot $\frac{1}{x-t}$=\frac{x-t}{xt(x-t)}$= $\frac{1}{xt}$
Lastly, we will use the property of exponents which states that, in general, $$\frac{1}{a}$=a^{-1}$.
$$\frac{1}{xt}$=(xt)^{-1}$
The answer is $(xt)^{-1}$.