We appreciate your visit to Morgan made a mistake when subtracting the rational expressions below tex frac 3t 2 4t 1 t 3 frac t 2 2t 2 t 3. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
We start with the subtraction of the two rational expressions:
[tex]$$
\frac{3t^2-4t+1}{t+3} - \frac{t^2+2t+2}{t+3}.
$$[/tex]
Since the denominators are the same, we can combine the expressions by subtracting the numerators:
[tex]$$
\frac{(3t^2-4t+1) - (t^2+2t+2)}{t+3}.
$$[/tex]
The key step is to distribute the negative sign across every term in the second numerator. Properly distributing the negative sign gives:
[tex]$$
3t^2 - 4t + 1 - t^2 - 2t - 2.
$$[/tex]
Next, we combine like terms:
1. Combine the [tex]$t^2$[/tex] terms:
[tex]$$
3t^2 - t^2 = 2t^2.
$$[/tex]
2. Combine the [tex]$t$[/tex] terms:
[tex]$$
-4t - 2t = -6t.
$$[/tex]
3. Combine the constant terms:
[tex]$$
1 - 2 = -1.
$$[/tex]
Thus, the correct numerator is:
[tex]$$
2t^2 - 6t - 1.
$$[/tex]
So the correct expression is:
[tex]$$
\frac{2t^2 - 6t - 1}{t+3}.
$$[/tex]
In the mistake given, the final result was:
[tex]$$
\frac{2t^2 - 2t + 3}{t+3}.
$$[/tex]
This incorrect result happens because the negative sign was not fully applied to every term inside the second numerator. In other words, Morgan forgot to distribute the negative sign to two of the terms in the second expression.
Therefore, the error was that Morgan forgot to distribute the negative sign to two of the terms in the second expression.
[tex]$$
\frac{3t^2-4t+1}{t+3} - \frac{t^2+2t+2}{t+3}.
$$[/tex]
Since the denominators are the same, we can combine the expressions by subtracting the numerators:
[tex]$$
\frac{(3t^2-4t+1) - (t^2+2t+2)}{t+3}.
$$[/tex]
The key step is to distribute the negative sign across every term in the second numerator. Properly distributing the negative sign gives:
[tex]$$
3t^2 - 4t + 1 - t^2 - 2t - 2.
$$[/tex]
Next, we combine like terms:
1. Combine the [tex]$t^2$[/tex] terms:
[tex]$$
3t^2 - t^2 = 2t^2.
$$[/tex]
2. Combine the [tex]$t$[/tex] terms:
[tex]$$
-4t - 2t = -6t.
$$[/tex]
3. Combine the constant terms:
[tex]$$
1 - 2 = -1.
$$[/tex]
Thus, the correct numerator is:
[tex]$$
2t^2 - 6t - 1.
$$[/tex]
So the correct expression is:
[tex]$$
\frac{2t^2 - 6t - 1}{t+3}.
$$[/tex]
In the mistake given, the final result was:
[tex]$$
\frac{2t^2 - 2t + 3}{t+3}.
$$[/tex]
This incorrect result happens because the negative sign was not fully applied to every term inside the second numerator. In other words, Morgan forgot to distribute the negative sign to two of the terms in the second expression.
Therefore, the error was that Morgan forgot to distribute the negative sign to two of the terms in the second expression.
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