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pychu [463]
3 years ago
13

What is -2 1/2 + 1 1/3 in fraction form?

Mathematics
1 answer:
olga nikolaevna [1]3 years ago
5 0

-2\frac{1}{2} +1\frac{1}{3}=\frac{-1}{6}

Solution:

Given expression is -2\frac{1}{2} +1\frac{1}{3}.

Let us first convert mixed fraction into improper fraction.

-2\frac{1}{2} +1\frac{1}{3}=\frac{(-2\times 2) +1}{2} +\frac{(1\times 3) + 1}{3}

              =\frac{-4 +1}{2} +\frac{3 + 1}{3}

              =\frac{-3}{2} +\frac{4}{3}

Take LCM for the denominators (LCM of 2, 3 = 6) and make the same.

              =\frac{-3\times3}{2\times3} +\frac{4\times2}{3\times2}

              =\frac{-9}{6} +\frac{8}{6}

              =\frac{-1}{6}

-2\frac{1}{2} +1\frac{1}{3}=\frac{-1}{6}

Hence the fraction form of -2\frac{1}{2} +1\frac{1}{3} is \frac{-1}{6}.

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That is, they are equivalent expressions. Two expressions are said to be equivalent if they have the same value irrespective of the value of the variable(s) in them. Example 1: Are the two expressions 2y+5y−5+8 and 7y+3 equivalent?

Step-by-step explanation:

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Morgan made a mistake when subtracting the rational expressions below 3t^2-4t+1/t+3-t^2+2t+2/t+3=2t^2-2t+3/t+3. What was Morgan'
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<h3>Answer: Choice D. </h3>

Morgan forgot to distribute the negative sign to two of the terms in the second expression.

=============================================================

Explanation:

Focus on the numerators.

We have (3t^2-4t+1) as the first numerator and we subtract off (t^2+2t+2) as the second numerator.

Morgan needs to simplify (3t^2-4t+1)-(t^2+2t+2) for the numerator.

Mistakenly, she had these steps

(3t^2-4t+1)-(t^2+2t+2)

3t^2-4t+1-t^2+2t+2 .... her mistake made here

(3t^2-t^2)+(-4t+2t)+(1+2)

2t^2-2t+3

All of this applies to the numerator. The denominator stays at t+3 the entire time. So effectively we can ignore it on a temporary basis.

Here's what Morgan should have for her steps when simplifying the numerator.

(3t^2-4t+1)-(t^2+2t+2)

3t^2-4t+1-t^2-2t-2 ..... distribute the negative

(3t^2-t^2)+(-4t-2t)+(1-2)

2t^2-6t-1

Note in the second step, the negative outside flips the sign of each term in the second parenthesis.

Therefore,

\frac{3t^2-4t+1}{t+3}-\frac{t^2+2t+2}{t+3}\\\\\frac{(3t^2-4t+1)-(t^2+2t+2)}{t+3}\\\\\frac{3t^2-4t+1-t^2-2t-2}{t+3}\\\\\frac{2t^2-6t-1}{t+3}\\\\

which means \frac{3t^2-4t+1}{t+3}-\frac{t^2+2t+2}{t+3}=\frac{2t^2-6t-1}{t+3}, \ \ \text{ where } t \ne -3\\\\

Side notes:

  • The fractions can only be subtracted since the denominators are the same.
  • We have t \ne -3 to avoid a division by zero error.
  • Rational expressions are a fraction, or ratio, of two polynomials.
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Step-by-step explanation:

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