How do you solve #a+ 1\frac { 2} { 3} = 2\frac { 5} { 6}#?

Answer 1

See a solution process below:

First, convert each of the mixed numbers into an improper fraction to make the problem easier to work with:

#1 2/3 = 1 + 2/3 = (3/3 xx 1) + 2/3 = 3/3 + 2/3 = (3 + 2)/3 = 5/3#
#2 5/6 = 2 + 5/6 = (6/6 xx 2) + 5/6 = 12/6 + 5/6 = (12 + 5)/6 = 17/6#

Next, use the incorrect fractions to rewrite the equation as follows:

#a + 5/3 = 17/6#
Now, subtract #color(red)(5/3)# from each side of the equation to solve for #a# while keeping the equation balanced:
#a + 5/3 - color(red)(5/3) = 17/6 - color(red)(5/3)#
#a + 0 = 17/6 - (2/2 xx color(red)(5/3))#
#a = 17/6 - color(red)(10/6)#
#a = (17 - color(red)(10))/6#
#a = 7/6#

We can change this incorrect fraction into a mixed number if needed:

#7/6 = (6 + 1)/6 = 6/6 + 1/6 = 1 + 1/6 = 1 1/6#

Thus, another way to write the solution is as follows:

#a = 1 1/6#
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Answer 2

To solve the equation ( a + \frac{1}{2} = 2\frac{5}{6} ), first, subtract ( \frac{1}{2} ) from ( 2\frac{5}{6} ), then:

[ 2\frac{5}{6} - \frac{1}{2} = 2 + \frac{5}{6} - \frac{1}{2} ] [ = 2 + \frac{5}{6} - \frac{3}{6} ] [ = 2 + \frac{2}{6} ] [ = 2 + \frac{1}{3} ] [ = 2\frac{1}{3} ]

So, the equation becomes ( a + \frac{1}{2} = 2\frac{1}{3} ).

To isolate ( a ), subtract ( \frac{1}{2} ) from both sides:

[ a + \frac{1}{2} - \frac{1}{2} = 2\frac{1}{3} - \frac{1}{2} ] [ a = 2\frac{1}{3} - \frac{1}{2} ]

Now, convert both fractions to have the same denominator:

[ a = \frac{2 \times 2}{3 \times 2} - \frac{1 \times 3}{2 \times 3} ] [ a = \frac{4}{6} - \frac{3}{6} ] [ a = \frac{4 - 3}{6} ] [ a = \frac{1}{6} ]

So, the solution is ( a = \frac{1}{6} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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