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Sagot :
Sure, let's find the number step-by-step.
Given that:
1. When [tex]\(\frac{3}{4}\)[/tex] of a number [tex]\(x\)[/tex] is added to [tex]\(3 \frac{1}{2}\)[/tex], the sum equals the situation when [tex]\(\frac{2}{3}\)[/tex] of the number [tex]\(x\)[/tex] is subtracted from [tex]\(6 \frac{1}{2}\)[/tex].
We can start by expressing the given conditions mathematically. Let's break it down.
First, let's convert the mixed fractions to improper fractions:
[tex]\[ 3 \frac{1}{2} = 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2} \][/tex]
[tex]\[ 6 \frac{1}{2} = 6 + \frac{1}{2} = \frac{12}{2} + \frac{1}{2} = \frac{13}{2} \][/tex]
Next, represent the conditions given in the problem using [tex]\(x\)[/tex]:
[tex]\[ \frac{3}{4}x + \frac{7}{2} = \frac{13}{2} - \frac{2}{3}x \][/tex]
Now we solve for [tex]\(x\)[/tex]. First, eliminate the fractions by finding a common denominator. Here, the common multiple of 4, 2, and 3 is 12.
Multiply every term by 12:
[tex]\[ 12 \left(\frac{3}{4}x\right) + 12 \left(\frac{7}{2}\right) = 12 \left(\frac{13}{2}\right) - 12 \left(\frac{2}{3}x\right) \][/tex]
[tex]\[ 9x + 42 = 78 - 8x \][/tex]
Next, combine the[tex]\(x\)[/tex] terms on one side and constant terms on the other side:
[tex]\[ 9x + 8x = 78 - 42 \][/tex]
[tex]\[ 17x = 36 \][/tex]
Finally, divide both sides by 17 to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{36}{17} \][/tex]
[tex]\[ x \approx 2.11764705882\dots \][/tex]
Thus, the number such that when [tex]\(\frac{3}{4}\)[/tex] of it is added to [tex]\(3 \frac{1}{2}\)[/tex], the sum is the same as when [tex]\(\frac{2}{3}\)[/tex] of it is subtracted from [tex]\(6 \frac{1}{2}\)[/tex] is approximately [tex]\(2.11764705882353\)[/tex].
Given that:
1. When [tex]\(\frac{3}{4}\)[/tex] of a number [tex]\(x\)[/tex] is added to [tex]\(3 \frac{1}{2}\)[/tex], the sum equals the situation when [tex]\(\frac{2}{3}\)[/tex] of the number [tex]\(x\)[/tex] is subtracted from [tex]\(6 \frac{1}{2}\)[/tex].
We can start by expressing the given conditions mathematically. Let's break it down.
First, let's convert the mixed fractions to improper fractions:
[tex]\[ 3 \frac{1}{2} = 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2} \][/tex]
[tex]\[ 6 \frac{1}{2} = 6 + \frac{1}{2} = \frac{12}{2} + \frac{1}{2} = \frac{13}{2} \][/tex]
Next, represent the conditions given in the problem using [tex]\(x\)[/tex]:
[tex]\[ \frac{3}{4}x + \frac{7}{2} = \frac{13}{2} - \frac{2}{3}x \][/tex]
Now we solve for [tex]\(x\)[/tex]. First, eliminate the fractions by finding a common denominator. Here, the common multiple of 4, 2, and 3 is 12.
Multiply every term by 12:
[tex]\[ 12 \left(\frac{3}{4}x\right) + 12 \left(\frac{7}{2}\right) = 12 \left(\frac{13}{2}\right) - 12 \left(\frac{2}{3}x\right) \][/tex]
[tex]\[ 9x + 42 = 78 - 8x \][/tex]
Next, combine the[tex]\(x\)[/tex] terms on one side and constant terms on the other side:
[tex]\[ 9x + 8x = 78 - 42 \][/tex]
[tex]\[ 17x = 36 \][/tex]
Finally, divide both sides by 17 to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{36}{17} \][/tex]
[tex]\[ x \approx 2.11764705882\dots \][/tex]
Thus, the number such that when [tex]\(\frac{3}{4}\)[/tex] of it is added to [tex]\(3 \frac{1}{2}\)[/tex], the sum is the same as when [tex]\(\frac{2}{3}\)[/tex] of it is subtracted from [tex]\(6 \frac{1}{2}\)[/tex] is approximately [tex]\(2.11764705882353\)[/tex].
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