Both the numerator and the denominator of 3/4 are increased by a positive integer, x, and those of 15/17 are decreased by the same integer. This operation results in the same value for both the fractions. What is the value of 𝑥?
3
Fraction Problem Analysis
This problem asks us to find the value of a positive integer, x, that affects two different fractions in specific ways, resulting in them having the same value. We are given two fractions: \( \frac{3}{4} \) and \( \frac{15}{17} \).
- For the first fraction, \( \frac{3}{4} \), both its numerator and denominator are increased by the positive integer x.
- For the second fraction, \( \frac{15}{17} \), both its numerator and denominator are decreased by the same integer x.
- The key condition is that after these operations, both modified fractions become equal in value.
Setting Up the Equation
Let's represent the modified fractions based on the problem description.
-
First Fraction:
Original fraction: \( \frac{3}{4} \)
Numerator increased by x: \( 3 + x \)
Denominator increased by x: \( 4 + x \)
New fraction: \( \frac{3+x}{4+x} \) -
Second Fraction:
Original fraction: \( \frac{15}{17} \)
Numerator decreased by x: \( 15 - x \)
Denominator decreased by x: \( 17 - x \)
New fraction: \( \frac{15-x}{17-x} \)
According to the problem, these two new fractions are equal. So, we can set up the equation:
\[ \frac{3+x}{4+x} = \frac{15-x}{17-x} \]Solving for the Value of x
To find the value of x, we need to solve this algebraic equation. We can do this by cross-multiplication:
\[ (3+x)(17-x) = (15-x)(4+x) \]Now, let's expand both sides of the equation:
Left side expansion:
\[ 3(17) + 3(-x) + x(17) + x(-x) \] \[ 51 - 3x + 17x - x^2 \] \[ 51 + 14x - x^2 \]Right side expansion:
\[ 15(4) + 15(x) + (-x)(4) + (-x)(x) \] \[ 60 + 15x - 4x - x^2 \] \[ 60 + 11x - x^2 \]Now, set the expanded left side equal to the expanded right side:
\[ 51 + 14x - x^2 = 60 + 11x - x^2 \]Notice that the \( -x^2 \) term appears on both sides of the equation. We can cancel them out:
\[ 51 + 14x = 60 + 11x \]Next, we want to gather all terms involving x on one side and constant terms on the other side. Let's subtract \( 11x \) from both sides and subtract \( 51 \) from both sides:
\[ 14x - 11x = 60 - 51 \] \[ 3x = 9 \]Finally, divide by 3 to find the value of x:
\[ x = \frac{9}{3} \] \[ x = 3 \]Verification of the Value of x
Let's verify if \( x = 3 \) indeed results in the same value for both fractions:
First Fraction (3/4 increased by x):
\[ \frac{3+x}{4+x} = \frac{3+3}{4+3} = \frac{6}{7} \]Second Fraction (15/17 decreased by x):
\[ \frac{15-x}{17-x} = \frac{15-3}{17-3} = \frac{12}{14} \]Simplifying \( \frac{12}{14} \) by dividing both numerator and denominator by 2:
\[ \frac{12 \div 2}{14 \div 2} = \frac{6}{7} \]Both fractions result in \( \frac{6}{7} \), which confirms our calculated value of \( x = 3 \). Also, \( x=3 \) is a positive integer, as required by the problem.
Conclusion
The value of the positive integer x is 3.
Important Questions from Fractions
Which fraction among the following is the least ?
\(\frac{5}{11}, \frac{7}{12}, \frac{8}{13}, \frac{9}{17}\)
Find the value of the following expression:
\(\frac{{3 \div 1 \times 2 + 5 - 2}}{{3 \times 3 - 2}}\)
Simplify the expression 441 ÷ \(\left[270 \div \frac{3}{7}+\left(17\div \frac{1}{3}\right)-\left(8\frac{1}{2}-\frac{5}{2}\right)\right]\)
If the sum of two positive numbers is 65 and the square root of their product is 26, then the sum of their reciprocals is:
The value of \(9 \div [\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\div(\frac{3}{4}-\frac{1}{3})\;of\;\frac{2}{9}]\) is:
