When the numerator and denominator of a fraction are each increased by 5, the
value of the fraction becomes 3/5
When the numerator and denominator of that same
fraction are each decreased by 5, the fraction is then 1/5
Find the original fraction.
(Hint: Let the fraction be p/q) Write in simultaneous equations
help me in math trish
The denominator of a rational number is greater than the numerator by 4. If the numerator is increased by 2 and the denominator is decreased by 1, the number becomes 7/8. Find the original rational number.
To solve this problem, we can set up a system of simultaneous equations based on the given information.
Let's assume the original fraction is p/q, where p is the numerator and q is the denominator.
According to the first sentence, when the numerator and denominator are each increased by 5, the value of the fraction becomes 3/5. So we can write the first equation as:
(p+5)/(q+5) = 3/5
Similarly, according to the second sentence, when the numerator and denominator are each decreased by 5, the fraction becomes 1/5. So we can write the second equation as:
(p-5)/(q-5) = 1/5
Now we have a system of two equations:
(p+5)/(q+5) = 3/5 ...(Equation 1)
(p-5)/(q-5) = 1/5 ...(Equation 2)
To solve this system of equations, we can use the method of substitution. Let's solve the second equation for p:
(p-5)/(q-5) = 1/5
Cross-multiplying, we get:
5(p-5) = q-5
5p - 25 = q - 5
5p = q + 20
p = (q + 20)/5 ...(Equation 3)
Now, substitute equation 3 into equation 1:
((q + 20)/5 + 5)/(q + 5) = 3/5
Simplify this equation to get:
(q + 20 + 25)/(5(q + 5)) = 3/5
(q + 45)/(5q + 25) = 3/5
Cross-multiplying, we get:
5(q + 45) = 3(5q + 25)
5q + 225 = 15q + 75
15q - 5q = 225 - 75
10q = 150
q = 15
Now substitute the value of q back into equation 3 to find p:
p = (q + 20)/5
p = (15 + 20)/5
p = 35/5
p = 7
Therefore, the original fraction is 7/15.
first equation:
(p+5)/(q+5) = 3/5 or
5p + 25 = 3q + 15
5p - 3q = -10
second equation:
(p-5)/(q-5) = 1/5 or
5p - 25 = q - 5
5p - q = 20
solve the two equations by adding them