Find a fraction which reduces to 3/4 when the numerator and denominator are each decreased by 1, and which reduces to 4/5 when the numerator and denominator are each increased by 1.
Let the original fraction be x/y
equation #1 : (x-1)/(y-1) = 3/4
4x - 4 = 3y - 3
4x - 3y = 1
equation #2 : (x+1)/*(y+1) = 4/5
5x + 5 = 4y + 4
5x - 4y = -1
#1 times 4 :16x - 12y = 4
#2 times 3 : 15x - 12y = -3
subtract them ...
x = 7
back in #1
28 - 3y = 1
-3y = -27
y = 9
the fraction was 7/9
check:
6/8 = 3/4 ✔
8/10 = 4/5 ✔
Let's solve the problem step by step:
Step 1: Let's assume the fraction as x/y.
Step 2: According to the given condition, when the numerator and denominator are each decreased by 1, the fraction becomes (x-1)/(y-1). We know that this is equal to 3/4. So we can write the equation as:
(x-1)/(y-1) = 3/4
Step 3: Similarly, when the numerator and denominator are each increased by 1, the fraction becomes (x+1)/(y+1), which is equal to 4/5. So we can write the equation as:
(x+1)/(y+1) = 4/5
Step 4: Let's solve these two equations simultaneously to find the values of x and y.
From equation 1: (x-1)/(y-1) = 3/4, we can cross-multiply and get:
4(x-1) = 3(y-1)
Expanding this equation, we have:
4x - 4 = 3y - 3
Simplifying further:
4x - 3y = 1 -----------(Equation A)
From equation 2: (x+1)/(y+1) = 4/5, we can cross-multiply and get:
5(x+1) = 4(y+1)
Expanding this equation, we have:
5x + 5 = 4y + 4
Simplifying further:
5x - 4y = -1 -----------(Equation B)
Step 5: Solving equations A and B:
Multiplying equation A by 4 and equation B by 5, we have:
16x - 12y = 4 ------------(Equation C)
25x - 20y = -5 ------------(Equation D)
Step 6: Now, we can solve these two equations simultaneously to find the values of x and y. We can use substitution or elimination method to solve.
If we multiply equation C by 5 and equation D by 4, we get:
80x - 60y = 20
100x - 80y = -20
Now, let's subtract equation C from equation D:
(100x - 80y) - (80x - 60y) = -20 - 20
Simplifying this equation, we have:
20x - 20y = -40
Dividing this equation by 20, we get:
x - y = -2
Step 7: Now, let's solve equation A again:
4x - 3y = 1
If we multiply equation A by 3 and equation B by 4, we get:
12x - 9y = 3
20x - 16y = -4
Now, let's subtract equation A from equation B:
(20x - 16y) - (12x - 9y) = -4 - 3
Simplifying this equation, we have:
8x - 7y = -7
Step 8: Let's solve the equations x - y = -2 and 8x - 7y = -7 simultaneously.
We can solve these equations using substitution or elimination method. Let's use elimination method to solve.
Multiply equation 1 by 8 and equation 2 by 1, we get:
8(x - y) = 8*(-2)
8x - 8y = -16
1(8x - 7y) = 1*(-7)
8x - 7y = -7
Now, let's subtract equation 2 from equation 1:
(8x - 8y) - (8x - 7y) = -16 - (-7)
Simplifying this equation, we have:
-y = -9
Dividing both sides by -1, we get:
y = 9
Step 9: Now, let's substitute the value of y in equation 1 or equation 2. Let's substitute it in equation 2:
8x - 7(9) = -7
8x - 63 = -7
Adding 63 to both sides, we get:
8x = 56
Dividing both sides by 8, we get:
x = 7
Step 10: Therefore, the fraction is 7/9.
To find the fraction, let's assume the fraction is written as "a/b".
According to the given information, when the numerator and denominator are each decreased by 1, the fraction reduces to 3/4. This can be written as:
(a-1)/(b-1) = 3/4
Similarly, when the numerator and denominator are each increased by 1, the fraction reduces to 4/5. This can be written as:
(a+1)/(b+1) = 4/5
Now, we have a system of equations. We can solve it to find the values of 'a' and 'b' that satisfy both equations.
Let's solve for 'a' in terms of 'b' in the first equation:
(a-1)/(b-1) = 3/4
Cross-multiplying, we have:
4(a-1) = 3(b-1)
Expanding, we get:
4a - 4 = 3b - 3
Simplifying further:
4a - 3b = 1 -- Equation 1
Now, let's solve for 'a' in terms of 'b' in the second equation:
(a+1)/(b+1) = 4/5
Cross-multiplying, we have:
5(a+1) = 4(b+1)
Expanding, we get:
5a + 5 = 4b + 4
Simplifying further:
5a - 4b = -1 -- Equation 2
We now have a system of linear equations:
4a - 3b = 1 -- Equation 1
5a - 4b = -1 -- Equation 2
We can solve it using the method of substitution or elimination to find the values of 'a' and 'b'.
Let's use the method of elimination:
Multiplying Equation 1 by 4 and Equation 2 by 3, we have:
16a - 12b = 4 -- Equation 3
15a - 12b = -3 -- Equation 4
Subtracting Equation 4 from Equation 3, we get:
(16a - 15a) - (12b - (-12b)) = 4 - (-3)
Simplifying further:
a = 7
Now, substitute the value of 'a' back into Equation 1 or Equation 2 to solve for 'b'. Let's use Equation 1:
4a - 3b = 1
Substituting 'a=7', we have:
4(7) - 3b = 1
28 - 3b = 1
Subtracting 28 from both sides:
-3b = 1 - 28
-3b = -27
Dividing by -3:
b = 9
Therefore, the fraction that satisfies the given conditions is 7/9.