Answer :
To solve the problem, let's break it down step-by-step:
Step 1: Setup the initial ratio equation
We are given that two numbers are in the ratio of 5:7. Let's denote these numbers as \( 5x \) and \( 7x \), where \( x \) is a common multiplier.
Step 2: Setup the modified ratio equation
It is stated that when 15 is added to each of these numbers, the new ratio becomes 5:6. So, the first number becomes \( 5x + 15 \) and the second number becomes \( 7x + 15 \). Therefore, the new ratio can be written as:
[tex]\[ \frac{5x + 15}{7x + 15} = \frac{5}{6} \][/tex]
Step 3: Formulate the equation
To find \( x \), we need to solve the proportion:
[tex]\[ \frac{5x + 15}{7x + 15} = \frac{5}{6} \][/tex]
We can solve this by cross-multiplying:
[tex]\[ 6(5x + 15) = 5(7x + 15) \][/tex]
Expanding both sides, we get:
[tex]\[ 30x + 90 = 35x + 75 \][/tex]
Step 4: Solve for \( x \)
To isolate \( x \), we will move all the terms involving \( x \) to one side and the constant terms to the other side:
[tex]\[ 30x + 90 = 35x + 75 \][/tex]
Subtract \( 30x \) from both sides:
[tex]\[ 90 = 5x + 75 \][/tex]
Next, subtract 75 from both sides:
[tex]\[ 15 = 5x \][/tex]
Now, divide both sides by 5:
[tex]\[ x = 3 \][/tex]
Step 5: Find the two original numbers
Using the value of \( x \) we found, we can now determine the two original numbers:
[tex]\[ 5x = 5 \cdot 3 = 15 \][/tex]
[tex]\[ 7x = 7 \cdot 3 = 21 \][/tex]
So, the two numbers are 15 and 21.
Step 1: Setup the initial ratio equation
We are given that two numbers are in the ratio of 5:7. Let's denote these numbers as \( 5x \) and \( 7x \), where \( x \) is a common multiplier.
Step 2: Setup the modified ratio equation
It is stated that when 15 is added to each of these numbers, the new ratio becomes 5:6. So, the first number becomes \( 5x + 15 \) and the second number becomes \( 7x + 15 \). Therefore, the new ratio can be written as:
[tex]\[ \frac{5x + 15}{7x + 15} = \frac{5}{6} \][/tex]
Step 3: Formulate the equation
To find \( x \), we need to solve the proportion:
[tex]\[ \frac{5x + 15}{7x + 15} = \frac{5}{6} \][/tex]
We can solve this by cross-multiplying:
[tex]\[ 6(5x + 15) = 5(7x + 15) \][/tex]
Expanding both sides, we get:
[tex]\[ 30x + 90 = 35x + 75 \][/tex]
Step 4: Solve for \( x \)
To isolate \( x \), we will move all the terms involving \( x \) to one side and the constant terms to the other side:
[tex]\[ 30x + 90 = 35x + 75 \][/tex]
Subtract \( 30x \) from both sides:
[tex]\[ 90 = 5x + 75 \][/tex]
Next, subtract 75 from both sides:
[tex]\[ 15 = 5x \][/tex]
Now, divide both sides by 5:
[tex]\[ x = 3 \][/tex]
Step 5: Find the two original numbers
Using the value of \( x \) we found, we can now determine the two original numbers:
[tex]\[ 5x = 5 \cdot 3 = 15 \][/tex]
[tex]\[ 7x = 7 \cdot 3 = 21 \][/tex]
So, the two numbers are 15 and 21.
