Solving Proportional Equations with Multiple Variables

The problem provides us with a set of equations relating six variables: a, b, c, d, e, and f. These equations are given as: \({2a = 3b = 6c = 9d = 12e = 18f}\). We need to find the value of the expression \({ (a + b)/(c + d + e + f) }\).

To solve this type of problem, where multiple quantities are equal, we can set their common value equal to a constant. Let's call this constant \({ k }\).

So, we have:

• \({2a = k \implies a = k/2}\)
• \({3b = k \implies b = k/3}\)
• \({6c = k \implies c = k/6}\)
• \({9d = k \implies d = k/9}\)
• \({12e = k \implies e = k/12}\)
• \({18f = k \implies f = k/18}\)

Now that we have expressed each variable in terms of \({ k }\), we can substitute these expressions into the expression we need to evaluate: \({ (a + b)/(c + d + e + f) }\).

Step 1: Calculate the Numerator (a + b)

Substitute the values of \({ a }\) and \({ b }\) in terms of \({ k }\):

\({ a + b = k/2 + k/3 }\)

To add these fractions, we find a common denominator, which is 6:

\({ a + b = (3k)/6 + (2k)/6 = (3k + 2k)/6 = 5k/6 }\)

Step 2: Calculate the Denominator (c + d + e + f)

Substitute the values of \({ c }\), \({ d }\), \({ e }\), and \({ f }\) in terms of \({ k }\):

\({ c + d + e + f = k/6 + k/9 + k/12 + k/18 }\)

To add these fractions, we need to find a common denominator for 6, 9, 12, and 18. The least common multiple (LCM) of these numbers is 36.

Convert each fraction to have a denominator of 36:

• \({ k/6 = (k \times 6)/(6 \times 6) = 6k/36 }\)
• \({ k/9 = (k \times 4)/(9 \times 4) = 4k/36 }\)
• \({ k/12 = (k \times 3)/(12 \times 3) = 3k/36 }\)
• \({ k/18 = (k \times 2)/(18 \times 2) = 2k/36 }\)

Now, add the converted fractions:

\({ c + d + e + f = 6k/36 + 4k/36 + 3k/36 + 2k/36 = (6k + 4k + 3k + 2k)/36 = 15k/36 }\)

We can simplify the fraction \({ 15k/36 }\) by dividing the numerator and denominator by their greatest common divisor, which is 3:

\({ 15k/36 = (15k \div 3)/(36 \div 3) = 5k/12 }\)

Step 3: Evaluate the Expression (a + b)/(c + d + e + f)

Now we divide the result from Step 1 by the result from Step 2:

\({ (a + b)/(c + d + e + f) = (5k/6) / (5k/12) }\)

Dividing by a fraction is the same as multiplying by its reciprocal:

\({ (5k/6) \times (12/5k) }\)

We can cancel out the \({ 5k }\) term (assuming \({ k \neq 0 }\), which is necessary for the initial equalities to define non-zero variables):

\({ (5k/6) \times (12/5k) = 12/6 }\)

\({ 12/6 = 2 }\)

Therefore, the value of \({ (a + b)/(c + d + e + f) }\) is 2.

Revision Table: Key Calculations

Additional Information: Proportions and Ratios

This problem deals with proportions, which are statements of equality between two ratios. When multiple quantities are related by a chain of equalities like \({ 2a = 3b = ... = 18f }\), it implies that the variables are inversely proportional to their respective coefficients. For example, from \({ 2a = 3b }\), we can write \({ a/b = 3/2 }\), or \({ a:b = 3:2 }\). From \({ 2a = k }\), we get \({ a = k/2 }\). This shows that \({ a }\) is proportional to \({ k }\) and inversely proportional to the coefficient 2.

Using a constant like \({ k }\) is a standard technique to solve problems involving such chained equalities. It allows us to express all variables in terms of a single unknown, simplifying the overall expression we need to evaluate. The constant \({ k }\) represents the common value of all the expressions \({ 2a, 3b, \dots, 18f }\). As long as \({ k \neq 0 }\), the value of the ratio of sums will not depend on the specific value of \({ k }\).