Surprisingly Sufficient Information

• Coefficients of first equation are in A.P with a common difference of (-2).
• Coefficients of second equation are also in A.P with a common difference of 3.
• Multiplying first equation by the absolute value of the Common difference of the second equation.
• Multiplying second equation by the absolute value of the Common difference of the first equation.
• We get,
• $15a + 9b + 3c = 51$
• $2a + 8b + 14c = 42$
• Adding both equations we get
• $17a + 17b + 17c = 93$
• $a + b + c = \dfrac{93}{17}$
• So,
• $93 + 17 = \boxed{110}$

We multiply the first equation by $m$ and the second equation by $n$ to get $\begin{cases} 5ma+3mb+mc=17m \\ na+4nb+7nc=21n \end{cases}$ Adding these two equations yields $(5m+n)a+(3m+4n)b+(m+7n)c=17m+21n$ We seek a value of $m$ and $n$ such that the above equation will be of the form $1a + 1b + 1c = \frac{x}{y}$ Comparison gives the equations $\begin{cases} 5m+n=1 \\ 3m+4n=1 \\ m+7n=1 \\ 17m+21n=\frac{x}{y}\end{cases}$ Solving $5m+n=1$ and $3m+4n=1$ simultaneously, we get $m=\frac{3}{17}, n = \frac{2}{17}$ To check if such combination is possible, we verify if this satisfies $m+7n=1$ $m+7n=\frac{3}{17}+7\cdot \frac{2}{17}=1$ which checks. Therefore, the obtained values of $m$ and $n$ are what we desire. Hence, $\frac{x}{y}=17m+21n=17\cdot \frac{3}{17}+21 \cdot \frac{2}{17}=\frac{93}{17}$ So, $93+17=\boxed{110}$ .

This is a very useful method for solving this type of equations. In order to get an equation of the form $a+b+c=\frac{x}{y}$ , we must add up j times of the first equation and k times of the second equation.

Thus, we have $5j+k=3j+4k=j+7k$ Solving, we have $j=1.5k$ . Now, let k be any random integer (not 0) and j be the corresponding value. Multiply the first equation by j, the second equation by k and add them up. Then, the answer can be obtained.

THE FOLLOWING IS AN OUTLINE

In a three variable equation with only two equations given, we can assume any one of the variables to be zero ( say c=0)

This reduces the problem to solving a two variable, two equation problem.

Solve.

I just made a+b+c=d be the third equation and solved for d (=x/y)