Assertion (A): Solution of 217x + 131y = 913 and 131x + 217y = 827 is x = 3, y = 2.

Reason (R): To solve equations of the type ax + by = c and bx + ay = d where a ≠ b, we add them to obtain x + y = $\dfrac{c + d}{a + b}$ and subtract them to obtain x - y = $\dfrac{c - d}{a - b}$ .

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

Both A and R are true.

Explanation

Given,

217x + 131y = 913

131x + 217y = 827

Subtracting second equation from first equation, we get:

$\begin{matrix} & 217x & + & 131y & = & 913 \ & 131x & + & 217y & = & 827 \ & - & - & & & - \ \hline & 86x & - & 86y & = & 86 \ \end{matrix}$

⇒ 86x - 86y = 86

⇒ x - y = 1 ……………(1)

Adding both equation, we get:

$\begin{matrix} & 217x & + & 131y & = & 913 \ & 131x & + & 217y & = & 827 \ & + & + & & & + \ \hline & 348x & + & 348y & = & 913 + 827 \ \end{matrix}$

⇒ 348x + 348y = 1,740

⇒ x + y = 5 ……………(2)

Adding equation (1) and (2), we get:

$\begin{matrix} & x & - & y & = & 1 \ & x & + & y & = & 5 \\hline & 2x & & & = & 1 + 5 \ \end{matrix}$

⇒ 2x = 6

⇒ x = $\dfrac{6}{2}$

⇒ x = 3

Putting x = 3 in equation (1),

⇒ 3 - y = 1

⇒ y = 3 - 1

⇒ y = 2

x = 3, y = 2

∴ Assertion (A) is true.

Given,

ax + by = c

bx + ay = d

Adding both equation, we get:

$\begin{matrix} & ax & + & by & = & c \ & bx & + & ay & = & d \\hline & (a + b)x & + & (b + a)y & = & c + d \ \end{matrix}$

⇒ (a + b)x + (a + b)y = c + d

⇒ (a + b)(x + y) = c + d

⇒ x + y = $\dfrac{c + d}{a + b}$

And, subtracting both equation, we get:

$\begin{matrix} & ax & + & by & = & c \ & bx & + & ay & = & d \ & - & - & & & - \ \hline & (a - b)x & - & (b - a)y & = & c - d \ \end{matrix}$

⇒ (a - b)x - (b - a)y = c - d

⇒ (a - b)(x - y) = c - d

⇒ x - y = $\dfrac{c - d}{a - b}$

∴Reason (R) is true.

Hence, both Assertion (A) and Reason (R) are true.