Question

25x+21y=26 ; 21x+25y=20​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of Linear equations are

$\rm :\longmapsto\:25x + 21y = 26 - - - - (1)$

and

$\rm :\longmapsto\:21x + 25y = 20 - - - - (2)$

Since, diagonally coefficients are same, so to solve such type of linear equations, we first add given equations and then subtract given equations to get two linear equations in simplest form and then use elimination method to solve the equations.

So, On adding equation (1) and (2), we get

$\rm :\longmapsto\:46x + 46y = 46$

$\rm :\longmapsto\:46(x + y) = 46$

$\green{\bf\implies \boxed{\bf{ \:x + y = 1} - - - - (3)}}$

On Subtracting equation (2) from equation (1), we get

$\rm :\longmapsto\:4x - 4y = 6$

$\rm :\longmapsto\:4(x - y) = 6$

$\rm\implies \:\boxed{\tt{ x - y = \frac{3}{2}}} - - - - (4)$

Now, On adding equation (3) and (4), we get

$\rm :\longmapsto\:2x = 1 + \dfrac{3}{2}$

$\rm :\longmapsto\:2x = \dfrac{2 + 3}{2}$

$\rm :\longmapsto\:2x = \dfrac{5}{2}$

$\bf\implies \:x = \dfrac{5}{4}$

Now, On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:\dfrac{5}{4} + y = 1$

$\rm :\longmapsto\:y = 1 - \dfrac{5}{4}$

$\rm :\longmapsto\:y = \dfrac{4 - 5}{4}$

$\bf\implies \:y = - \dfrac{1}{4}$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\bf{x \: = \: \dfrac{5}{4} } \\ \\ &\bf{y \: = \: - \: \dfrac{1}{4} } \end{cases}\end{gathered}\end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given pair of Linear equations are

$\rm :\longmapsto\:25x + 21y = 26 - - - - (1)$

and

$\rm :\longmapsto\:21x + 25y = 20 - - - - (2)$

Since, diagonally coefficients are same, so to solve such type of linear equations, we first add given equations and then subtract given equations to get two linear equations in simplest form and then use elimination method to solve the equations.

So, On adding equation (1) and (2), we get

$\rm :\longmapsto\:46x + 46y = 46$

$\rm :\longmapsto\:46(x + y) = 46$

$\green{\bf\implies \boxed{\bf{ \:x + y = 1} - - - - (3)}}$

On Subtracting equation (2) from equation (1), we get

$\rm :\longmapsto\:4x - 4y = 6$

$\rm :\longmapsto\:4(x - y) = 6$

$\rm\implies \:\boxed{\tt{ x - y = \frac{3}{2}}} - - - - (4)$

Now, On adding equation (3) and (4), we get

$\rm :\longmapsto\:2x = 1 + \dfrac{3}{2}$

$\rm :\longmapsto\:2x = \dfrac{2 + 3}{2}$

$\rm :\longmapsto\:2x = \dfrac{5}{2}$

$\bf\implies \:x = \dfrac{5}{4}$

Now, On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:\dfrac{5}{4} + y = 1$

$\rm :\longmapsto\:y = 1 - \dfrac{5}{4}$

$\rm :\longmapsto\:y = \dfrac{4 - 5}{4}$

$\bf\implies \:y = - \dfrac{1}{4}$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\bf{x \: = \: \dfrac{5}{4} } \\ \\ &\bf{y \: = \: - \: \dfrac{1}{4} } \end{cases}\end{gathered}\end{gathered}$