Question

$\frac{x + 1}{2} + \frac{y - 1}{3 } = 8 \\ \frac{x - 1}{3} + \frac{y + 1}{2} = 9$
solve the above system using elimination method​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

Solve these equation using elimination method,

• (x+1)/2 + (y-1)/3 = 8
• (x-1)/3 + (y+1)/2 = 9

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

Equation is

• (x+1)/2 + (y-1)/3 = 8

➠ 3(x+1)+2(y-1) = 6×8

➠ 3x + 3 + 2y -2 = 48

➠ 3x + 2y = 48 - 1

➠ 3x + 2y = 47 ...............(1)

And

• (x-1)/3 + (y+1)/2 = 9

➠ 2(x-1)+3(y+1) = 9 × 6

➠ 2x - 2 + 3y +3 = 54

➠ 2x + 3y = 54 - 1

➠ 2x + 3y = 53 ..................(2)

Multiply by 2 in equ(1) and 3 in equ(2)

• 6x + 4y = 94
• 6x + 9y = 159

____________Subtract it's (eliminate x )

➠ (4y-9y) = (94-159)

➠ -5y = - 65

➠ y = -65/(-5)

➠ y = 13

Keep value of equ(1),

➠ 3x + 2×13 = 47

➠ 3x = 47 - 26

➠ 3x = 21

➠ x = 21/3

➠ x = 7

Thus:-

• Value of x = 7
• Value of y = 13

$\Large{\underline{\mathfrak{\bf{\red{Answer\:verification}}}}}$

Case(1):-

keep value of x and y in equ(1)

➠ 3 × (7) + 2 × (13 ) = 47

➠ 21 + 26 = 47

➠ 47 = 47

L.H.S = R.H.S

Case(2):-

keep value of x and y in equ(2)

➠ 2 × 7 + 3 × 13 = 53

➠ 14 + 39 = 53

➠ 53 = 53

L.H.S. = R.H.S

that's proved

Hence, we can say that our solution is right .

Answer 2

$\huge{ \underline{ \underline{ \green{ \sf{ SoLuTiOn :-}}}}}$

By elimination method :-

$\frac{x + 1}{2} + \frac{y - 1}{3} = 8 \\ \\ \implies \frac{3(x + 1) + 2(y - 1)}{6} = 8 \\ \\ \implies \frac{3x + 3 + 2y - 2}{6} = 8 \\ \\\implies 3x + 2y + 1 = 48 \\ \\ \implies 3x + 2y = 48 - 1\\ \\\implies 3x + 2y = 47 - (1) \\ \\ \frac{ x - 1}{3} + \frac{y + 1}{2} = 9 \\ \\ \implies \frac{2(x - 1) + 3(y + 1)}{6} = 9 \\ \\ \implies \frac{2x - 2 + 3y + 3}{6} = 9 \\ \\\implies 2x + 3y + 1 = 54 \\ \\ \implies 2x + 3y = 54 - 1 \\ \\\implies 2x + 3y = 53 - (2)$

Multiply by 2 in equ. (1) and 3 in equ. (2) =>

6x + 4y = 94 -------------(3)

6x + 9y = 159 ------------(4)

Subtracting equ. (3) and (4) =>

$\implies 6x + 4y - (6x + 9y) = 94 - 159 \\ \\ \implies 6x + 4y - 6x - 9y = - 65 \\ \\ \implies - 5y = - 65 \\ \\\implies 5y = 65 \\ \\\implies y = \frac{65}{5} \\ \\ y = 13$

Putting y = 13 in equ. (1) =>

$\implies 3x + 2y = 47 \\ \\\implies 3x + 2 \times 13 = 47 \\ \\ \implies 3x + 26 = 47 \\ \\\implies 3x = 47 - 26 \\ \\\implies 3x = 21 \\ \\ \implies x = \frac{21}{3} \\ \\\implies x = 7$

Answer :-

${\purple{\boxed{\large{\bold{x = 7 \:and \: y = 13}}}}}$