Question

x+1/2+y-1/3=8 x-1/3+y-1/2=9 simultaneous linear equation

Answer 1

therefore the value of x is 23/5 and y is 83/5

Answer 2

Given that,

Simultaneous linear equations are

$(\dfrac{x+1}{2})+(\dfrac{y-1}{3})=8$....(I)

$(\dfrac{x-1}{3})+(\dfrac{y-1}{2})=9$.....(II)

We need to make a equation

Using equation (I)

$(\dfrac{x+1}{2})+(\dfrac{y-1}{3})=8$

$\dfrac{3(x+1)+2(y-1)}{6}=8$

$\dfrac{3x+3+2y-2}{6}=8$

$3x+2y+1=48$

$3x+2y=47$.....(III)

We need to make a equation

Using equation (II)

$(\dfrac{x-1}{3})+(\dfrac{y-1}{2})=9$

$\dfrac{2(x-1)+3(y-1)}{6}=9$

$\dfrac{2x-2+3y-3}{6}=9$

$2x+3y-5=54$

$2x+3y=59$.....(IV)

We need to calculate the value of x and y

Using equation (III) and (IV)

$3x+2y=47$

$2x+3y=59$

Solving by elimination method

Multiply by 2 of equation (III) and multiply by 3 of equation (IV)

$6x+4y=94$

$6x+9y= 177$

After solving

$-5y=-83$

$y=\dfrac{83}{5}$

Put the value of y in equation (I)

$3x+2\times\dfrac{83}{5}=47$

$3x+\dfrac{166}{5}=47$

$x=\dfrac{235-166}{15}$

$x=\dfrac{23}{5}$

Hence, The value of x and y is $\dfrac{23}{5}$ and $\dfrac{83}{5}$