Question

Anyone plz solve this question......​

Answer 1

hope it helps

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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}

I hope it will help you.