Question

Solve the following pairs of equation by reducing them to linears from using elimination method.​

Answer 1

Answer:

The value of 'x' and 'y' is 4 and 5 respectively.

Step-by-step explanation:

5/(x-1) + 1/(y-2)=2

6/(x-1) -3/(y-2)=1.

firstly we,let 1/(x-1)and1/(y-2) be u,and v respectively.

1/(x-1)=u , and 1/(y-2)=v.

then,we find two new equation

5u+1v=2, and 6u-3v=1

then,

we equal u or v units in both equation by multiplying.

So, we equal unit'v' in both equation by multiplying:-

(5u+1v=2)×3 ----equation(1).

(6u-3v=1)×1 --------equation(2).

then,

15u+3v=6. --------equation (1)

6u-3v=1. ------- equation (2).

then,we adding both equation:-

like this---. 15u+3v=6

+ 6u-3v=1

then,3v is cancelled out in both the equation--

we find that,

21u=7.

u=7/21

u=1/3.

and,

then we putting the value of'u' in equation 2

6u-3v=1

(6×1/3)-3v=1

2-3v=1

then, -3v=-2+1

-3v=-1

then, minus is cancelled out in equation.

3v=1

so,v=1/3.

then,

we know that, u=1/(x-1) and v=1/(y-2).

then,we putting the value of u and v in upper equation.

1/3=1/(x-1)= x-1=3

x=3+1, x=4.

and,1/3=1/(y-2)

y-2=3, y=3+2

y=5.

So,I find the value of x and y by the elemination method.

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