Question

solve by elimination method 2x+y=10 and 3x+27=17 eliminattion method

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Answer 1

Step-by-step explanation:

2x+3y=17 and 3x+5y=27

Let, 2x+3y=17 ...(1)

and 3x+5y=27 ...(2)

First of all we have to check how many solution(or solutions) it have. Let constants of eqn (1) be A,B and C. And constants of eqn (2) be a,b, and c. We know that-

If 1) A/a=B/b=C/c then the pair have infinitely many solutions.

2) A/a=B/b≠C/c then the pair have no solution.

3)A/a≠B/b then the pair have a unique solution.

Here, A/a≠B/b

or 2/3≠3/5

Hence this pair have a unique situation

We have,

2x+3y=17 ….(1)

3x+5y=27 …(2)

According to elimination method….

multiplying by 3 to eqn (1) and multiplying by 2 to eqn (2) then we get

6x+9y=51 …(3)

And 6x+10y=54 …(4)

Now subtracting eqn (3) by eqn (4) we get

y=3 and on putting this value of y in eqn (1) -

2x+3(3)=17

or, 2x+9=17

or, 2x=8

Therefore, x=4

Hence for this given pair of linear equation in two variables x=4 and y=3 (ans.)

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Answer 2

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