Question

the point of intersection of the straight lines 2x-y+3=0 and 3x-7y+10=0 lies in which quadrant​

Answer 1

6x -3y = -9 ......1

6x -14y = -20 ......2

subtract 2 from 1 .... we get

y = 1

put the Value of y in eq. 1

x = - 1

lies in second quadrant

Answer 2

2x-y+3=0------>1

3x-7y+10=0----->2

by multiplying eq1by 3 and eq 2 by 2

eq1-eq2

6x-3y+9=0

6x-14y+20=0

___________

11y-11=0

y=1

substitute y=1 in eq 1

2x-y+3=0

2x-1+3=0

2x=-2

x=-1

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