solve in substitution method root2x+root3y=0,root3x+root8y=0
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root2x+ root3y=0 -(1)
root3x+root8y=0 -(2)
by substitution,
x= 0- root3y/2
x= -root3y/2
put this value of x in (2)
root3(-root3y/2)+root8y=0
-3y/root2+root8y=0
-3y+root16y/root2=0
-3y+root16y=0*root2
-3y+4y=0
y=0
x=-root3(0)/2
x=0/2
x=0
#plz mark this answer as the brainliest answer ^_^
Anonymous:
ok anvi what u like that u do
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