1f (25×√5)=(3√5)^y+1
find the value of y.
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Answer:
Given equation, 2x+3y=k
x=2,y=1 is solution of resultant equation.
2(2)+3(1)=k
∴k=7
∴ Resultant equation is 2x+3y=7
At x=0,2(0)+3y=7
y=
3
7
At y=0,2x+3(0)=7
x=
2
7
∴(0,
3
7
) and (
2
7
,0) are two more solutions of resultant equations.
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