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ANSWER
In the given fig. (i), the solutions of the equation are (−1,1),(0,0) and (1,−1).
∴ the equation which satisfies these solutions is the correct equation.
Equation (ii) x+y=0, satisfies these solutions.
Proof:
If we put the value of x=−1 and y=1 in the equation x+y=0
=x+y=−1+1=0
∴L.H.S=R.H.S
if we put the the value of x = 0 and y = 0
=x+y=0+0=0
∴L.H.S=R.H.S
If we put the value of x=1 and y=−1
=x+y=1+(−1)=1–1=0
L.H.S=R.H.S
Hence, option (ii) x+y=0 is correct.
In the given Fig.(ii) the solutions of the equation are (−1,3),(0,2) and (2,0).
∴The equation which satisfies these solutions is the correct equation.
Equation (iii) y=−x+2 , satisfies these solutions.
Proof:
If we put the value of x=−1 and y=3 in the equation y=−x+2
y=−x+2
3=−(−1)+2
3=3
∴L.H.S=R.H.S
if we put the the value of x=0 and y=2
y=−x+2
2=−0+2
2=2
∴L.H.S=R.H.S
If we put the value of x=2 and y=0
y=−x+2
0=−2+2
0=0
L.H.S=R.H.S
Hence, option (iii) y=−x+2 is correct.
Answer:
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