y x < 1 3y > x + 6 In the xy-plane if the point with coordinates (a,
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07 Jun 2024, 14:55
If we convert the inequalities into a standard equation. remember this is a MUST be true question
we do have that
\(y-x<1\)
\(y-x=1\)
\(y=x+1\)
Same for \(3y>x+6\)
\(3y=x+6\)
\(y=x/3+2\)
Now to the observation
1) 𝑦-intercept is 1, i.e. (0,1) and slope is positive. and 𝑦-intercept is 2 i.e. (0,2) and slope is positive so we do know that both are in the I quadrant and must be positive BOTH Therefore D and E are suddenly out
2) y=x+1 and in A a> b should follow that x>y but this is NOT true because we do know that x+1 is Y. For example 3=2+1; x can not be > y; So A is out
3) in B a<b which means that x<y this could be true or not. From the second equality we do know that if x=1 we do have that y=1/3+6=6.3 and in this case a<b BUT if we do have that x is 9 we have that y is also 9 and x=y
The only thing that must be true s that both x and y are positive. The only answer choice correct is C