I'm stuck.... help?
Any guess ??
No... I did the best that I could
repitation is coming in it
If any point is on x axis then its y co ordinate is always zero If any point is on y axis then its x co ordinate is always zero getting ???
C is point on quadrant II
for the last question y= -x + 1 you can test each pair: replace the "x" with the first number from the pair. simplify and see if you get the second number. If you do, you found the answer
you are not right for option c
you can do the something for 2x -3y= -6 replace both the x (with the first number of the pair), and the y (from the second number) simplify. If you get -6, you found the correct pair.
2x-3y=-6 is (-6, 3)
let's see: 2*-6 - 3*3 = -12-9 = -21 but you want -6.
please check this
Then it's not (-6, 7) either
you just have to grit your teeth and test each pair.
2*-6 - 3*7= -12-21=33
That's not the answer.
try (3,4)
btw you meant 2*-6 - 3*7= -12-21= -33
That does equal -6...
Wow, that was easy
can you do the last question y = -x + 1
How would I solve for y=-x+1?
replace x and y with the first number and second number of a pair simplify the right side, and see if you get the same number on both sides. test the pairs that are not already an answer for the other questions
So I have (-6, 3) and (-6, 7) left...
3=-6+1 = 3=-5 7=-6+1 = 7=-5
you have to be careful about the signs y = -x + 1 try (-6,3) (notice x is -6) y = - (-6) + 1 y = 6+1 y=7 it looks like if x= -6 you get y= 7
so (-6,7) is the answer
oh, i didn't put the - in parenthesis
I put the parens in to make it more clear. but you could write it 7 = - -6 + 1 and get 7= 6+1 7=7 the important thing is to not confuse y = -x + 1 and y = x + 1
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