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CCSS Math: HSA.REI.B.3

Solve for y. We have 3y plus 7 is less
than 2y and 4y plus 8 is greater than negative 48. So we have to find
all the y's that meet both of these constraints. So let's just solve for y
in each of the constraints and just remember that
this "and" is here. So we have 3y plus
7 is less than 2y. So let's isolate the y's
on the left-hand side. So let's get rid of this
2y on the right-hand side, and we can do that by
subtracting 2y from both sides. So we're going to subtract
2y from both sides. The left-hand side, we have
3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there. That's just going to be 0. And then we can
get rid of this 7 here by subtracting
7 from both sides. So let's subtract
7 from both sides. Left-hand side,
y plus 7 minus 7. Those cancel out. We just have y is less than 0
minus 7, which is negative 7. So that's one of
the constraints. That's this constraint
right over here. Now let's work on
this constraint. We have 4y plus 8 is
greater than negative 48. So let's get rid of the 8
from the left-hand side. So we can subtract
8 from both sides. The left-hand side,
we're just left with a 4y because these guys cancel out. 4y is greater than
negative 48 minus 8. So we're going to go
another 8 negative. So 48 plus 8 would
be a 56, so this is going to be negative 56. And now to isolate the y,
we can divide both sides by positive 4, and we don't
have to swap the inequality since we're dividing
by a positive number. So it's divide both
sides by 4 over here. So we get y is greater than--
what is 56/4, or negative 56/4? Let's see. 40 is 10 times 4, and then we
have another 16 to worry about. So it's 14 times 4. So y is greater
than negative 14. Is that right? 4 times 10 is 40,
4 times 4 is 16. Yep, 56. So y is greater than negative
14 and-- let's remember, we have this "and" here-- and
y is less than negative 7. So we have to meet both of
these constraints over here. So let's draw them
on the number line. So I have my number
line over here. And let's say negative
14 is over here. So you have negative
13, 12, 11, 10, 9, 8, 7-- that's negative 7-- and
then negative 6, 5, 4, 3, 2, 1. This would be 0,
and then you could keep going up more positive. And so we're looking
for all of the y's that are less than negative 7. So let's look at this,
less than negative 7. So not including
negative 7, so we'll do an open circle
around negative 7, and less than negative 7. And if that was the
only constraint, we would keep going to the left. But we have this other
constraint-- and y has to be greater
than negative 14. So you make a circle
around negative 14, and everything that's
greater than that. And if you didn't have
this other constraint, you would keep going. But the y's that
satisfy both of them are all of the y's in between. These are the y's that are
both less than negative 7 and greater than negative 14. And we can verify
that things here work. So let's try some values out. So a value that
would work, well, let me just do
negative 10 is right here, 8, 9, this is negative 10. That should work. So let's try it out. So we'd have 3 times
negative 10 plus 7 should be less than 2
times negative 10. So this is negative 30 plus
7 is negative 23, which is indeed less than negative 20. So that works. And negative 10 has to
work for this one as well. So you have 4 times negative 10,
which is negative 40, plus 8. Negative 40 plus 8 should
be greater than negative 48. Well, negative 40
plus 8 is negative 32. We're going 8 in the
positive direction, so we're getting less negative. And negative 32 is
greater than negative 48. It's less negative. So this works. So negative 10 works. Now, let's just verify some
things that shouldn't work. So 0 should not work. It's not in the solution set. So let's try it out. We've got 3 times is 0 plus 7. That would be 7. And 7 is not less than 0. So it would violate this
condition right over here if we put a 0 over here. If you put a negative
15 over here, it should violate this
condition right over here because it wasn't in
this guy's solution set. Anyway, hopefully you
found that useful.