Can someone please help me with inequalities?

Question

I just don't undertsand how to figure out math inequalities such as k/9>7 or any inequality. My final exam is tomorrow so please explain to me how to fugure out inequalities! Thanks!!!

Answer 1

just pretend like the inequality is like an equal sign, but with ONE DIFFERENCE: if you multiply or divide both sides by a negative number, you MUST reverse the inequality sign.
EXAMPLES:
k/9>7...multiply both sides by 9 and get k>63
-2x>4...divide both sides by -2 and DON'T FORGET TO SWITCH the sign...and get x<-2
now, if you have to graph, first pretend the the inequality sign is an equal sign again. then you have to shade the side/part of the graph where the points satisfy the inequality.
example:
take x<1. this will produce a dotted line (use a dotted line for greater/less than, and a solid line for greater and/or equal to/less than and or equal to) at x=1. then, pick a point on one of the sides (the point (0,0) is usually the easiest) and plug it in. if it is true, shade that side. if not, shade the other side. so plug (0,0) into x<1 and get 0<1, which is true, so shade that side.

for the most part when solving, just treat it as if the inequality was an equal sign and REMEMBER to change the sign if you multiply or divide by a negative number or constant.

Answer 2

First, good luck on your exam. An inequality is used to show a portion of an axis (in one dimension) or an area (in two dimension) in which some expression would be valid. To determine the inequality in ONE dimension, say x/9 >7, momentarily replace the inequality by an equal sign, and do any math to have x = something. In your example, x = 63. Then replace the equal sign by the inequality. In this case, the region is limited to x above, but not equal, to 63. If x was allowed to be equal to 63, you would see a >/= combination.

The real hassle occurs when you have absolute values, such as | x-7| <3. Expect to have one central region defined or two external regions defined. In this case, if you remove the absolute sign and replace the "less than" with an "=", you can see that x must be "less than" 10. However, when x is 4, the value of the term (x-7) is -3, and its absolute value will be +3, so x must be greater than 4. Below 4, the value of term (x-7) gets more negative.

Answer 3

Alright, for k/9 > 7, if you're really having trouble, you can imagine that the > sign is an = sign. Solve it like you would any algebraic equation:
k/9 > 7
k/9 = 7
k = 63
And put the > sign back in for the = sign.
k > 63.

Remember that if you multiply or divide both sides of an inequality by a negative number, you have to change the inequality. For example:
k/-9 > 7
k/-9 = 7
k = -63
k < -63 ___ see, because you multiplied both sides by (-9) the > sign needed to be flipped.. so now it's a < sign.

Hope this helped & good luck on your exam! If you need any extra help, see if you can meet with your teacher some time before the test. He or she can do a better job of explaining this than I can.

Answer 4

If the greater-than or less-than symbol is giving you trouble, then imagine it's an equal sign for now.

k / 9 = 7
k = 63

Now put the greater-than sign back in, so it becomes k > 63. On a graph, you would not include 63 because it is not the greater-than-or-equal-to symbol, so you use a circle around it. Then shade the part of the graph (or number line) that is greater than 63.

If it had been greater-than-or-equal-to, then you would fill in the circle to include 63, and shade the side greater than 63 also.

Answer 5

it may be easier explained that k* 1/9 > 7, because there are no negatives, just do basic algebra.

Multiply both sides by nine and you get K * (1/9 *9) > 7*9 which equals K > 63

THerfore K can equal anything greater then 63 but not be equal to 63,
Understand?

Answer 6

Treat it the same as an EQUALITY.

k/9 = 7, multiply each side by 9 to get:
k = 63!!
so, k/9>7 => k>63

Answer 7

for k/9>7, you do this:

k/9>7
(k/9)*9>7(9)
then you flip the sign.
k<63

sorry about earlier... i divided by accident


Hope this helped.

Answer 8

k/9 > 7

isolate the variable

multiply both sides by 9

k > 63



o-------------->
63

Answer 9

I'm no longer an knowledgeable in this, however I performed with it for amusing. a^b + b^a > one million ln(a^b + b^a) > ln(one million) ln(a^b) + ln(b^a) > zero b*ln(a) + a*ln(b) > zero b*ln(b) > -a*ln(a) (expr one million): b*ln(b)/a*ln(a) > -one million Note that: ln(x) = zero for x = one million ln(x) > zero for x > one million ln(x) < zero for x >= zero < one million ln(x) ways -infinity while x ways zero ln(x) is imaginary for x < zero So if a and b are > one million then expr one million is right And if a and b are > zero < one million then expr one million is right. And if a and b = zero then expr one million is right. And if a and b = one million then expr one million is right And if b = one million and a > zero expr one million is right And if a or b < zero then expr one million is fake if b > one million and a >=zero < one million it will get elaborate: if b*ln(b) >= one million and -one million < a*ln(a) < zero then expr one million might be fake If b >= one million.763 and a < one million then expr one million is fake So: if b > one million.763 and a > one million then expr one million is right It turns out as despite the fact that there are countless probabilities and a few extraordinary levels of values that allows you to paintings. I wager I'm no longer certain what "belonging to (zero,one million)" approach precisely. Update: If a and b are > zero < one million then expr one million is right seeing that the LHS of expr one million will consistently be optimistic and as a result better than RHS of expr one million (-one million).