Inequalities - PSAT Math

The equation simplifies to x > -1.41. -1 is the answer.

Recall that with absolute values and "less than" inequalities, we have to hold the following:

12x + 3y < 15

AND

12x + 3y > –15

Otherwise written, this is:

–15 < 12x + 3y < 15

In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:

–15 – 12x < 3y < 15 – 12x

Now, we have to divide each element by 3:

(–15 – 12x)/3 < y < (15 – 12x)/3

This simplifies to:

–5 – 4x < y < 5 – 4x

This inequality could be rewritten as:

4x + 14 > 30 OR 4x + 14 < –30

Solve each for x:

4x + 14 > 30; 4x > 16; x > 4

4x + 14 < –30; 4x < –44; x < –11

Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.

For this problem, we must take into account the absolute value.

First, we solve for 2_x_ – 2 > 20. But we must also solve for 2_x_ – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).

First step:

2_x_ – 2 > 20

2_x_ > 22

x > 11

Second step:

2_x_ – 2 < –20

2_x_ < –18

x < –9

Therefore, x > 11 and x < –9.

A possible value for x would be –10 since that is less than –9.

Note: the value 11 would not be a possible value for x because the inequality sign given does not include an equal sign.

-2y+7>-7+ySubtract from both sides:

Subtract 7 from both sides:

Divide both sides by -3:

Remember to switch the inequality when dividing by a negative number:

Since y<$\frac{14}{3}$ is not an answer, we must find an answer that, at the very least, does not contradict the fact that is less than (approximately) 4.67. Since any number that is less than 4.67 is also less than any number that is bigger than 4.67, we can be sure that is less than 5.

Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.

* Notice that when we multiply or divide both sides by a negative number the ineqaulity sign changes orientation.

Absolute value problems always have two sides: one positive and one negative.

First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.

Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).

We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.

3_w_/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.

To solve this problem, add the two equations together:

x+1<4

y-2<-1

x+1+y-2<4-1

x+y-1<3

x+y<4

The only answer choice that satisfies this equation is 0, because 0 is less than 4.

In order to solve this inequality, you must isolate on one side of the equation.

Therefore, the only option that solves the inequality is .

First, solve the inequality :

Since we are dealing with absolute value, must also be true; therefore:

Choice a is equivalent because we can say that technically we are multiplying two fractions together: (xy)/z and (5(x + y))/1. We multiply the numerators together and the denominators together and end up with xy (5x + 5y)/z. xy (5y + 5x)/z is also equivalent because it is only simplifying what is inside the parentheses and switching the order- the commutative property tells us this is still the same expression. 5x²y + 5xy²/z is equivalent as it is just a simplified version when the numerators are multiplied out. Choice 5x² + y²/z is not equivalent because it does not account for all the variables that were in the given expression and it does not use FOIL correctly.

First, we need to find all of the values that are in the set S, and then we need to find the values in T. Once we do this, we must find the numbers in the intersection of S and T, which means we must find the values contained in BOTH sets S and T.

S contains all of the values of x such that 2x + 4 < 8. We need to solve this inequality.

2x + 4 < 8

Subtract 4 from both sides.

2x < 4

Divide by 2.

x < 2

Thus, S contains all of the values of x that are less than (but not equal to) 2.

Now, we need to do the same thing to find the values contained in T.

-2x + 3 < 8

Subtract 3 from both sides.

-2x < 5

Divide both sides by -2. Remember, when multiplying or dividing an inequality by a negative number, we must switch the sign.

x > -5/2

Therefore, T contains all of the values of x that are greater than -5/2, or -2.5.

Next, we must find the values that are contained in both S and T. In order to be in both sets, these numbers must be less than 2, but also greater than -2.5. Thus, the intersection of S and T consists of all numbers between -2.5 and 2.

The question asks us to find the sum of the integers in the intersection of S and T. This means we must find all of the integers between -2.5 and 2.

The integers between -2.5 and 2 are the following: -2, -1, 0, and 1. We cannot include 2, because the values in S are LESS than but not equal to 2.

Lastly, we add up the values -2, -1, 0, and 1. The sum of these is -2.

The answer is -2.

We simplify this inequality similarly to how we would simplify an equation

3x+8-8<35-8

$\frac{3x}{3}$<$\frac{27}{3}$

Thus x<9

Let be the number of pizzas made and sold. Each pizza will require $2 worth of ingredients, so the ingredients in total will cost . Add this to the cost to rent the equipment and the cost will be .

The pizzas will cost $5.50 each, so the money raised will be .

The profit will be the difference between the revenue and the cost -

The Spanish club wants a profit of at least $600, so we set up and solve the inequality:

The Spanish club must sell at least 258 pizzas to earn a profit.

First, multiplying each side of the equality by gives . Next, dividing each side of the inequality by will solve for ; .

In order to find the solution set, we solve as we would an equation:

Therefore, the solution set is any value of .

Begin by moving all of the values to the left side of the inequality:

becomes

Next, move the to the right side:

Finally, divide both sides by :