Inequalities - SAT Math

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:

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

First, solve the inequality :

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

To solve , isolate .

Divide by three on both sides.

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

Subtract on both sides.

Divide on both sides.

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

Subtract on both sides.

Divide on both sides. Remember to flip the sign.

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

Subtract on both sides.

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

We need to set-up two equations since its absolute value.

Subtract on both sides.

Divide on both sides which flips the sign.

Subtract on both sides.

Since we have the 's being either greater than or less than the values, we can combine them to get .

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

We need to set-up two equations since it's absolute value.

Subtract on both sides.

Distribute the negative sign to each term in the parenthesis.

Add and subtract on both sides.

Divide on both sides.

We must check each answer. Let's try .

This is true therefore is a correct answer. Let's next try .

This is not true therefore is not correct.

Final answer is just .

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

We need to set-up two equations since it's absolute value.

Subtract on both sides.

Divide on both sides.

Distribute the negative sign to each term in the parenthesis.

Add and subtract on both sides.

Divide on both sides.

We must check each answer. Let's try .

This is true therefore is a correct answer. Let's next try .

This is not true therefore is not correct.

Final answer is just .

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

Thus

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

Therefore, the solution set is any value of .

The inequality that is presented in the problem is:

Start by moving your variables to one side of the inequality and all other numbers to the other side:

Divide both sides of the equation by . Remember to flip the direction of the inequality's sign since you are dividing by a negative number!

Reduce:

The only answer choice with a value greater than is .

To get the lowest value, you need the lowest numerator and the highest denominator. That would be or reduced to be . For the highest value, you need the highest numerator and the lowest denominator. That would be or .

The inequality can be rewritten as the three-part inequality

Isolate the in the middle expression by performing the same operations in all three expressions. Subtract 32 from each expression:

Divide each expression by , switching the direction of the inequality symbols:

This can be rewritten in interval notation as .

In an absolute value inequality, the absolute value expression must be isolated first, as follows:

Adding 12 to both sides:

Multiplying both sides by , and switching the inequality symbol due to multiplication by a negative number:

We do not need to go further. An absolute value expression must always be greater than or equal to 0; it is impossible for the expression to be less than any negative number. The inequality has no solution.