All PSAT Math Resources
Example Question #1 : How To Find The Solution To An Inequality With Addition
What values of x make the following statement true?
|x – 3| < 9
x < 12
–6 < x < 12
6 < x < 12
–12 < x < 6
–3 < x < 9
–6 < x < 12
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.
Example Question #192 : Equations / Inequalities
If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?
3w/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.
Example Question #53 : Equations / Inequalities
Solve for .
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.
Example Question #54 : Equations / Inequalities
If and , then what is the value of ?
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.
Example Question #2 : How To Find The Solution To An Inequality With Addition
If , which of the following could be a value of ?
In order to solve this inequality, you must isolate on one side of the equation.
Therefore, the only option that solves the inequality is .
Example Question #21 : Inequalities
What values of make the statement true?
First, solve the inequality :
Since we are dealing with absolute value, must also be true; therefore: