PSAT Math : PSAT Mathematics
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Example Questions
Example Question #3 : Creating Equations With Whole Numbers
What is the solution of 
We first multiply the second equation by 4.
So our resulting equation is:
Then we subtract the first equation from the second new equation.
Left Hand Side:
Right Hand Side:
Resulting Equation:
We divide both sides by -15
Left Hand Side:
Right Hand Side:
Our result is:
Example Question #12 : How To Find The Solution For A System Of Equations
Find the solutions for the following set of equations:
If we multiply both sides of our bottom equation by 
Example Question #11 : How To Find The Solution For A System Of Equations
Give the solution to the system of equations below.
No solution
Solve the second equation for 
Substitute for 
Now, substitute for 
Now we can write the solution in the notation 
Example Question #1 : How To Find The Solution To An Inequality With Subtraction
|12x + 3y| < 15
What is the range of values for y, expressed in terms of x?
–5 – 4x < y < 5 – 4x
y > 15 – 12x
5 – 4x < y < 5 + 4x
5 + 4x < y < 5 – 4x
y < 5 – 4x
–5 – 4x < y < 5 – 4x
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
Example Question #1 : Inequalities
|4x + 14| > 30
What is a possible valid value of x?
1
7
4
–3
–11
7
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.
Example Question #2 : Inequalities
Given the inequality, |2x – 2| > 20,
what is a possible value for x?
11
10
0
–8
–10
–10
For this problem, we must take into account the absolute value.
First, we solve for 2x – 2 > 20. But we must also solve for 2x – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).
First step:
2x – 2 > 20
2x > 22
x > 11
Second step:
2x – 2 < –20
2x < –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.
Example Question #3 : Inequalities
Solve for 
Move +5 using subtraction rule which will give you
Divide both sides by 2 (using division rule) and you will get 
Example Question #4 : Inequalities
If 
I.
II.
III.
I only
I and II only
III only
II only
I, II, and III
I only
Subtract 5 from both sides of the inequality:
Multiply both sides by 5:
Therefore only I must be true.
Example Question #5 : Inequalities
Which of the following is equivalent to 
Solve for both x – 3 < 2 and –(x – 3) < 2.
x – 3 < 2 and –x + 3 < 2
x < 2 + 3 and –x < 2 – 3
x < 5 and –x < –1
x < 5 and x > 1
The results are x < 5 and x > 1.
Combine the two inequalities to get 1 < x < 5
Example Question #6 : Inequalities
Which of the following is a possible set of solutions to 
Manipulate the inequality until 
For this equation, 
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