All SAT Math Resources
Example Questions
Example Question #1 : Simplifying Expressions
If x + y = 4, what is the value of x + y – 6?
2
0
6
4
–2
–2
Substitute 4 for x + y in the expression given.
4 minus 6 equals –2.
Example Question #2 : Simplifying Expressions
If 6 less than the product of 9 and a number is equal to 48, what is the number?
3
4
5
6
6
Write an equation for the written expression: 9x – 6 = 48. When we solve for x we get x = 6.
Example Question #3 : Simplifying Expressions
If x y = (5x - 4y)/y , find the value of y if 6 y = 2.
5
10
2
4
5
If we substitute 6 in for x in the given equation and set our answer to 2, we can solve for y algebraically. 30 minus 4y divided by y equals 2 -->2y =30 -4y --> 6y =30 --> y=5. We could also work from the answers and substitute each answer in and solve.
Example Question #4 : Simplifying Expressions
Evaluate: (2x + 4)(x2 – 2x + 4)
2x3 + 16
2x3 + 8x2 – 16x – 16
2x3 – 4x2 + 8x
4x2 + 16x + 16
2x3 – 8x2 + 16x + 16
2x3 + 16
Multiply each term of the first factor by each term of the second factor and then combine like terms.
(2x + 4)(x2 – 2x + 4) = 2x3 – 4x2 + 8x + 4x2 – 8x + 16 = 2x3 + 16
Example Question #5 : Simplifying Expressions
Which of the following is equivalent to ?
ab5c
ab/c
a2/(b5c)
abc
b5/(ac)
b5/(ac)
First, we can use the property of exponents that xy/xz = xy–z
Then we can use the property of exponents that states x–y = 1/xy
a–1b5c–1 = b5/ac
Example Question #6 : Simplifying Expressions
Solve for x: 2y/3b = 5x/7a
15b/14ay
6ab/7y
7ab/6y
5by/3a
14ay/15b
14ay/15b
Cross multiply to get 14ay = 15bx, then divide by 15b to get x by itself.
Example Question #1 : How To Simplify An Expression
Three consecutive positive integers are added together. If the largest of the three numbers is m, find the sum of the three numbers in terms of m.
3m – 3
3m
3m + 6
3m + 3
3m – 6
3m – 3
Three consecutive positive integers are added together. If the largest of the three numbers is m, find the sum of the three numbers in terms of m.
If m is the largest of three consecutive positive integers, then the integers must be:
m – 2, m – 1, and m, where m > 2.
The sum of these three numbers is:
m - 2 + m – 1 + m = 3m – 3
Example Question #3 : How To Do Distance Problems
Sophie travels f miles in g hours. She must drive another 30 miles at the same rate. Find the total number of hours, in terms of f and g, that the trip will take.
g + f
g + f + 30
Using d = rt, we know that first part of the trip can be represented by f = rg. The second part of the trip can be represented by 30 = rx, where x is some unknown number of hours. Note that the rate r is in both equations because Sophie is traveling at the same rate as mentioned in the problem.
Solve each equation for the time (g in equation 1, x in equation 2).
g = f/r
x = 30/r
The total time is the sum of these two times
Note that, from equation 1, r = f/g, so
=
Example Question #771 : Algebra
If a + b = 10 and b + c = 15, then what is the value of (c – a)/(a + 2b + c)?
150
5
2/3
1/5
3/2
1/5
Add the two equations:
a + b = 10
b + c = 15
------------
a + b + b + c = 10 + 15
a + 2b + c = 25 (this is the denominator of the answer)
Subtract the two equations:
b + c = 15
a + b = 10
------------
b + c – (a + b) = 15 – 10
c – a = 5 (this is the numerator of the answer)
5/25 = 1/5
Example Question #772 : Algebra
If a = 2b, 3b = c, and 2c = 3d, what is the value of d/a?
2/3
3
1
2
3/2
1
Eq 1: a = 2b
Eq 2: 3b = c
Eq 3: 2c = 3d
Rewrite Eq. 3 substituting using Eq. 2.
2(3b) = 3d (because c = 3b)
6b = 3d (simplify)
2b = d (divide by 3)
Since a and d both equal 2b, a = d. Therefore, d/a = 1.
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