Algebra dominates Praxis Core Math. Even questions related to geometry, statistics, and numbers/quantity may have an algebraic component. So it’s good to know exactly what to expect when it comes to Praxis Core algebra.
There are three types of algebra questions on the Praxis Core math exam. The most complex algebra question type is the story problem. In Core Math algebra story problems, you’ll need to read about a situation involving numbers, come up with an algebra equation that works, and calculate the answer.
The other two categories of algebra questions are somewhat simpler. You can expect to see problems related to equivalent expressions and solving algebraic equations for a single variable. In both cases, the equation that you start with is provided, and you must carry out the correct algebraic steps to arrive at the correct answer.
Below is a practice set of Praxis Core Math algebra problems that can help you prepare for the exam. An answer key appears at the very bottom of this post. Once you finish the problems, check the answer key and see how you did.
Practice algebraic story problem
The pictograph above shows the number of new textbooks a local public school must purchase for their fifth grade classes before the next school year begins. Math books cost $25 each, English books cost $15 each, and the total cost of the textbooks is $5700. If each history book costs c dollars, what is the value of c?
 A) 35
 B) 30
 C) 25
 D) 20
 E) 15
Practice equivalent expressions problem
Which of the following expressions is equivalent to 193x for all values of x?
A) 3 – 2(8 –x)
B) 22 – (3 – 3x)
C) 4 + 3(5 –x)
D) 3(2 –x) 13
E) 19(1 –x) – 16x
Practice algebra equation problem
If 14(2x + 1) = 7(x + 9) + 1, what is the value of x? Give your answer as a fraction.
(Note: This question is in one of the Praxis Core’s alternative question formats for Math.)
Answer Key:
 1) B
 2) C
 3) 50/21
Notes about the answers:
Story problem:The correct starting equation uses the three variables Math, English, and History. These variables can be expressed respectively as M, E, and H. The equation you start with will multiply the number of each type of book (20 times the number of icons on the chart) with its unit price, to equal a total of $5700. From there, you plug in the numbers you have to solve for unknown variable d, the cost of an individual history textbook. The steps look like this:

1) 20*25*M+ 20*15*E+ 20*d*H = 5700
2) 20*25*3 + 20*15*2 + 20*d*6 = 5700
3) 1500 + 600 + 120d= 5700
4) 2100 +120d = 5700
5) 2100 + 120d– 2100 = 5700 – 2100
6) 120d= 3600
7) (120d)/120 = 3600/120
8) d= 3600/120
9) d= 30
Equivalent expressions problem: To solve this, use the commutative property of multiplication to “factor out” everything in the parentheses in each answer choice. From there, rewrite each answer choice in its simplest form. You’ll see that only answer C becomes 193x after being factored out and simplified.
Algebra equation problem: Starting with the given equation, the steps are as follows:

1) 14(2x+ 1) = 7(x+ 9) + 1
2) 28x+ 14 = 7x+ 63 +1
3) 28x+ 14 = 7x+ 64
4) 28x+ 14 – 14 = 7x+ 64 – 14
5) 28x= 7x+ 50
6) 28x– 7x= 7x+ 50 – 7x
7) 21x= 50
8) 21x/21 = 50/21
9) x = 50/21
#3 on the Algebra should be 50/21
I also got 50/21
50/21 is correct. I made a mistake in my original post. (Embarrassing, but it happens.) Thanks to everyone who brought this to my attention. 🙂
Pamela (above) is correct. Here are the steps:
1) 14(2x + 1) = 7(x + 9) + 1
2) 28x + 14 = 7x + 63 +1
3) 28x + 14 = 7x + 64
4) 28x + 14 – 14 = 7x + 64 – 14
5) 28x = 7x + 50
6) 28x – 7x = 7x + 50 – 7x
7.) 21X = 50
8) x – 50/21
Ted– You’re absolutely correct– not quite sure how I made that mistake in my original post. But apologies from me, and thanks to you bringing it to my attention.
I’ve corrected the steps for the algebra equation problem, thanks to your feedback. I added one extra step between your steps 7 and 8, just to make the whole process as clear as possible. And I also fixed some troublesome formatting issues that were making this post hard to read. 😉