Algebra Equations for Beginners – Exercises
Here are some exercises on math, with algebra equations for beginners. If you want to see the examples first, please scroll down.
Instructions: Find the value of the variable in the following equations. Your answer may be a whole number, a decimal, or a fraction.
1) 5(x – 1) = 20
2) 4x + 3 = 9
3) \(\frac{3x – 5}{2} = 10\)
4) x3 + 5 = 32
5) \(\sqrt{x} + 7 = 15\)
Algebra Equations for Beginners – Answers
Answer 1
1) The correct answer is: x = 5
Expand the equation:
5(x – 1) = 20
(5 × x) + (5 × –1) = 20
5x – 5 = 20
Deal with the integer.
5x – 5 + 5 = 20 + 5
5x = 25
Divide by the coefficient to solve.
5x ÷ 5 = 25 ÷ 5
x = 5
Answer 2
2) The correct answer is: x = 3/2 or x = 1.5
Deal with the integer.
4x + 3 = 9
4x + 3 – 3 = 9 – 3
4x = 6
Divide by the coefficient to solve.
4x ÷ 4 = 6 ÷ 4
\(\text x = \frac{6}{4} = \frac{3}{2} = 1.5\)
Answer 3
3) The correct answer is: x = 25/3
Eliminate the fraction by multiplying each side by the denominator.
\(\frac{3x – 5}{2} = 10\)
\(\frac{3x – 5}{2} \times 2 = 10 \times 2\)
3x – 5 = 20
Deal with the integer.
3x – 5 + 5 = 20 + 5
3x = 25
Divide by the coefficient to solve.
3x ÷ 3 = 25 ÷ 3
\(\text x = \frac{25}{3}\)
Answer 4
4) The correct answer is: x = 3
Deal with the integer.
x3 + 5 = 32
x3 + 5 – 5 = 32 – 5
x3 = 27
Find the cube root of each side to solve.
\(\sqrt[3]{x^3} = \sqrt[3]{27}\)
3 × 3 × 3 = 27
x = 3
Answer 5
5) The correct answer is: x = 64
Deal with the integer.
\(\sqrt{x} + 7 = 15\)
\(\sqrt{x} + 7 – 7 = 15 – 7\)
\(\sqrt{x} = 8\)
Square each side to solve
\(\sqrt{x}^2 = 8^2\)
82 = 64
x = 64
Algebra equations for beginners, both with and without parentheses, are included on almost all college placement tests and on many educator licensing examinations.
You should study the algebra equations with parentheses on this page if you need more help on that part of your math test.
What are Algebra Equations?
There are many different types of algebra equations.
To see practice problems on one-sided equations, please go to the bottom of this page.
One-Sided Algebra Equations – Example
To solve one-sided algebra equations, we need to isolate the variable to one side of the equation by performing operations on the other numbers.
You may need to add, subtract, divide, or multiply the other numbers in a one-sided algebra equation.
The basic concept is that you can perform any operation on one side of the equation if you perform the exact same operation on the other side of the equation.
Example: If 3x – 7 = 8, what is the value of x?
Answer: x = 5
STEP 1: Deal with the integers first.
Add 7 to each side:
3x – 7 = 8
3x – 7 + 7 = 8 + 7
3x = 15
STEP 2: Deal with the coefficients (the numbers in front of the variables) to solve.
Divide each side by 3:
3x ÷ 3 = 15 ÷ 3
x = 5
To see examples of how to solve equations that have a variable on each side, please see our post entitled “double-sided equations.”
Algebra Equations – Problem Solving Methods
This post has shown only some very brief examples on solving equations on the algebra part of your test.
Our blog also has posts on how to deal with binomials, trinomials, and other types of polynomials.
You will also see posts on how to set up equations, as well as on how to solve equations.
We provide examples of different problem solving methods for practical problems on your exam.
For help with algebra equations with parentheses, please see our posts on:
You can also try our pre-algebra review and our algebra exercises.
