Algebraic equations are the backbone of algebra. In this article, we will discuss the definition of algebraic equations, their key terms, and algebraic equations examples with answers, as well as practice questions to improve your understanding of this topic.
What is an algebraic equation?
An algebraic equation is a mathematical statement that shows two algebraic expressions are equal and contains one or more variables.
For example 2x +3y = 5
First of all, the word equation comes from equal, and the word algebraic comes from algebra. It means two algebraic expressions are equal, and we find the value of the variables.
Key terms of algebraic equations
Here we will explain different terms used in algebraic equations, so that you can easily understand the topic of algebraic equations and can solve its examples with ease.
1. Variable
A term whose value is unknown. For example
2x +3y = 5
Here, x and y are variables because we don’t know their real value.
2.Constant
A term whose value is known is called a constant.
For example 2x +3y = 5
2, 3, and 5 are the constants.
3.Coefficient
The constant terms used at the beginning of the variable are called the coefficient.
2x +3y = 5
2 and 3 are coefficients here.
4. Equal sign
The most important term is the equal sign used in algebraic equations. This is the main thing that differentiates algebraic equations from algebraic expressions.
This sign equates two different terms.
These are the basic concepts used in algebraic equation examples.
How to solve algebraic equations
When you have given the algebraic equation to solve, keep the following points in mind.
1. Collect the same terms
In this step, you should collect the same terms on one side and the other terms on the other side. For example, it means taking x’s terms on one side and y’s terms on the other side.
2. Simplify
Then the second step is to solve the algebraic equation carefully. Use the operations that are given in the equation.
3. Get your answer and check
After simplifying the equation, you will get your answer; substitute the answer into the original equation.
If both sides become equal after putting the answer, then your answer is correct.
4. Keep balance
The most important thing is to maintain balance when solving an equation. The same operations should be performed on both sides of the equation to balance it.
Difference Between Algebraic Equations and Expressions
| Algebraic Expression | Algebraic Equation |
|---|---|
| A combination of numbers, variables, and operations. | A statement showing that two expressions are equal. |
| Does not contain an equals sign (=). | Contains an equals sign (=). |
| Cannot be solved. | Can be solved to find the value of the variable. |
| Example: (3x + 5) | Example: 3x + 5 = 6 |
Related Articles:
Algebraic Expressions with Examples
You can also visit algebraic equations on Wikipedia.
Algebraic Equations Examples with Answers
Linear Equation Examples with Answers
These linear algebraic equations examples with answers have the highest degree of 1 of the variable.
Q1. 5x + 6 =10
Solution:
5x + 6 – 6 = 10 – 6
5x = 4
x = ⅘
Checking:
5x + 6 =10
By putting x = 4 / 5
5 (4/5)+ 6 =10
4 + 6 = 10
10 = 10
Q2. 4x – 5 = 3x + 6
Solution:
4x – 5 = 3x + 6
4x – 3x – 5 = 6
x – 5 + 5 = 6 + 5
x = 11
Checking:
4x – 5 = 3x + 6
Put x = 11
4(11) – 5 = 3(11) + 6
44 – 5 = 33 + 6
39 = 39
Q3. y / 3=15
Solution:
y / 3=15
y=3(15)
y=45
Checking:
y / 3=15
45 / 3 = 15
15 = 15
Quadratic Equation Examples with Answers
These quadratic algebraic equations examples with answers have the highest degree of 2 of the variable.
Q4. x2+ 6 = 10
Sol:
x2+ 6 = 10
x2+ 6 – 6 = 10 – 6
x2 = 4
By taking the square root on both sides:
x = ± 2
Checking:
x2+ 6 = 10
By putting x = ± 2
(± 2)2+ 6 = 10
4 + 6 = 10
10 = 10
Q5. x2 + 2x + 1 = 16
Solution:
x2 + 2x + 1 =16
As we know that
x2 + 2x + 1 =(x+1)2
So,
(x+1)2 =16
By taking the square root of both sides,
x + 1 = ± 4
x = 4 -1
x = 3
And,
x = -4 – 1
x = -5
Checking:
x2 + 2x + 1 = 16
putting x = -5
x2 + 2x + 1 = 16
(-5)2 + 2(-5) + 1 = 16
25 – 10 + 1 = 16
15 + 1 = 16
16 = 16
putting x = 3
x2 + 2x + 1 = 16
(3)2 + 2(3) + 1 = 16
9 + 6 + 1 = 16
16 = 16
Q6. y2 + 3/4 = 19/4
Solution:
y2 + 3/4 – 3/4 = 19/4 – 3/4
y2 + 3/4 – 3/4 = (19-3)/4
y2+0=16/4
y2 =4
By taking the square root of both sides
y = ± 2
Checking:
y2 + 3/4 = 19/4
Put y = 2, (2)2 + 3/4 = 19/4
4 + 3/4 = 19/4
(16+3)/4 = 19/4
19/4 = 19/4
Also, check the answer by putting y= -2 into the question.
Cubic Algebraic Equations Examples with Answers
These cubic algebraic equations examples with answers have the highest degree of 3 of the variable.
Q7. x3 = 27
Solution:
x3 = 27
By taking the cube root on both sides:
\sqrt[3]{x3} = \sqrt[3]{27}
x = 3
Checking:
x3 = 27
(3)3 = 27
27 = 27
Q8. y3 + 15 = 16
Solution:
y3 + 15 = 16
y3 + 15 – 15 = 16 – 15
y3 = 1
By taking the cube root on both sides:
y = 1
Checking:
y3 + 15 = 16
(1)3 + 15 = 16
1 + 15 = 16
Q9. 4/3 z3– 3/16 = ⅜
Solution:
4/3 z3– 3/16 = 3/8
4/3 z3 – 3/16 + 3/16 = 3/8+ 3/16
4/3 z3 = 9/16
z3 = (9)(3)/(16)(4)
z3 = 27/ 64
By taking the cube root on both sides:
z = 3/4
Checking:
4/3 z3– 3/16 = ⅜
By putting z = 3/4
4/3 (3/4)3– 3/16 = ⅜
4/3 (27/64) – 3/16 = ⅜
9/16 – 3/16 = 3/8
(9-3)/16 = 3/8
6/16 = 3/8
3/8 = 3/8
Algebraic Equations Examples with Answers Related to Square Root
Q10. \sqrt{x} = 4
Solution:
\sqrt{x}= 4
By taking the square on both sides:
x = 16
Checking:
\sqrt{x} = 4
By putting x = 16
\sqrt{16} = 4
4 = 4
Q11. \sqrt{x2-1} = 0
Solution:
\sqrt{x2-1}= 0
By taking the square on both sides:
x2-1= 0
x2 = 1
By taking the square root on both sides:
x = ± 1
Since square roots are non-negative, checking is recommended.
Checking:
\sqrt{x2-1} = 0
By putting x = ± 1
\sqrt{(± 1)2-1} = 0
\sqrt{1 – 1} = 0
0 = 0
Q12. \sqrt{4y2+1} = \sqrt{3y2+5}
Solution:
\sqrt{4y2+1} = \sqrt{3y2+5}
By taking the square on both sides:
4y2+1 = 3y2+5
4y2– 3y2= 5 – 1
y2= 4
By taking the square root on both sides:
y = ± 2
Checking:
\sqrt{4y2+1} = \sqrt{3y2+5}
By putting y = ± 2
\sqrt{4(± 2)2+1} = \sqrt{3(± 2)2+5}
\sqrt{4(4)+1} = \sqrt{3(4)+5}
\sqrt{16 + 1} = \sqrt{12+5}
\sqrt{17} = \sqrt{17}
4.123 = 4.123
Real-life Algebra Problems
Q13. If a number is increased by 7 and equals 20. Find the number.
Solution:
Let the number be x, according to the condition
x + 7 = 20
x = 20 – 7
x = 13
Checking:
x + 7 = 20
13 + 7 = 20
20 = 20
Q14. The perimeter of a rectangle is 50 cm. If the length is twice its width, then find its length and width.
Solution:
Let the width be x, and the length be 2x.
According to the condition,
2(length + width ) = Perimeter
2(2x + x ) = 50
2(3x) = 50
6x = 50
x = 50/6
x = 8. 333 cm
The width is 8.33 cm, and the length is 16.66 cm.
Checking:
2(length + width ) = Perimeter
2(16.66 + 8.33) = 50
2(25) = 50
50 = 50
Q15. The father’s age is 3 times his son’s age. If the sum of their ages is 60 years, find the age of the son and the father.
Solution:
Let the son’s age be x, and the father’s age is 3x.
According to the condition
x + 3x = 60
4x = 60
x = 60/4
x = Son’s age = 15 years
Father’s age = 3x = 3(15)
Father’s age = 45 years
Checking:
x + 3x = 60
Putting x = 15
15 + 3(15) = 60
15 + 45 = 60
60 = 60
Practice Questions:
Q1. 2x = 8
Q2. 5x = 4x – 6
Q3. x^2 = 4
Q4. 2y^2 + 6 = 12
Q5. x^ 3 = 64
Q6. ⅓ z^ 3 = 9/8
Q7.\sqrt{x} = 5
Q8. y^2 = 4
Q9. /sqrt{x+1} = 4
Q10. ⅔ x = ⅚
Answers:
- x = 4
- x = -6
- x = ± 2
- y = ±3
- x = 4
- z = 3/2
- x = 25
- y = ± 2
- x = 15
- x = 5/4
FAQs
Q1. What is an algebraic equation?
An equation that contains terms of algebra is called an algebraic equation.
Q2. What is the purpose of algebraic equations?
These equations are helpful in both academic and daily life.
Q3. How to solve algebraic equations examples with answers step by step?
If you follow all the above steps to solve equations, it will be easier for you.
Q4. Are all equations algebraic equations?
All algebraic equations are equations, but not all equations are algebraic.
Conclusion:
In this article, there are various algebraic equation examples with answers, as well as their method of solving for your convenience. And also practice questions related to all algebraic equation types so that every student can have a grip on this article.
