Algebraic equations questions with solutions are provided here to practice solving algebraic equations. An algebraic equation is a statement that equates two algebraic expressions.
The difference between an algebraic equation and an algebraic expression is:
Algebraic Expression | Algebraic Equation |
An algebraic equation is a mathematical statement formed by constants, variables and algebraic operators. | An algebraic equation is a mathematical statement that equates two algebraic expressions. |
An algebraic expression is a kind of rule and is used to find any value. | An algebraic equation is always equal to some value. |
Algebraic expressions are true for all values of the variables. | Algebraic equations are true for specific values of the variables. |
We find zeros of an algebraic expression. | We find the root or solution of an algebraic equation. |
For example, x^{2} + 2, 2x + 3y etc. | For example, x^{2} + 2 = 7, 2x + 3y = x + 1. |
A general form of an algebraic equation:
To learn the types of algebraic equations and how to solve them, click here.
Video Lesson on Solving Algebraic Equations
Algebraic Equations Questions with Solutions
Let us solve some questions on algebraic equations.
Question 1:
Solve for x: 2x + 22 = x + 1
Solution:
Given,
2x + 22 = x + 1
⇒ 2x + 22 = x + 1
⇒ 2x – x = –22 + 1
⇒ x = –21
Question 2:
Solve for x: x^{2} – 8x – 33 = 0
Solution:
Given,
x^{2} – 8x – 33 = 0
⇒ x^{2} – 11x + 3x – 33 = 0
⇒ (x – 11)(x + 3) = 0
⇒ x = 11 and x = 3
Refer: how to solve a quadratic equation?
Question 3:
Solve: (5m – 3)/4 + (3m – 2)/5 = (m + 6)/7
Solution:
\(\begin{array}{l}\frac{5m-3}{4}+\frac{3m-2}{5}=\frac{m+6}{7}\end{array} \)
\(\begin{array}{l}\Rightarrow\frac{25m-15}{20}+\frac{12m-8}{20}=\frac{m+6}{7}\end{array} \)
\(\begin{array}{l}\Rightarrow\frac{37m-23}{20}=\frac{m+6}{7}\end{array} \)
⇒ 259m – 161 = 20m + 120
⇒ 259m – 20m = 161 + 120
⇒ 239m = 281
⇒ m = 281/239.
Question 4:
The sum of two numbers is 82 and their product is 1456, find the two numbers.
Solution:
Let the two numbers be x and y, then as per the question
x + y = 82
⇒ y = 82 – x ………(i)
Again, x × y = 1456
Putting the value of y, we get
x × (82 – x) = 1456
⇒ 82x – x^{2} = 1456
⇒ x^{2} – 82x + 1456
⇒ x^{2} – 26x – 56x + 1456
⇒ (x – 26)(x – 56) = 0
x = 26 and 56
∴ the two numbers are 26 and 56.
Question 5:
Solve: (y^{2} – 4)/3 = 20.
Solution:
Given,
(y^{2} – 4)/3 = 20
⇒ y^{2 }– 4 = 60
⇒ y^{2} = 60 + 4
⇒ y = –8 and 8
Question 6:
Solve the following system of algebraic equations:
2x + y = 8
3y = 4 + 4x
Solution:
Given,
2x + y = 8 ….(i)
3y = 4 + 4x
⇒ 4x – 3y = –4 ….(ii)
Multiply (i) by 2 on both sides, then subtracting (ii) from (i), we get
2y + 3y = 16 + 4
⇒ 5y = 20
⇒ y = 4
Putting the value of y in (i), we get x = 2.
∴ the solution of the given equations is x = 2 and y = 4
Question 7:
Find the values of x and y, if
\(\begin{array}{l}\frac{2}{x}+\frac{3}{y}=\frac{9}{xy}\:\:and\:\: \frac{4}{x}+\frac{9}{y}=\frac{21}{xy};x\neq0, y\neq0\end{array} \)
Solution:
Given,
\(\begin{array}{l}\frac{2}{x}+\frac{3}{y}=\frac{9}{xy}\end{array} \)
= 3x + 2y = 9 ……..(i)
\(\begin{array}{l} \frac{4}{x}+\frac{9}{y}=\frac{21}{xy}\end{array} \)
= 9x + 4y = 21 ……..(ii)
Multiply (i) by 3 on both sides, then subtracting (ii) from (i), we get
6y – 4y = 27 – 21
⇒ 2y = 6
⇒ y = 3
Putting the value of y in (i), we get x = 1.
∴ the solution of the given equations is x = 1 and y = 3.
Question 8:
A man said, “Each of my sons has as many brothers as sisters and each of my daughters has twice as many brothers as sisters”. How many sons and daughters does the man have?
Solution:
Let the man have x boys and y girls, then as per the question
x – 1 = y
⇒ x – y = 1 …..(i)
and 2(y – 1) = x
⇒ x – 2y = –1 …….(ii)
Subtracting (i) from (ii), we get
–2y + y = –1 –1
⇒ –y = –2
⇒ y = 2
Putting the value of y in (i), we get x = 3.
∴ the man has 2 sons and 3 daughters.
Question 9:
A man bought 4 cows and 9 hens for ₹ 13400. If he sells the cows at 10% profit and hens at 20% profit, then he earns a total profit of ₹ 1880. What is the cost of a cow and a hen?
Solution:
Let x be the cost of one cow and y be the cost of a hen.
Then 4x + 9y = 13400 ……(i)
Also, as there is 10% profit on selling cows and 20% profit on selling hens,
∴ 10/100 × 4x + 20/100 × 9y = 1880
⇒ 4x + 18y = 18800 ……(ii)
Subtracting equation (i) from (ii), we get
9y = 5400 ⇒ y = 600
Putting the value of y in (i), we get x = 2000
∴ the man bought a cow at ₹ 2000 and a hen at ₹ 600.
Question 10:
50 kg of an alloy of lead and tin contains 60% of lead. How much lead must be melted into it to make the alloy of 75% lead?
Solution:
Let x kg lead is required to make the alloy contain 75% lead.
Now, the amount of lead in 50 kg of alloy = 50 × 60/100 = 30 kg
The weight of lead after melting x kg of lead = 30 + x
New weight of the alloy = 50 + x kg
Percentage of lead in the new alloy
(30 + x)/(50 + x) × 100% = 75%
⇒ (30 + x)/(50 + x) = 75/100
⇒ (30 + x)/(50 + x) = ¾
⇒ 4(30 + x) = 3(50 + x)
⇒ 120 + 4x = 150 + 3x
⇒ x = 30
∴ 30 kg of lead is need to be melted into the alloy.
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Practice Questions on Algebraic Equations
1. Solve the following:
(i) 1/(a + 2) + 3/(a + 23) = –4/(2a + 10)
(ii) (m + 2)/(m – 2) = 1
(iii) 4/x + 5y = 7 and 3/x + 4y = 5
(iv) 4√3x^{2} + 5x – 2√3 = 0
2. If 2^{a} + 3^{b} = 17 and 2^{a + 2} – 3^{b + 1} = 5, then find the value of a and b.
3. The average of two numbers is 6 and four times the difference between them is 16. Find the numbers.
4. Find the area of a square whose perimeter is 240 cm.
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