**Linear Equations In One Variable**

A statement of equality of two algebraic expressions, which involve one or more unknown quantities is known as an equation.

A linear equation is an equation which involves linear polynomials.

A value of the variable which makes the two sides of the equation equal is called the solution of the equation.

Same quantity can be added/subtracted to/from both the sides of an equation without changing the equality.

Both the sides of an equation can be multiplied/divided by the same non-zero number without changing the equality.

**GENERAL FORM OF LINEAR EQUATION IN TWO VARIABLES**

ax + by + c = 0, a ≠ 0, b ≠ 0 or any one from a & b can zero.

**Read More:**

- What is a Linear Equation
- Linear Equations In Two Variables
- Graphical Method Of Solving Linear Equations In Two Variables
- RS Aggarwal Class 10 Solutions Linear Equations In Two Variables
- RS Aggarwal Class 9 Solutions Linear Equations In Two Variables
- RS Aggarwal Class 8 Solutions Linear Equations
- RS Aggarwal Class 7 Solutions Linear Equations in One Variable
- RS Aggarwal Class 6 Solutions Linear Equation In One Variable

**General Form Of Linear Equation In Two Variables Example ****Problems With Solutions**

**Example 1: **Express the following linear equations in general form and identify coefficients of x, y and constant term.

**Solution:**

**Make linear equation by the following statements:**

**Example 2: ** The cost of 2kg of apples and 1 kg of grapes on a day was found to be 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is 300. Represent the situation algebraically.

**Solution: **Let cost of per kg apples & grapes are x & y respectively then by Ist condition:

2x + y = 160 ……(i)

& by IInd condition: 4x + 2y = 300 …..(ii)

**Example 3: **The coach of a cricket team buys 3 bats and 6 balls for 3900. Later, she buys another bat and 3 more balls of the same kind for 1300. Represent this situation algebraically.

**Solution: ** Let cost of a bat and a ball are x & y respectively. According to questions

3x + 6y = 3900 …..(i)

& x + 3y = 1300 …..(ii)

**Example 4: **10 students of class IX took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys.

**Solution: **Let no. of boys and girls are x & y then according to question

x + y = 10 ……(i)

& y = x + 4 ……(ii)

**Example 5: **Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is

36 m.

**Solution: ** Let length & breadth are x m and y m.

∴ according to question 1/2 perimeter = 36

1/2 [2(l + b)] = 36

⇒ x + y = 36 …..(i)

also length = 4 + breadth

x = 4 + y ..…(ii)

**Example 6: ** The difference between two numbers is 26 and one number is three times the other.

**Solution: **Let the numbers are x and y & x > y

∴ x – y = 26 ……(i)

and x = 3y ……(ii)

**Example 7: **The larger of two supplementary angles exceeds the smaller by 18 degrees.

**Solution: **Sol. Let the two supplementary angles are x and y & x > y

Then x + y = 180° ……(i)

and x = y + 18° ……(ii)

**Example 8: **A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6.

**Solution: **Let fraction is x/y

Now according to question

⇒ 11x + 22 = 9y + 18

⇒ 11x – 9y = – 4 …..(i)

and

⇒ 6x + 18 = 5y + 15

⇒ 6x – 5y = –3 ….(ii)

**Example 9: **Five years hence, the age of Sachin will be three times that of his son. Five years ago, Sachin’s age was seven times that of his son.

**Solution: **Let present ages of Sachin & his son are

x years and y years.

Five years hence,

age of Sachin = (x + 5) years & his son’s age = (y + 5) years

according to question (x + 5) = 3(y + 5)

⇒ x + 5 = 3y + 15

⇒ x – 3y = 10 ……(i)

and 5 years ago age of both were (x – 5) years and (y – 5) years respectively

according to question (x – 5) = 7(y – 5)

⇒ x – 5 = 7y – 35

⇒ x – 7y = –30 .…(ii)

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