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Linear equations in two variables are said to be an equation represented as ax+by+c = 0, where a, b, and c are referred to as the real numbers.
- The coefficient of x and y means that a and b are unequal to 0.
- Linear equations in two variables give a straight line when plotted on the graph.
- They are the equations of first order, i.e., the highest power of the variable is one.
- The equation will form a unique solution, no solutions, or infinitely many solutions.
- Linear equations in two variables can be converted into the form y = mx + b.
- The cost of hiring a taxi while going on a vacation is a real-life application of these equations.
- It can be represented as:
ax+by+c = 0
Key Terms: Linear equations in two variables, Linear Equations, Variables, Graph, Infinite Solutions, Real Numbers, Elimination Method, Substitution Method, Cross Multiplication Method
What are Linear Equation in Two Variables?
The linear equation in two variables is a type of equation in which two variables are of order 1. These equations are also known as simultaneous linear equations.
- The linear equation in two variables is written in the form of the following:
ax+by+c=0,
- Where a, b, and c are real numbers and constants.
- Coefficients of variables x and y are a and b, which cannot be equal to zero.
- It can be explained with the help of the geometry of lines or the graph of two lines.
- The graphic method and algebraic method are two ways of solving these equations.
There are three conditions which are applicable in the case of linear equation in two variables if we are provided two equations as
a1x + b1y + c1 = 0 ……………(1)
a2x + b2y + c2 = 0 ……………(2)
- Then conditions are as follows:
- If the pair of linear equations form consistent solution, then:
a1/a2 ≠ b1/b2
- If the pair of linear equations form inconsistent solution, then
a1/a2 = b1/b2 ≠ c1/c2
- If the pair of linear equations form dependent and consistent solution, then
a1/a2 = b1/b2 = c1/c2
Example of What are Linear Equation in Two Variables?Example: x+2y= 6; 2x+y=7. |
linear equation in two variables
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Linear Equation in Two Variables Detailed Video Explanation:
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Solution of Linear Equation in Two Variables
A linear equation in two variables can be written in different forms by rearranging the order of constants. In the graphical method, the pair of lines will either intersect with each other or parallel to each other.
- There are cases when two lines coincide with each other.
Example of Solution of Linear Equation in Two VariablesExample: Suppose x+2y=6 is the linear equation in two variables.
On plotting we get,  |
Observations from the graphs are as follows,
- Every value of x and y that satisfies the equation lies on the line.
- Every point (e.g. x= 0, y= 3) on the line gives the solution of the equation.
- Any point that does not coincide with the line is not the solution of the equation.
To find the solution of any equation with one variable, one equation is required; similarly, to find the solution for the equation with two variables, two equations are required.
- Two equations will form two straight lines on the graph.
- These two straight lines could be intersecting, parallel, or coincident lines.
Consider a system of two linear equations, a1x+b1y+c1=0 and a2x+b2y+c2=0. Let’s plot the graph for these two equations under different conditions.
Condition I: When a1/a2 ≠ b1/b2
Under this condition x and y will have only one value i.e. the equation will have a unique solution. The lines will intersect at one point. The coordinates of that point will give the solution.
Condition II: When a1/a2 = b1/b2 ≠ c1/c2
Under this condition the lines will be parallel to each other and the equations will have no solution.
Condition III: When a1/a2 = b1/b2 = c1/c2
Under this condition, the two lines will coincide with each other. Since every intersecting point gives the solution of the equation, coinciding lines will have infinitely many solutions.
Methods of Solving Linear Equation in Two Variables
Linear equations in two variables can be solved in four different ways, that are as follows.
Graphical Method
The steps followed to solve linear equation in two variables with graphical method are as follows:
- To draw the graph of two equations, first convert them in the form of y=mx+b.
- Now solve for different values of x, such as x = 0,1, 2 and so on.
- Mark the points where both lines of equations meet.
- The solution of a linear equation in two variables is the point of intersection.
Example of Graphical MethodExample: Let’s consider the two linear equations in two variables as 2x-y+1=0 and x-2y+8=0.
y= 2x+1 can be expressed in the form of a table by writing the different values for y corresponding to different values for x.
y=(x+8)/2 can be expressed in the form of a table by writing the different values for y corresponding to different values for x.
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Substitution Method
The steps followed to solve linear equation in two variables with substitution method are as follows:
- First, select one of the equations from two sets of equations.
- Express the selected equation in terms of another variable.
- Substitute the value of the variable obtained in the above step in the second equation.
- Calculate the remaining values.
Example of Substitution MethodExample: Let’s consider the same example with two equations as 2x-y+1=0 and x-2y+8=0. Let’s solve the equation 2x-y+1=0 for y first. 2x-y+1=0 2x+1=y Now we have the value of y in terms of y, on substituting this value of y in other equation x-2y+8=0 x-2(2x+1)+8=0 x-4x-2+8=0 x-4x+6=0 -3x+6=0 3x=6 x=2 From here we get the value of x, on substituting this value of x=2, in the equation 2x+1=y, we get 2x+1=y 2(2)+1=y 4+1=y y=5 |
Cross Multiplication Method
Cross Multiplication Method is the simplest way to solve linear equations in two variables. There are three ways to solve a linear equation using the cross multiplication method:
- Multiply the fraction value of the numerator on the right-hand side with the fraction value of the denominator on the left side.
- Secondly, reverse the position in the first method.
- Thirdly, calculate LHS equal to RHS.
Example of Cross Multiplication MethodExample: Considering the same example with 2x-y+1=0 and x-2y+8=0 equations.
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Method of Elimination
The steps followed to solve the linear equation in two variables with the elimination method are as follows:
- First, arrange both the equations in two variables in the form of axe + by + c = 0.
- Add or subtract the equation so as to cancel similar terms.
- Use the value obtained to determine the value of another variable.
Example of Method of EliminationExample: Let’s solve 2x-y+1=0 and x-2y+8=0 equations by method of elimination. Make any one of the coefficients (either of x or y) equal for the two equations. Let’s make the coefficient of x equal in two equations. In equation 2x-y+1=0, the coefficient of x is 2 and in equation x-2y+8=0, coefficient of x is 1. To make them equal we can multiply the whole equation x-2y+8=0 with 2.
On substituting the value of y=5, in any of the equations we will get the value for x.
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Things to Remember
- Linear equation in two variables can be expressed in terms of ax+by+c=0.
- The Coefficients of variables x and y are a and b, which cannot be equal to zero.
- It may have one solution, infinite solutions, or no solution.
- The graphical method, substitution method, and cross-multiplication method are different ways to solve an equation.
- The equations can be used in the fields of business and economics.
- To plot the graph for a linear equation in two variables, the value of two coordinates is enough.
- However, it is advisable to plot more than two points to check the correctness of the graph.
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| Class 9 Maths Related Concepts | ||
|---|---|---|
| Vertex | Types Of Triangles | Linear Equations Applications |
| Properties of Triangle | Discriminant of an Equation | Properties of Parallel Lines |
Sample Questions
Ques. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (2 marks)
Ans. Let the cost of the copy be x and cost of the pen be y.
As per the question, Cost of a notebook= 2 x Cost of a pen
x= 2y
x-2y=0
This is the required equation.
Ques. Draw the graph for the equation x+y=4. (3 marks)
Ans. The equation x+y= 4 can be expressed in the form of a table by writing the different values for y corresponding to different values for x.
| x | 0 | 1 | 2 | 3 | 4 |
| y | 4 | 3 | 2 | 1 | 0 |
Ques. The taxi fare in a city is as follows: For the first kilometer, the fare Is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs.y, write a linear equation for this Information, and draw Its graph. (4 marks)
Ans. As per the question, total distance= x and total fare=y
- Fare for first 1km= 8
- Fare for the remaining distance= 5 x remaining distance
- Remaining distance after covering 1km= x-1
- Therefore, fare for the remaining distance= 5(x-1)
- Total fare= 8 + 5(x-1) = y
- 8+5x-5=y
- 5x+3=y
The equation x+y= 4 can be expressed in the form of a table by writing the different values for y corresponding to different values for x.
| x | 0 | 1 | 2 |
| y | 3 | 8 | 13 |
Ques. Solve y-4x=1 and 6x-5y=9 for x and y. (5 marks)
Ans. We have two equations y-4x=1 and 6x-5y=9
- Let’s use the elimination method to solve this.
- To make the coefficients of x equal in the two equations we can multiply equation y-4x=1 with -5.
- -5 x (y-4x=1)
- -5y+20x=-5
Now since the coefficients are equal, let’s subtract -5y+20x=-5 by 6x-5y=9
- (6x-5y) - (-5y+20x)= 9- (-5)
- 6x-5y+5y-20x= 14
- -14x=14
- x= 14/-14
- x= -1
- Substituting the value of x in y-4x=1, we get
- y -4x=1
- y-4(-1)=1
- y+4=1
- y=1-4
- y=-3
Ques. Solve x+y=5 and 2x-3y=4 using the substitution method. (3 marks)
Ans. From the equation x+5=y, we get
- x=5-y
- Substituting this value of x in the equation 2x-3y=4, we get
- 2 (5-y)-3y=4
- 10-2y-3y=4
- 10-5y=4
- 10-4=5y
- 6=5y
- y= 6/5
- On substituting y= 6/5 in x+y=5, we get
- X+ 6/5 =5
- x= 5 - 6/5 = (25-6)/5
- x= 19/5
Ques. For what value of k system of equations x+2y=3 and 5x+ky+7=0 has a unique solution? (2 marks)
Ans. We know that when a1/a2 ≠ b1/b2, equations have one unique solution.
Here a1=1, a2=5, b1=2, b2=k
For a unique solution a1/a2 ≠ b1/b2
1/5 ≠ 2/k
k ≠ 10
Thus for any real number other than 10, equations x+2y=3 and 5x+ky+7=0 will have a unique solution.
Ques. Sum of the ages of father and son is 40 years. If the father's age is three times that of his son, then find their ages. (3 marks)
Ans. Let the age of the father and son be x and y.
As per the question,
- Father’s age + Son’s age = 40
- x+y=40
- Also,
- Father’s age= 3 x son’s age
- x=3y
- On substituting the value of x=3y in x+y=40, we get
- 3y+y=40
- 4y=40
- y=10
- Again on substituting the value of y=10 in x+y=40, we get
- x+10=40
- x=40-10
- x=30
Ques. The cost of a notebook is five times the cost of a pen. Write a linear equation in two variables to represent this statement. (2 marks)
Ans. Let the cost of the copy be x and cost of the pen be y.
As per the question, Cost of a notebook= 5 x Cost of a pen
- x= 5y
- x-5y=0
This is the required equation.
Ques. Raj's piggy bank has 12 coins (only quarters or dimes) that have a total value of $2.85. How many dimes and quarters does the piggy bank has. (3 marks)
Ans. Let us assume that the number of dimes be x and the number of quarters be y in the piggy bank. Let us form linear equations in two variables based on the given information.
- Since there are 12 coins in total, x+y=12 ⇒ y=12-x.
- We know that, 1 dime = 10 cents and 1 quarter = 25 cents.
- The total value of the money in the piggy bank is $2.85 (285 cents).
- Thus we get the equation 10x + 25y = 285
- 10x + 25(12-x) = 285 (as y = 12-x)
- 10x + 300 - 25x =285
- -15x =-15
- x = 1
Ques. Sum of the ages of father and son is 90 years. If the father's age is five times that of his son, then find their ages. (3 marks)
Ans. Let the age of the father and son be x and y.
As per the question,
- Father’s age + Son’s age = 90
- x+y=90
- Also,
- Father’s age= 5 x son’s age
- x=5y
- On substituting the value of x=5y in x+y=90, we get
- 5y+y=90
- 6y=90
- y=15
- Again on substituting the value of y=15 in x+y=90, we get
- x+15=90
- x=90-15
- x=75
Ques. The sum of a two-digit number and the number obtained by reversing the digits is 99. If the digits of the number differ by 3, find the number. How many such numbers are there. (3 marks)
Ans. Let the ten’s and the unit’s digits in the first number be x and y, respectively.
So, the first number may be written as 10 x + y in the expanded form (for example, 56 = 10(5) + 6).
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation, is 10y + x (for example, when 56 is reversed, we get 65 = 10(6) + 5).
According to the given condition.
(10x + y) + (10y + x) = 99
i.e., 11(x + y) = 99
i.e., x + y = 9
We are also given that the digits differ by 3, therefore,
either x – y = 3 (2) or y – x = 3 (3) If x – y = 3,
then solving (1) and (2) by elimination, we get x = 6 and y = 3. In this case, we get the number 63.
If y – x = 3, then solving (1) and (3) by elimination, we get x = 3 and y = 6. In this case, we get the number 36. Thus, there are two such numbers 36 and 63.
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