Linear Equation in Two Variables: Graph, Elimination and Substitution

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Linear equations in two variables is said to be an equation represented as ax+by+c = 0, where a, b, and c are referred to as the real numbers and the coefficient of x and y which means that a and b are unequal to 0. Equation that gives a straight line when plotted on the graph is known as a linear equation. Linear equations may have more than one variable. When a linear equation has two variables, it is known as a linear equation in two variables. They are the equations of first order i.e. highest power of the variable is one.

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Linear Equation in Two Variables Definition

An equation in the form of ax+by+c=0, where a, b, and c are real numbers, and constants. The Coefficients of variables x and y are a and b which cannot be equal to zero.

Example: x+2y= 6; 2x+y=7.

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Graphical Representation of Linear Equation in Two Variables

A linear equation in two variables can be written in different forms by rearranging order of constants. Suppose x+2y=6 is the linear equation in two variables.

This equation can be expressed in the form of a table by writing the different values for y corresponding to different values for x.

x

0

2

4

6

..

y

3

2

1

0

..

On plotting we get,

Graphical Representation of Linear Equation in Two Variables
Graphical Representation of Linear Equation in Two Variables

Observation from the graphs are as follows,

  • Every value of x and y that satisfies the equation lies on the line.
  • Every point (eg. 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 lines, parallel lines, and 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 I- When a1/a2 ? b1/b2
Condition I- When a1/a2 ≠ b1/b2
  • 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 II- When a1/a2 = b1/b2 ? c1/c2
Condition II- When a1/a2 = b1/b2 ≠ c1/c2
  • 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.

Condition III- When a1/a2 = b1/b2 = c1/c2
Condition III- When a1/a2 = b1/b2 = c1/c2

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Methods of Solving Linear Equation in Two Variables

Linear equations in two variables can be solved in 4 different ways, that are as follows.

  • Graphical Method
  • Substitution Method
  • Cross Multiplication Method
  • Elimination Method

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Graphical Method

To solve the equations having two variables, two equations are essential.

Let’s consider the two linear equations in two variables as 2x-y+1=0 and x-2y+8=0.

On converting these equations in the form of y=mx+c, we get

y= 2x+1

y=(x+8)/2

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.

x

0

1

2

y

1

3

5

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.

x

0

2

4

y

4

5

6

A Graph can be plotted with these coordinates.

Graphical Method
Graphical Method

There is one point where lines intersect, x=2 and y=5. This is the solution for the considered equations.

Lines may come out to be parallel or coinciding as well. In the case of parallel lines, equations will have no solution and in case of coinciding lines, the equation will have infinitely many solutions.

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Substitution Method

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

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Cross Multiplication Method

Considering the same example with 2x-y+1=0 and x-2y+8=0 equations.

Let’s write the coefficient of x and y, and constants for equations.

Cross Multiplication Method
Cross Multiplication Method

Now these coefficients and constants are cross multiplied in the following way.

Cross Multiplication Method
Cross Multiplication Method

Multiplied terms are subtracted and written in the following way in order to form the desired equation.

Cross Multiplication Method
Cross Multiplication Method

Method of Elimination

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.

(x-2y+8=0) x 2= 2x-4y+16=0

Now the equations 2x-y+1=0 and 2x-4y+16=0 have equal coefficients of x i.e. 2.

On subtracting equation 2x-y+1=0 by the equation 2x-4y+16=0, we get

(2x-4y+8) - (2x-y+1) = 0

2x-4y+16-2x+y-1=0 (This step will eliminate the variable x from the equation)

-3y+15=0

3y=15

y= 15/3

y=5

On substituting the value of y=5, in any of the equations we will get the value for x.

Let’s substitute y=5 in 2x-y+1=0, we get

2x-(5)+1=0

2x-5-1=0

2x-6=0

2x=6

x=3

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Things to Remember

  • Linear equation in two variables is an equation in the form of ax+by+c=0, where a, b, and c are real numbers and constants. The Coefficients of variables x and y are a and b which cannot be equal to zero.
  •  Linear equations in two variables may have one solution, infinite solutions, or no solution.
  1. When a1/a2 ≠ b1/b2 equations have one unique solution.
  2. When a1/a2 = b1/b2 ≠ c1/c2 equations have no solution.
  3. When a1/a2 = b1/b2 = c1/c2 equations have infinitely many solutions.
  • There are four ways for solving linear equations in two variables.
  • Graphical Method
  • Substitution Method
  • Cross Multiplication Method
  • Elimination Method
  • 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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Sample Questions

Ques 1. 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 2. 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

raw the graph for the equation x+y=4
graph for the equation x+y=4

Ques 3. 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

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
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

Ques 4. 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 5. 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 6. 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/b1, equations have one unique solution.

Here a1=1, a2=5, b1=2, b2=k

For a unique solution a1/a1 ≠ 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 7. 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

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CBSE X Related Questions

  • 1.
    \(17 \times 11 \times 13 + 11\) is

      • a prime number.
      • multiple of 17.
      • a composite number.
      • an odd number.

    • 2.
      Find the H.C.F. and L.C.M. of 408 and 312.


        • 3.
          The distance between the points \(P(-2, 5)\) and \(Q(5, -2)\) is

            • \(7\sqrt{2}\)
            • 14
            • \(2\sqrt{7}\)
            • 7

          • 4.
            Three coins are tossed together. The probability of getting exactly two tails is

              • \(\frac{2}{8}\)
              • \(\frac{1}{2}\)
              • \(\frac{3}{8}\)
              • 1

            • 5.
              In the given figure, two triangles ABC and PQR are shown such that \(\angle A = \angle P\) and \(\angle C = \angle R\). If \(AD \perp BC\) and \(PS \perp QR\), then prove that (i) \(\Delta ADB \sim \Delta PSQ\) (ii) \(AD \times QS = BD \times PS\).


                • 6.
                  Prove that : \(\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\tan A}{\sec A + 1}\).

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