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# Linear Equation In Two Variable

#### Algebraic Solutions of a System of Linear Equations

Sometimes, graphical method does not give an accurate answer. While reading the coordinates of a point on a graph paper, we are likely to make an error. So, we require some precise method to obtain accurate result. Algebraic methods given below yield accurate answers.

1. Method of elimination by substitution.
2. Method of elimination by equating the coefficients.
3. Method of cross multiplication.

SUBSTITUTION METHOD

In this method, we first find the value of one variable (y) in terms of another variable (x) from one equation. Substitute this value of y in the second equation. Second equation becomes a linear equation in x only and it can be solved for x.

• Putting the value of x in the first equation, we can find the value of y.
• This method of solving a system of linear equations is known as the method of elimination by substitution.
• ‘Elimination’, because we get rid of y or ‘eliminate’ y from the second equation. ‘Substitution’, because we ‘substitute’ the value of y in the second equation.

Working rule :

Let the two equations be a1x + b1y + c1 = 0 ....(1) and a2x + b2y + c2 = 0 ....(2)
Step I : Find the value of one variable, say y, in terms of the other i.e., x from any equation, say (1).
Step II : Substitute the value of y obtained in step 1 in the other equation i.e., equation (2). This equation becomes equation in one variable x only.
Step III : Solve the equation obtained in step II to get the value of x.
Step IV : Substitute the value of x from step II to the equation obtained in step I. From this equation, we get the value of y. In this way, we get the solution i.e. values of x and y.

Verification process is a must to check the answer.

Example: Solve each of the following system of equations by eliminating x (by substitution) :
(i) x + y = 7 and 2x – 3y = 11, (ii) x + y = 7 and 12x + 5y = 7, (iii) 2x – 7y = 1 and 4x + 3y = 15, (iv) 3x – 5y = 1 and 5x + 2y = 19, (v) 5x + 8y = 9 and 2x + 3y = 4

Solution: (i) We have system of equations x + y = 7 ....(1) and 2x – 3y = 11 ....(2)
We shall eliminate x by substituting its value from one equation into the other. from equaton (1), we get ;
x + y = 7 ⇒ x = 7 – y
Substituting the value of x in equation (2), we get ;
2 × (7 – y) – 3y = 11
⇒ 14 – 2y – 3y = 11
⇒ –5y = – 3 or, y = 3/5.
Now, substituting the value of y in equation (1), we get;
x + 3/5 = 7 ⇒ x = 32/5.
Hence, x = 32/5 and y = 3/5.

(ii) We have, system of equations are x + y = 7 ....(1) and 12x + 5y = 7 ....(2)
From equation (1), we have;
x + y = 7
⇒ x = 7 – y
Substituting the value of y in equation (2), we get ;
⇒ 12(7 –­ y­) + 5y = 7
⇒ 84 – 12y + 5y = 7
⇒ –7y = – 77
⇒ y = 11
Now, Substituting the value of y in equation (1), we get ;
x + 11 = 7 ⇒ x = – 4
Hence, x = – 4, y = 11.

(iii) We have; system of equations are 2x – 7y = 1 ....(1) and 4x + 3y = 15 ....(2)
From equation (1), we get
2x – 7y = 1 ⇒ x =
Substituting the value of x in equation (2), we get ;
⇒ 4× + 3y = 15
+ 3y = 15
⇒ 28y+4+ 6y = 30
⇒ 34y = 26 ⇒ y = 26/34 = 13/17
Now, substituting the value of y in equation (1), we get;
2x – 7× 13/17= 1
⇒ 2x = 1 + 91/17= 108/17 ⇒ x = 108/34 = 54/17
Hence, x = 54/17, y = 13/17

(iv) We have ; equations are 3x – 5y = 1 .... (1) and 5x + 2y = 19 .... (2)
From equation (1), we get;
3x – 5y = 1 ⇒ x =
Substituing the value of x in equation (2), we get ;
⇒ 5 × + 2y = 19
⇒ 25y + 5 + 6y = 57 ⇒ 31y = 52
Thus, y = 52/31
Now, substituting the value of y in equation (1), we get ;
3x – 5 × 52/31 = 1
⇒ 3x – 260/31= 1 ⇒ 3x = 291/31
⇒ x = 291/(31×3) = 97/31
Hence, x = 97/31, y = 52/31

(v) We have, system of linear equations are 5x + 8y = 9 ....(1) and 2x + 3y = 4 ....(2)
From equation (1), we get;
5x + 8y = 9 ⇒ x =
Substituting the value of x in equation (2), we get ;
⇒ 2 × + 3y = 4
⇒ 18 – 16y + 15y = 20
⇒ –y = 2 or y = – 2
Now, substituting the value of y in equation (1), we get ;
5x + 8 (–2) = 9
⇒ 5x = 25 ⇒ x = 5
Hence, x = 5, y = – 2.

Example: Solve the following systems of equations by eliminating ‘y’ (by substitution) :
(i) 3x – y = 3 and 7x + 2y = 20 (ii) 7x + 11y – 3 = 0 and 8x + y – 15 = 0 (iii) 2x + y – 17 = 0 and 17x – 11y – 8 = 0

Sol. (i) We have the equations are 3x – y = 3 ....(1) and 7x + 2y = 20 ....(2)
From equation (1), we get ;
3x – y = 3 ⇒ y = 3x – 3
Substituting the value of ‘y’ in equation (2), we get ;
⇒ 7x + 2 × (3x – 3) = 20
⇒ 7x + 6x – 6 = 20
⇒ 13x = 26 ⇒ x = 2
Now, substituting x = 2 in equation (1), we get;
3 × 2 – y = 3
⇒ y = 3
Hence, x = 2, y = 3.

(ii) We have; the equations are 7x + 11y – 3 = 0 ....(1) and 8x + y – 15 = 0 .....(2)
From equation (1), we get;
7x + 11y = 3
⇒ y =
Substituting the value of ‘y’ in equation (2), we get;
⇒ 8x + = 15
⇒ 88x + 3 – 7x = 165
⇒ 81x = 162
⇒ x = 2
Now, substituting, x = 2 in the equation (2), we get ;
8 × 2 + y = 15
⇒ y = – 1
Hence, x = 2, y = – 1.

(iii) We have the system of equations are 2x + y = 17 ....(1) and 17x – 11y = 8 ....(2)
From equation (1), we get;
2x + y = 17 ⇒ y = 17 – 2x
Substituting the value of ‘y’ in equation (2), we get ;
17x – 11 (17 – 2x) = 8
⇒ 17x – 187 + 22x = 8
⇒ 39x = 195
⇒ x = 5
Now, substituting the value of ‘x’ in equation (1), we get ;
2 × 5 + y = 17 ⇒ 10 + y = 17
⇒ y = 7
Hence, x = 5, y = 7.

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#### NTSE Mathematics (Class X)

• Real Numbers
• Polynomials
• Linear Equation in Two Variables
• Trigonometry
• Similar Triangles
• Statistics
• Quadratic Equation
• Arithmetic Progressions
• Application of Trigonometry
• Circle
• Co-ordinate Geometry
• Area related to Circle
• Surface Area & Volume
• Constructions
• Probability

#### NTSE Mathematics (Class IX)

• Real Numbers
• Polynomials
• Linear Equation in Two Variables
• Trigonometry
• Similar Triangles
• Statistics
• Quadratic Equation
• Arithmetic Progressions
• Application of Trigonometry
• Circle
• Co-ordinate Geometry
• Area related to Circle
• Surface Area & Volume
• Constructions
• Probability

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