Solving Systems Of Equations By 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)
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.
Remark : Verification is a must to check the answer.
Substitution Method Examples
Example 1: Solve each of the following system of equations by eliminating x (by substitution) :
(i) x + y = 7 (ii) x + y = 7 (iii) 2x – 7y = 1
2x – 3y = 11 12x + 5y = 7 4x + 3y = 15
(iv) 3x – 5y = 1 (v) 5x + 8y = 9
5x + 2y = 19 2x + 3y = 4
Sol. (i) We have,
x + y = 7 ….(1)
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,
x + y = 7 ….(1)
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,
2x – 7y = 1 ….(1)
4x + 3y = 15 ….(2)
From equation (1), we get
2x – 7y = 1 ⇒ x = \(\frac{7y+1}{2}\)
Substituting the value of x in equation (2), we get ;
\(\Rightarrow 4\times \frac{7y+1}{2}+3y=15\)
\(\Rightarrow \frac{28y+4}{2}+3y=15\)
⇒ 28y + 4 + 6y = 30
⇒ 34y = 26 ⇒ y = \(\frac{13}{17}\)
Now, substituting the value of y in equation (1), we get
2x – 7 × \(\frac{13}{17}\) = 1
⇒ 2x = 1 + \(\frac{91}{17}\) = \(\frac{108}{17}\) ⇒ x = \(\frac{108}{34}\) = \(\frac{54}{17}\)
Hence, x = \(\frac{54}{17}\) , y = \(\frac{13}{17}\)
(iv) We have,
3x – 5y = 1 …. (1)
5x + 2y = 19 …. (2)
From equation (1), we get;
3x – 5y = 1 ⇒ x = \(\frac{5y+1}{3}\)
Substituing the value of x in equation (2), we get
⇒ 5 × \(\frac{5y+1}{3}\) + 2y = 19
⇒ 25y + 5 + 6y = 57 ⇒ 31y = 52
Thus, y = \(\frac{52}{31}\)
Now, substituting the value of y in equation (1), we get
3x – 5 × \(\frac{52}{31}\) = 1
⇒ 3x – \(\frac{260}{31}\) = 1 ⇒ 3x = \(\frac{291}{31}\)
⇒ x = \(\frac{97}{31}\)
Hence, x = \(\frac{97}{31}\) , y = \(\frac{52}{31}\)
(v) We have,
5x + 8y = 9 ….(1)
2x + 3y = 4 ….(2)
From equation (1), we get
5x + 8y = 9 ⇒ x = \(\frac{9-8y}{5}\)
Substituting the value of x in equation (2), we get
⇒ 2 × \(\frac{9-8y}{5}\) + 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 2: Solve the following systems of equations by eliminating ‘y’ (by substitution) :
(i) 3x – y = 3 (ii) 7x + 11y – 3 = 0 (iii) 2x + y – 17 = 0
7x + 2y = 20 8x + y – 15 = 0 17x – 11y – 8 = 0
Sol. (i) We have,
3x – y = 3 ….(1)
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,
7x + 11y – 3 = 0 ….(1)
8x + y – 15 = 0 …..(2)
From equation (1), we get;
7x + 11y = 3
⇒ y = \(\frac{3-7x}{11}\)
Substituting the value of ‘y’ in equation (2), we get
⇒ 8x + \(\frac{3-7x}{11}\) = 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,
2x + y = 17 ….(1)
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
⇒ y = 7
Hence, x = 5, y = 7.
Example 3: Solve the following systems of equations,
(i) \(\frac{15}{u}\) + \(\frac{2}{v}\) = 17 (ii)\(\frac{11}{v}\) – \(\frac{7}{u}\) = 1
\(\frac{1}{u}\) + \(\frac{1}{v}\) = \(\frac{36}{5}\) \(\frac{9}{v}\) + \(\frac{4}{u}\) = 6
Sol. (i) The given system of equation is
\(\frac{15}{u}\) + \(\frac{2}{v}\) = 17 ….(1)
\(\frac{1}{u}\) + \(\frac{1}{v}\) = \(\frac{36}{5}\) ….(2)
Considering 1/u = x, 1/v = y, the above system of linear equations can be written as
15x + 2y = 17 ….(3)
x + y = \(\frac{36}{5}\) ….(4)
Multiplying (4) by 15 and (iii) by 1, we get
15x + 2y = 17 ….(5)
15x + 15y = \(\frac{36}{5}\) × 15 = 108 ….(6)
Subtracting (6) form (5), we get
–13y = – 91 ⇒ y = 7
Substituting y = 7 in (4), we get
x + 7 = \(\frac{36}{5}\) ⇒ x = \(\frac{36}{5}\) – 7 = \(\frac{1}{5}\)
But, y = \(\frac{1}{v}\) = 7 ⇒ v = \(\frac{1}{7}\)
and, x = \(\frac{1}{u}\) = \(\frac{1}{5}\) ⇒ u = 5
Hence, the required solution of the given system is u = 5, v = 1/7.
(ii) The given system of equation is
\(\frac{11}{v}\) – \(\frac{7}{u}\) = 1; \(\frac{9}{v}\) + \(\frac{4}{u}\) = 6
Taking 1/n = x and 1/u = y, the above system of equations can be written as
11x – 7y = 1 ….(1)
9x – 4y = 6 ….(2)
Multiplying (1) by 4 and (2) by 7, we get,
44x – 28y = 4 ….(3)
63x – 28y = 42 ….(4)
Subtracting (4) from (3) we get,
– 19x = –38 ⇒ x = 2
Substituting the above value of x in (2), we get;
9 × 2 – 4y = 6 ⇒ –4y = – 12
⇒ y = 3
But, x = \(\frac{1}{v}\)= 2 ⇒ v = \(\frac{1}{2}\)
and, y = \(\frac{1}{u}\) = 3
⇒ u = \(\frac{1}{3}\)
Hence, the required solution of the given system of the equation is
v = \(\frac{1}{2}\), u = \(\frac{1}{3}\)
Example 4: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which
y = mx + 3.
Sol. We have,
2x + 3y = 11 ….(1)
2x – 4y = – 24 ….(2)
From (1), we have 2x = 11 – 3y
Substituting 2x = 11 – 3y in (2), we get
11 – 3y – 4y = –24
⇒ –7y = – 24 – 11
⇒ –7y = – 35
⇒ y = 5
Putting y = 5 in (1), we get
2x + 3 × 5 = 11
2x = 11 – 15
⇒ x = –4/2 = – 2
Hence, x = – 2 and y = 5
Again putting x = – 2 and y = 5 in y = mx + 3, we get
5x = m(–2) + 3
⇒ –2m = 5 – 3
⇒ m = – 1