There are different methods for solving simultaneous linear Equations:
I. Elimination of a variable
II. Substitution
III. Crossmultiplication
IV. Evaluation of proportional value of variables
This topic is purely based upon numerical examples. So, let us solve some examples based upon solving linear equations in two variables.
I: Solved example on simultaneous linear Equations using elimination method:
Solve for ‘x’ and ‘y’:
3x + 2y = 18.
4x + 5y = 25.
Solution:
3x + 2y = 18 ............. (i)
4x + 5y = 25 ............. (ii)
Let us multiply equation (i) by 4 on both sides and equation (ii) by 3 on both sides, so as to make coefficients of ‘x’ equal.
On multiplying, we get;
4(3x + 2y) = 4 ∙ 18
or, 12x + 8y = 72 ............. (iii)
and
3(4x + 5y) = 3 ∙ 25
or, 12x + 15y = 75 ............. (iv)
Subtracting (iii) from (iv), we get;
12x + 15y  (12x + 8y) = 75  72
or, 12x + 15y  12x  8y = 3
or, 7y = 3
or, y = \(\frac{3}{7}\).
Substituting value of ‘y’ in equation (i), we get;
3x + 2(\(\frac{3}{7}\)) = 18.
or, 3x + \(\frac{6}{7}\) = 18
or, 3x = 18 – \(\frac{6}{7}\)
or, 3x = \(\frac{120}{7}\).
or, x = \(\frac{120}{21}\).
or, x = \(\frac{40}{7}\).
Hence, x = \(\frac{40}{7}\) and y = \(\frac{3}{7}\).
II: Solved examples on simultaneous linear Equations using substitution method:
1. Solve for ‘x’ and ‘y’:
x + 3y = 9
3x + 4y = 20
Solution:
x + 3y = 9 ............. (i)
3x + 4y = 20 ............. (ii)
Taking first equation in reference, i.e,
x + 3y = 9
x = 9  3y ............. (iii)
Substituting this value of ‘x’ from previous equation in 2nd equation, we get;
3x + 4y = 20.
or, 3(9  3y) + 4y = 20
or, 27  9y + 4y = 20
or, 5y = 20  27
or, 5y = 7
or, 5y = 7
or, y = \(\frac{7}{5}\)
Substituting this value of y into equation of x in (iii), we get;
x = 9  3y
or, x = 9  3\(\frac{7}{5}\)
or, x = 9 – \(\frac{21}{5}\)
or, x = \(\frac{247}{5}\).
Hence, x = \(\frac{247}{5}\) and y = \(\frac{7}{5}\).
2. Solve for ‘x’ and ‘y’;
x + y = 5
4x + y = 10
Solution:
x + y = 5 ............. (i)
4x + y = 10 ............. (ii)
From equation (i), we get value of y as:
y = 5  x
substituting this value of y in equation (ii), weget
4x + (5  x) = 10
or, 4x + 5  x = 10
or, 3x = 10  5
or, 3x = 5
x = \(\frac{5}{3}\).
Substituting this value of x as \(\frac{5}{3}\) in equation y = 5  x , we get;
y = 5  \(\frac{5}{3}\)
or, y = \(\frac{10}{3}\).
Hence, x = \(\frac{5}{3}\) and y = \(\frac{10}{3}\).
III. Solved example on simultaneous linear Equations using Crossmultiplication method:
Solve for ‘x’ and ‘y’;
3x + 5y  25 = 0.
5x + 3y – 35 = 0.
Solution:
Assume two linear equations be
A_{1 }x + B_{1}y + C_{1 }= 0, and
A_{2}x + B_{2}y + C_{2 }= 0.
The coefficients of x are: A_{1 }and A_{2.}
The coefficients of y are: B_{1} and B_{2.}
The constant terms are: C_{1} and C_{2}.
To solve the equations in a simplified way, we use following table:
\(\frac{x}{B_{1}C_{2}  B_{2}C_{1}} = \frac{y}{C_{1}A_{2}  C_{2}A_{1}} = \frac{1}{A_{1}B_{2}  A_{2}B_{1}}\)
In the given equations,
The coefficients of x are 3 and 5.
The coefficients of y are 5 and 3.
The constant terms are 25 and 35.
On substituting the respective values, we get
\(\frac{x}{5 × (35)  3 × (25)} = \frac{y}{(25) × 5  (35) × 3} = \frac{1}{3 × 3  5 × 5}\).
or, \(\frac{x}{ 175 + 75} = \frac{y}{125 + 105} = \frac{1}{9  25}\).
or, \(\frac{x}{100} = \frac{y}{20} = \frac{1}{16}\).
On equating x term with constant term, we get;
x = \(\frac{25}{4}\).
On equating y term with constant term, we get;
y = \(\frac{5}{4}\).
IV: Solved example on simultaneous linear Equations using evaluation method:
Solve for x and y:
2x + 3y = 4; 3x  5y = 2
Solution:
The given equations are
2x + 3y = 4 .......... (1)
3x  5y = 2.......... (2)
Multiplying equation (2) by 2, we get
6x  10y = 4.......... (3)
Now we add equations (1) and (3) we get
8x  7y = 0
or, 8x = 7y
or, \(\frac{x}{7}\) = \(\frac{y}{8}\) (= k)
Substituting x = 7k and y = 8k in equation (1) we get
2 ∙ 7k + 3 ∙ 8k = 4
or, 14k + 24k = 4
or, 38k = 4
or, K = \(\frac{4}{38}\)
or, k = \(\frac{2}{19}\)
Therefore, x = 7 ∙ \(\frac{2}{19}\) and y = 8 ∙ \(\frac{2}{19}\)
Thus, x = \(\frac{14}{19}\) and y = \(\frac{16}{19}\)
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