There are different methods for solving simultaneous linear Equations:
I. Elimination of a variable
II. Substitution
III. Cross-multiplication
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 Cross-multiplication 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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