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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 βxβ and βyβ:
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 βxβ equal.
On multiplying, we get;
4(3x + 2y) = 4 β 18
or, 12x + 8y = 72 ............. (iii)
and
3(4x + 5y) = 3 β 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 = 37.
Substituting value of βyβ in equation (i), we get;
3x + 2(37) = 18.
or, 3x + 67 = 18
or, 3x = 18 β 67
or, 3x = 1207.
or, x = 12021.
or, x = 407.
Hence, x = 407 and y = 37.
II: Solved examples on simultaneous linear Equations using substitution method:
1. Solve for βxβ and βyβ:
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 βxβ 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 = 75
Substituting this value of y into equation of x in (iii), we get;
x = 9 - 3y
or, x = 9 - 375
or, x = 9 β 215
or, x = 2475.
Hence, x = 2475 and y = 75.
2. Solve for βxβ and βyβ;
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 = 53.
Substituting this value of x as 53 in equation y = 5 - x , we get;
y = 5 - 53
or, y = 103.
Hence, x = 53 and y = 103.
III. Solved example on simultaneous linear Equations using Cross-multiplication method:
Solve for βxβ and βyβ;
3x + 5y - 25 = 0.
5x + 3y β 35 = 0.
Solution:
Assume two linear equations be
A1 x + B1y + C1 = 0, and
A2x + B2y + C2 = 0.
The coefficients of x are: A1 and A2.
The coefficients of y are: B1 and B2.
The constant terms are: C1 and C2.
To solve the equations in a simplified way, we use following table:
xB1C2βB2C1=yC1A2βC2A1=1A1B2βA2B1
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
x5Γ(β35)β3Γ(β25)=y(β25)Γ5β(β35)Γ3=13Γ3β5Γ5.
or, xβ175+75=yβ125+105=19β25.
or, xβ100=yβ20=1β16.
On equating x term with constant term, we get;
x = 254.
On equating y term with constant term, we get;
y = 54.
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, x7 = y8 (= k)
Substituting x = 7k and y = 8k in equation (1) we get
2 β 7k + 3 β 8k = 4
or, 14k + 24k = 4
or, 38k = 4
or, K = 438
or, k = 219
Therefore, x = 7 β 219 and y = 8 β 219
Thus, x = 1419 and y = 1619
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