Today our topic of discussion is Solution of simple simultaneous equations.
Solution of simple simultaneous equations
Solution of simple simultaneous equations
We shall discuss the solutions of only the consistent and independent simple simultaneous equations. Such system of equation has only one (unique) solution. Here, four methods of solutions are discussed:
- Method of substitution 2. Method of elimination 3. multiplication 4. Graphical method.
In class VIII, we have known how to solve by the methods of substitution and elimination. Here, examples of one for each of these two methods are given.
Example 2. Solve by the method of substitution:
2x + y = 8
3x – 2y = 5
Solution: Given equations are;
2x + y =8…(1)
3x – 2y =5…(2)
From equation (1), y = 8 – 2x (3)
Putting the value of y from equation (3) in equation (2) we get,
3x – 2(8 – 2x) = 5
or 0.3x – 16 + 4x = 5
or, 7x = 5 + 16
or, 7x = 21
or, x = 3
Putting the value of r in equation (3) we get,
y = 8 – 2 * 3
or, y = 8 – 6
or, y = 2
∴ solution (x, y) = (3, 2)
Substitution method: Conveniently from any of the two equations, value of one variable is expressed in terms of the other variable and putting the obtained value in the other equation, we get an equation in one variable. Solving this equation, value of the variable can be found.
This value can be put in any of the equations. But, if it is put in the equation in which one variable has been expressed in terms of the other variable, the solution will be easier. From this equation, value of the other variable will be found.
Example 3. Solve by the method of elimination:
2x + y = 8
3x – 2y = 5
Note: To show the difference between the methods of substitution and elimination, same equations of example 2 have been taken in this
example 3. Solution: Given equations are:
2x + y =8…(1)
3x – 2y =5…(2)
Multiplying both sides of equation (1) by 2, 4x + 2y = 16 (3)
Adding (2) and (3),
7x = 21
or, x = 3
Putting the value of x in (1) we get,
2 * 3 + y = 8
or, y = 8 – 6
or, y = 2
∴ solution (x, y) = (3, 2)
Elimination method: Conveniently one equation or both equations are multiplied by such a number so that after multiplication, absolute values of the coefficients of the same variable become equal. Then as per need, if the equations are added or subtracted, the variable with equal coefficient will be eliminated.
Then, solving the obtained equation, the value of the existing variable will be found. If that value is put conveniently in any of the given equations, value of the other variable will be found.
Cross multiplication method:
We consider the following two equations:
a₁*x + b₁*y + c₁ =0…(1) a₂*x + b₂*y + c₂ = 0 (2)
Multiplying equation (1) by b₂ and equation (2) by b₁ we get,
a₁*b₂*x + b₁*b₂*y + b₂*c₁ =0…(3)
a₂*b₁*x + b₁*b₂*y + b₁*c₂ =0…(4)
Subtracting equation (4) from equation (3) we get,
a₁*b₂ – a₂*b₁ * x + b₂*c₁ – b₁*c₂ = 0
or, a₁*b₂ – a₂*b₁ * x = b₁*c₂ – b₂*c₁
or, x/b₁*c₂ – b₂*c₁ = 1 a 1 b 2 -a 2 b 1 …(5)
Again, multiplying equation (1) by a₂ and equation (2) by a₁ we get,
a₁*a₂*x + a₂*b₁*y + c₁*a₂ =0…(6)
a₁*a₂*x + a₁*b₂*y + c₂*a₁ =0…(7)
Subtracting equation (7) from equation (6) we get,
a₂*b₁ – a₁*b₂ * y + c₁*a₂ – c₂*a₁ = 0
o*r₁ – a₁*b₂ – a₂*b₁* y = – (c₁*a₂ – c₂*a₁ or , y/(c₁*a₂ – c₂*a₁ = 1 a 1 b 2 -a 2 b 1 …(8)
From (5) and (8) we get, x/b₁*c₂ – b₂*c₁ = y/(c₁*a₂ – c₂*a₁ = 1/a₁*b₂ – a₂*b₁
From such relation between x and y, the technique of finding their values is called the method of cross- multiplication.
From the above relation between x and y we get, I
x/(b₁*c₂ – b₂*c₁) = 1/(a₁*b₂ – a₂*b₁) Again, – or, 1 x = (b₁*c₂ – b₂*c₁)/(a₁*b₂ – a₂*b₁) or, y = (c₁*a₂ – c₂*a₁/(a₁*b₂ – a₂*b₁
∴ The solution of the given equations: (x, y) = (b₁*c₂ – b₂*c₁)/(a₁* b₂ – a₂*b₁, (c₁*a₂ – c₂*a₁/a₁* b₂ – a₂*b₁)
Note: The method of cross multiplication can also be applied by keeping the constant terms of both equations on the right hand side. In that case, changes of sign will occur but the solution will remain the same.
Example 4. Solve by the method of cross multiplication:
6x – y = 1
3x + 2y = 13
Solution:
Making the right hand side of the equations 0 (zero) by transposition, we get,
6x – y – 1 = 0
3x+2y-130
respectively we get, a6, b₁ = – 1 c₁ = – 1 a₂ = 3 by 2, c₂ = -13
By the method of cross multiplication we get,
x/b₁*c₂ – b₂*c₁ = y/c₁*a₂ – c₂*a₁ = 1/a₁*b₂ – a₂*b₁
comparing the equations with
a₁*x + b₁*y + c₁ = 0 and
a₂*x + b₂*y + c₂ = 0
r, x/((- 1)(- 13) – 2(- 1)) = y/((- 1) * 3 – (- 13) * 6) = 1/(6 * 2 – 3(- 1))
or, x/(13 + 2) = y/(- 3 + 78) = 1/(12 + 3)
or, x/15 = y/75 = 1/15
Therefore, x/15 = 1/15 * epsilon , x = 15/15 = 1
Again, y/75 = 1/15
or, y = 75/15 = 5
∴ solution (x, y) = (1, 5)
a₂ b₂
[[6, – 1], [3, 2]]
Example 5. Solve by the method of cross multiplication:
3x – 4y = 0
2x – 3y = – 1
Solution: Given equations are:
3x-4y=0
or, 3x-4y+0=0
2x – 3y = – 1
By the method of cross multiplication, we get:
2x – 3y + 1 = 0
x/(- 4 * 1 – (- 3) * 0) = y/(0 * 2 – 1 * 3) = 1/(3(- 3) – 2(- 4))
or, x/(- 4 + 0) = y/(0 – 3) = 1/(- 9 + 8)
or, x/- 4 = y/- 3 = 1/- 1
3| matrix x&y&1&-4\\ 2 * det [[- 4, 0, 3, – 4], [2, 1, 2, – 3]]
or, x/4 = y/3 = 1/1
x/4 = 1/1 * or , x = 4
Again, y/3 = 1/1 * or y = 3
∴ solution (x, y) = (4, 3)
Example 6. Solve by the method of cross multiplication: x/2 + y/3 = 8 (5x)/4 – 3y = – 3
Solution: Arranging the given equations in the form ax + by + c = 0 we get,
x/2 + y/3 = 8
or, (3x + 2y)/6 = 8
or, 3x + 2y – 48 = 0
Again, (5x)/4 – 3y = – 3
6, (5x – 12y)/4 = – 3
or, 5x – 12y + 12 = 0
∴ the given equations are:
3x + 2y – 48 = 0
5x – 12y + 12 = 0
By the method of cross multiplication we get,
matrix x= y/(2 * 12 – (- 12)(- 48)) &=& 1 (-48)*5-12*3 &=& 1 3*(-12)-5*2 matrix or, x/(24 – 576) = y/(- 240 – 36) = 1/(- 36 – 10) I 1
or, x/- 552 = y/- 276 = 1/- 46
or, x/552 = y/276 = 1/46
x/552 = 1/46 * or x = 552/46 = 12
Again, y/276 = 1/46 or, y = 276/46 = 6
∴ solution: (x, y) = (12, 6)
Verification of the correctness of the solution:
Putting the values of ry in given equations,
In first equation, L.H. S = x/2 + y/3 = 12/2 + 6/3 = 6 + 2 = 8 =R.H.S
In second equation L.H. S = (5x)/4 – 3y = (5 * 12)/4 – 3 * 6 = 15 – 18 = – 3 =R.H.S.
∴the solution is correct.
Example 7. Solve by the method of cross multiplication: ax – by = ab = bx – ay
Solution: Given equations are:
ax – by = ab
or, ax – by – ab = 0 ,bx – ay – ab = 0
By the
bx – ay = ab
By the method of cross multiplication, we get,
x/((- b)(- a * b) – (- a)(- a * b)) = y/((- a * b) * b – (- a * b) * a) = a(- a) – b(- b)
x/(a * b² – a² * b) = y/(- a * b² + a² * b) = 1/(- a² + b²)
or, x/(- a * b(a – b)) = y/(ab(a – b)) = 1/(- (a + b) * (a – b)) a -ab -ab -b -a -a b
or, x/(ab(a – b)) = y/(- a * b(a – b)) = 1/((a + b)(a – b))
x/(ab(a – b)) = 1/((a + b)(a – b)),
or, x = (ab(a – b))/((a + b)(a – b)) = (ab)/(a + b)
Again, y/(- a * b(a – b)) = 1/((a + b)(a – b))
or, y = (- a * b(a – b))/((a + b)(a – b)) = (- a * b)/(a + b)
(x, y) = ((ab)/(a + b), (- a * b)/(a + b))
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