Write whether the following pair of linear equations is consistent or not.

x + y = 14
x - y = 4 


x + y = 14
x - y = 4

H e r e comma space space a subscript 1 over a subscript 2 equals 1 comma space space b subscript 1 over b subscript 2 equals negative 1
S o comma space space space space comma space space a subscript 1 over a subscript 2 space not equal to space b subscript 1 over b subscript 2

the equation have unique solution.
Pair of linear equations are consistent.

 
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Find the value of k so that the following system of equations has infinite solutions:

3x - y - 5 = 0;     6x - 2y + k = 0


Here, a1 = 3, b1 = -1 and c1 = -5
a2 = 6, b2 = -2 and c2 = k

therefore space straight a subscript 1 over straight a subscript 1 equals 3 over 6 equals 1 half comma space space space straight b subscript 1 over straight b subscript 2 equals fraction numerator negative 1 over denominator negative 2 end fraction equals 1 half comma space straight c subscript 1 over straight c subscript 2 equals fraction numerator negative 5 over denominator straight k end fraction
Now comma space space space
space space space space space space space space space space space space space space fraction numerator negative 5 over denominator straight k end fraction equals 1 half
rightwards double arrow space space space space straight k space space equals space minus 10
Hence the given system will have infinite no. of solution if k = -10.

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Given below are three equations. Two of them have infinite solutions and two of them have no solution. State the two pairs :

2x - 3y = 4; 4x - 6y = 7; 6x - 9y - 12.
Sol. Pair of infinite solutions :
2x - 3y = 4; 6x - 9y = 12
Here,    a1 = 2, b1 = -3 and c1 = 4
and    a2 = 6, b2 = -9 and c2 = 12
The given values satisfy the condition :

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So, given equations have infinite solutions pair of no solutions :
2x - 3y = 4; 4x - 6y = 7
Here,    a1 = 2, b1 = -3 and c1 = 4
and    a2 = 4, b2 = -6 and c2 = 7
The given values satisfy the condition:

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So, given equations have no solutions.

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Find the value of k for which the pair of equations kx - 4y = 3; 6x - 12y = 9 has an infinite number of solution.


We have,
kx - 4y = 3; 6x - 12y = 9
Here,    a1 = k, b = -4,c1 = 3
a2 = 6, b2 = - 12, c2 = 9
For infinitely many solutions,

straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2
rightwards double arrow space space space space space space space space space space space space space space straight k over 6 equals fraction numerator negative 4 over denominator negative 12 end fraction equals 3 over 9
rightwards double arrow space space space space space space space space space space space space space space space straight k over 6 equals 1 third equals 1 third
rightwards double arrow space space space space space space space space space space space space space space space space straight k over 6 equals 1 third rightwards double arrow 3 straight k space rightwards double arrow box enclose straight k equals 2 end enclose space space space
Hence, the given system of equation has infinitely many solution if, k = 2.

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Show that x = 1 and y = 1 is a solution of the pair of equations 2x + 3y = 5. 5x - 2y = 3.

The given pair of equation is
2x + 3y = 5
5x - 2y = 3
Putting x = 1,y = 1 in eq. (i) and (ii), we get
2x + 3y = 5
⇒ 2(1) + 3(1) = 5
⇒ 5 = 5 which is true.
Also, 5a - 2y = 3
⇒ 5(1) - 2(1) = 3
⇒    3 = 3 which is true.
Thus, x = 1, y - 1 is the solution of the given pair of equations.

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