Solve the following pair of linear equations by the elimination

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 Multiple Choice QuestionsShort Answer Type

111.

For what values of a and b does the following pair of linear equations have an enfinitc no. of solution :

2x + 3y = 7
a (x + y) -b (x - y) = 3a + b - 2

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112.

Solve the following pair of linear equations by the elimination method and the substitution method
x + y = 5 and 2x – 3y = 4

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113.

Solve the following pair of linear equations by the elimination method and the substitution method
3x + 4y = 10 and 2x – 2y = 2


3x + 4y = 10    ...(i)
2x - 2y = 2    ...(ii)
For making the coefficient of y in (i) and (ii) equal, we multiply (ii) by 2 and adding we get.


3 straight x plus 4 straight y equals 10
bottom enclose 4 straight x minus 4 straight y equals 4 space space end enclose
7 straight x space space space space space space space equals space 14

rightwards double arrow space space space space straight X space equals space 14 over 7 equals 2

Now, putting the value ofx in (i), we get
3x + 4y = 10
⇒ 3(2) + 4y = 10
⇒ 6 + 4y = 10
⇒ 4y = 4
⇒ y = 1
Hence, x = 2, y = 1
Substitution method :
We have,
3x + 4y = 10    ...(i)
2x - 2y = 2    ...(ii)
From (i), we have
3x + 3y = 10

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Substituting the value of (iii) in (ii), we get
2x - 2y = 2
rightwards double arrow space 2 open parentheses fraction numerator 10 minus 4 straight y over denominator 3 end fraction close parentheses minus 2 straight y equals 2
rightwards double arrow space fraction numerator 2 left parenthesis 10 minus 4 straight y right parenthesis minus 6 straight y over denominator 3 end fraction equals 2
rightwards double arrow space 20 minus 8 straight y minus 6 straight y equals 6
rightwards double arrow space minus 14 straight y equals 6 minus 20
rightwards double arrow negative 14 straight y equals negative 14
rightwards double arrow space straight y equals 1
Now, substituting the value ofy in (iii), we get

straight x equals fraction numerator 10 minus 4 straight y over denominator 3 end fraction equals fraction numerator 10 minus 4 left parenthesis 1 right parenthesis over denominator 3 end fraction equals fraction numerator 10 minus 4 over denominator 3 end fraction equals 6 over 3 equals 2
Hence,   x=2, y=1

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114.

Solve the following pair of linear equations by the elimination method and the substitution method
3x – 5y – 4 = 0 and 9x = 2y + 7

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115.

Solve the following pair of linear equations by the elimination method and the substitution method

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116.

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to It becomes   1 half if we only add I to the denominator. What is the fraction?

104 Views

117.

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

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118.

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

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119.

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

Meena went to a bank to withdraw ` 2000. She asked the cashier to give her ` 50 and ` 100 notes only. Meena got 25 notes in all. Find how many notes of ` 50 and ` 100 she received.

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120.

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ` 27 for a book kept for seven days, while Susy paid ` 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. 

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