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Pair of Linear Equations in Two Variables

Short Answer Type

131. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

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132. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

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133. Solve the following pairs of equations by reducing them to a pair of linear equations :

fraction numerator 1 over denominator 2 straight x end fraction plus fraction numerator 1 over denominator 3 straight y end fraction equals 2 fraction numerator 1 over denominator 3 straight x end fraction plus fraction numerator 1 over denominator 2 straight y end fraction equals 13 over 6

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134. Solve the following pairs of equations by reducing them to a pair of linear equations :

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fraction numerator 4 over denominator square root of straight x end fraction minus fraction numerator 9 over denominator square root of 9 end fraction equals negative 1

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135. Solve the following pairs of equations by reducing them to a pair of linear equations :

4 over straight x plus 3 straight y equals 14 3 over straight x minus 4 straight y equals 23

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136.

Solve the following pairs of equations by reducing them to a pair of linear equations:

$fraction numerator 5 over denominator x minus 1 end fraction plus fraction numerator 1 over denominator y minus 2 end fraction equals 2$

$fraction numerator 6 over denominator straight x minus 1 end fraction minus fraction numerator 3 over denominator straight y minus 2 end fraction equals 1$

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137.

Solve the following pairs of equations by reducing them to a pair of linear equations:

$space space space space space space space space space fraction numerator 7 straight x minus 2 straight y over denominator xy end fraction equals 5 space space space space space space space fraction numerator 8 straight x plus 7 straight x over denominator x y end fraction equals 15$

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138.

Solve the following pairs of equations by reducing them to a pair of linear equations:

6x + 3y = 6xy
2x + 4y = 5x

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139.

Solve the following pairs of equations by reducing them to a pair of linear equations:

$fraction numerator 10 over denominator straight x plus straight y end fraction plus fraction numerator 2 over denominator straight x minus straight y end fraction equals 4$

$fraction numerator 15 over denominator straight x plus straight y end fraction minus fraction numerator 5 over denominator straight x minus straight y end fraction equals negative 2$

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140.
Solve the following pairs of equations by reducing them to a pair of linear equations:

$fraction numerator 1 over denominator 3 x plus y end fraction plus fraction numerator 1 over denominator 3 x minus y end fraction space equals space 3 over 4 fraction numerator 1 over denominator 2 left parenthesis 3 x plus y right parenthesis end fraction plus fraction numerator 1 over denominator 2 left parenthesis 3 x minus y right parenthesis end fraction space equals space fraction numerator negative 1 over denominator space space 8 end fraction$

Let space space fraction numerator 1 over denominator 3 straight x plus straight y end fraction equals straight u space and space space fraction numerator 1 over denominator 3 straight x minus straight y end fraction space equals space straight v

Then the given system of equations become

space space space space space space space space space space space space straight u plus straight v space equals space 3 over 4 and space space space space space space space straight u over 2 minus straight v over 2 equals fraction numerator negative 1 over denominator 8 end fraction rightwards double arrow space space space space space space space space space straight u minus straight v equals fraction numerator negative 1 over denominator 4 end fraction space

Thus, we have following equations

straight u space plus space straight v equals 3 over 4 space space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis straight i right parenthesis straight u minus straight v equals fraction numerator negative 1 over denominator 4 end fraction space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis ii right parenthesis

Adding equation (i) and (ii), we get

2 straight u space equals space 3 over 4 plus open parentheses fraction numerator negative 1 over denominator 4 end fraction close parentheses 2 straight u space equals space fraction numerator 3 minus 1 over denominator 4 end fraction equals 2 over 4 rightwards double arrow space space space space space straight u space equals space 1 fourth

Putting the value of ‘u’ in eqn. (i), we get

Then the given system of equations becomeThus, we have following equa

Adding (iii) and (iv), we get
6x = 6
⇒ x = 1
Putting the value of ‘x’ in (iii), we get
3 x + y = 4
⇒ 3(1) + y = 4
⇒ y = 1
Hence, x = 1, y = 1 is the solution of the given equation.

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Previous Year Papers

Test Series

Test Yourself

Short Answer Type

140. Solve the following pairs of equations by reducing them to a pair of linear equations: