Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the val

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 Multiple Choice QuestionsShort Answer Type

41.

Solve the following pairs of linear equations by the substitution method.


space space space space space space space space fraction numerator 3 straight x over denominator 2 end fraction minus fraction numerator 5 straight y over denominator 3 end fraction equals negative 2

space space space space straight x over 3 plus straight y over 2 equals 13 over 6


    
     

 

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42.

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.


The given equations are

2x + 3y = 11    ...(i)

2x - 4y = - 24    ...(ii)

From (i), we have

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Putting the value of ‘x’ in (ii), we get  

open parentheses fraction numerator 11 minus 3 straight y over denominator 2 end fraction close parentheses minus 4 straight y equals negative 24
rightwards double arrow fraction numerator 22 minus 6 straight y over denominator 2 end fraction minus 4 straight y equals negative 24
rightwards double arrow space fraction numerator 22 minus 6 straight y minus 8 straight y over denominator 2 end fraction equals negative 24
rightwards double arrow space space 22 minus 14 straight y space equals space minus 48
rightwards double arrow space space space minus 14 straight y space equals space minus 48 space equals space minus 22
rightwards double arrow space space space space minus 14 straight y space equals space minus 70
rightwards double arrow space space space space space space space space straight y space equals space 5
Putting the value ofy in (iii), we get


space space space space straight x equals fraction numerator 11 minus 3 straight y over denominator 2 end fraction equals fraction numerator 11 minus 15 over denominator 2 end fraction
space space space rightwards double arrow space space space space straight x equals fraction numerator negative 4 over denominator 2 end fraction equals negative 2

Hence, the solution is x = -2,y = 5.
It is given that: y = mx + 3 Putting the values of.v andy in given condition we gel
5 = m(-2) + 3
⇒    5 = - 2m + 3
⇒    -2m = 2
⇒ m = -1.



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43. Form the pair of linear equations for the following problems and find their solutions by substitution method.

The difference between two numbers is 26 and one number is three times the other. Find them.
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44. Form the pair of linear equations for the following problems and find their solutions by substitution method.

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
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45.

Form the pair of linear equations for the following problems and find their solution by substitution method.

The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and each ball.

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46.

Form the pair of linear equations for the following problems and find their solution by substitution method.

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 105 and for a journey of 15 km, the charge paid is ` 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

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47.

Form the pair of linear equations for the following problems and find their solution by substitution method.

A fraction becomes 9 over 11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes space 5 over 6 . Find the fraction.

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48.

Form the pair of linear equations for the following problems and find their solution by substitution method.

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

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49.

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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50.

Verify that the following system of equations has a unique solution and find the solution
ax + by + m = 0; ax - cy - n = 0

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