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$a_1x +b_1y+ c_1 =0 $
$a_2x +b_2y+ c_2 = 0 $
$$ { a_1\over a_2}\ne{ b_1\over b_2} $$
$$ { a_1\over a_2}={ b_1\over b_2} = { c_1\over c_2} $$
$$ { a_1\over a_2}={ b_1\over b_2} \ne { c_1\over c_2}$$
Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5
2x − 3y = 4.
Solution :
Elimination method:
$$ x + y = 5 .... (1) $$
$$ 2x − 3y = 4..... (2) $$
Multiply equation (1) by 3 to make the coefficient of y equal and opposite to that in equation (2).
$$ 3x + 3y = 15 .... (3) $$
By Adding equation (3) to equation (2), The terms with y will cancel out.
$$ (3x + 3y) + ( 2x − 3y ) = 15 + 4 $$
$$ ⇒ 3x + 3y + 2x − 3y = 19 $$
$$ ⇒ 5x = 19 $$
$$ ⇒ x = { 19 \over 5}$$
Substituting this value $ x ={ 19 \over 5} $ in equation (1), we obtain
$$ x + y = 5 $$
$$ ⇒ { 19 \over 5} + y = 5 $$
$$ ⇒ y = 5 - { 19 \over 5}$$
$$ ⇒ y = {25 - 19 \over 5}$$
$$ ⇒ y = {6 \over 5}$$
Substitution method:
From equation (1), express x in terms of y
$$ x = 5 - y .... (4) $$
Substituting this value $ x = 5 - y $ in equation (2), we obtain
$$ 2x − 3y = 4 $$
$$ ⇒ 2(5 - y) − 3y = 4 $$
$$ ⇒ 10 - 2y -3y = 4 $$
$$ ⇒ -5y = 4 -10 $$
$$ ⇒ -5y = -6 $$
$$ ⇒ y = {6 \over 5}$$
Substituting this value $ y = {6 \over 5} $ in equation (4)
$$ x = 5 - y $$
$$ ⇒ x = 5 - ({6 \over 5}) $$
$$ ⇒ x = {25 - 6 \over 5} $$
$$ ⇒ x = {19 \over 5} $$
The solution for the pair of linear equations is $ x = {19 \over 5} , y= {6 \over 5}$ ;
Solve the following pair of linear equations by the elimination method and the substitution method :
(ii) 3x + 4y = 10
2x – 2y = 2.
Solution :
Elimination method:
$$ 3x + 4y = 10 .... (1) $$
$$ 2x – 2y = 2..... (2) $$
Multiply equation (2) by 2 to make the coefficient of y a negative 4.
$$ 4x - 4y = 4 .... (3) $$
By Adding equation (3) to equation (1), The terms with y will cancel out.
$$⇒ (3x + 4y) + ( 4x - 4y ) = 10 + 4 $$
$$⇒ 3x + 4y + 4x − 4y = 14 $$
$$⇒ 7x = 14 $$
$$⇒ x = { 14 \over 7}$$
$$⇒ x = 2 $$
Substituting this value $ x = 2 $ in equation (1), we obtain
$$ 3x + 4y = 10 $$
$$ ⇒ (3 × 2) + 4y = 10 $$
$$⇒ 4y = 10 - 6 $$
$$⇒ y = { 4 \over 4} $$
$$⇒ y = 1 $$
Substitution method:
From equation (1), express x in terms of y
$$ x = {{10 - 4y}\over 3 } .... (4) $$
Substituting this value $ x = {{10 - 4y}\over 3 } $ in equation (2), we obtain
$$ 2x – 2y = 2 $$
$$⇒ 2({{10 - 4y}\over 3 }) – 2y = 2 $$
$$⇒ {{20 - 8y}\over 3 } – 2y = 2 $$
$$⇒ {{20 - 8y -6y}\over 3 } = 2 $$
$$⇒ -14y = 6- 20 $$
$$⇒ -14y = - 14 $$
$$⇒ y = {14 \over 14}$$
$$⇒ y = 1 $$
Substituting this value $ y = {1} $ in equation (4)
$$⇒ x = {{10 - 4y}\over 3 } $$
$$⇒ x = {{10 - 4}\over 3 } $$
$$⇒ x = {{6}\over 3 } $$
$$⇒ x = 2 $$
The solution for the pair of linear equations is $ x = {2} , y= {1}$ ;
Solve the following pair of linear equations by the elimination method and the substitution method :
(iii) 3x – 5y – 4 = 0
9x = 2y + 7
Solution :
Elimination method:
$$ 3x – 5y = 4 .... (1) $$
$$ 9x – 2y = 7..... (2) $$
Multiplying equation (1) by 3
$$ 9x - 15y = 12 .... (3) $$
By subtracting equation (3) from equation (2), we obtain
$$ (9x – 2y) - ( 9x - 15y ) = 7 - 12 $$
$$⇒ 9x – 2y - 9x + 15y = -5 $$
$$⇒ 13y = -5 $$
$$⇒ y = { -5 \over 13}$$
Substituting this value $ y = { -5 \over 13} $ in equation (1), we obtain
$$ 3x – 5y = 4 $$
$$⇒ 3x – 5( { -5 \over 13}) = 4 $$
$$⇒ 3x +( { 25 \over 13}) = 4 $$
$$⇒ 3x = 4 - ( { 25 \over 13}) $$
$$⇒ 3x = {52 - 25 \over 13} $$
$$⇒ 3x = {27\over 13} $$
$$⇒ x = {27\over 13} × {1\over 3} $$
$$⇒ x = {9\over 13} $$
Substitution method:
From equation (2), express x in terms of y
$$ x = {{2y + 7}\over 9 } .... (4) $$
Substituting this value $ x = {{2y + 7}\over 9 } $ in equation (1), we obtain
$$ 3x – 5y = 4 $$
$$⇒ 3({{2y + 7}\over 9 }) – 5y = 4 $$
$$⇒ ({{2y + 7}\over 3 }) – 5y = 4 $$
$$⇒ {2y + 7– 15y = 12 }$$
$$⇒ { 7 – 13y = 12 } $$
$$ -13y = 12- 7 $$
$$⇒ -13y = 5 $$
$$ ⇒ y = -{5 \over 13}$$
Substituting this value $ y = -{5 \over 13} $ in equation (4)
$$ x = {{2y + 7}\over 9 } $$
$$⇒ x = {{2 × -{5 \over 13} + 7}\over 9 } $$
$$⇒ x = {{-{10 \over 13} + 7}\over 9 } $$
$$⇒ x = {{{{-10 + 91} \over 13}}\over 9 } $$
$$⇒ x = {{81 \over 13} ×{1 \over 9} } $$
$$⇒ x = {9 \over 13} $$
The solution for the pair of linear equations is $ x = {9 \over 13} , y= -{5 \over 13}$ ;
Solve the following pair of linear equations by the elimination method and the substitution method :
(iv) $$ { x \over 2} + { 2y\over 3} = - 1 $$ $$ x - { y\over 3} = 3 $$
Solution :
Elimination method:
$$ { x \over 2} + { 2y\over 3} = - 1 .... (1) $$
$$ x - { y\over 3} = 3..... (2) $$
Multiplying equation (1) by 6 and equation (2) by 3
$$ 6 ×[{ x \over 2} + { 2y\over 3}] = - 1 × 6 $$
$$⇒ { 3x + 4y = -6 }.... (3) $$
$$ 3 × [x - { y\over 3}] = 3 × 3 $$
$$⇒ 3x - y = 9 .... (4)$$
By subtracting equation (4) from equation (3), we obtain
$$ (3x + 4y) - ( 3x - y) = -6 - 9 $$
$$⇒ 3x +4y - 3x + y = -15 $$
$$⇒ 5y = -15 $$
$$⇒ y = { -3 }$$
Substituting this value $ y = { -3 }$ in equation (3), we obtain
$$ { 3x + 4y = -6 }$$
$$⇒ { 3x + 4(-3) = -6 }$$
$$⇒ { 3x - 12 = -6 }$$
$$⇒ { 3x = -6 + 12 }$$
$$⇒ { x = { 6 \over 3} }$$
$$⇒ x = 2 $$
Substitution method:
By solving the equation (2)
$$ x - { y\over 3} = 3 $$
$$⇒ x = 3 +{ y\over 3} $$
$$⇒ x = {9 + y\over 3} .... (5) $$
Putting this value $ x = {9 + y\over 3} $ in equation (1), we obtain
$$ { x \over 2} + { 2y\over 3} = - 1 $$
$$⇒ { 1\over 2}({9 + y\over 3} ) + { 2y\over 3} = - 1 $$
$$⇒ {9 + y\over 6} + { 2y\over 3} = - 1 $$
$$⇒ {9 + y + 4y\over 6} = - 1 $$
$$⇒ {9 + 5y} = - 6 $$
$$⇒ 5y = - 15 $$
$$⇒ y = - 3 $$
Substituting this value $ y = -3 $ in equation (5)
$$ x = {9 + y\over 3} $$
$$⇒ x = {9 -3\over 3} $$
$$ ⇒ x = {6\over 3 } $$
$$ ⇒ x = 2 $$
The solution for the pair of linear equations is $ x = {2} , y= -{3}$ ;
Form the pair of linear equations in the following problems, and find their Solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $ { 1 \over 2} $ if we only add 1 to the denominator. What is the fraction?
Solution :
Let If the Numerator is x
And denominator is y
then the Fraction is $ {x \over y} $
According to the question
Adding 1 to Numerator and subtract 1 from the denominator
$$ {x + 1\over {y -1}} = 1 $$
After cross multiplication,
$$ {x + 1 } = {y -1} $$
$$ x - y + 1 + 1 = 0 $$
$$ x - y + 2 = 0 .... (1)$$
When 1 is added to the denominator;
$$ {x \over {y + 1}} = { 1 \over 2} $$
After cross multiplication,
$$ {2x } = {y + 1} $$
$$ { 2x - y - 1 = 0 }.... (2)$$
Now, after subtracting equation (2) from equation (1), we obtain
$$ ( x - y + 2) - (2x - y - 1 )= 0 $$
$$ x - y + 2 - 2x + y + 1 = 0 $$
$$ -x + 3 = 0 $$
$$ x = 3 $$
Substituting this value $ x = 3 $ in equation (1), we obtain
$$ { x - y + 2 = 0 }$$
$$ { 3 - y + 2 = 0 }$$
$$ y = 5 $$
Therefore , The answer is $ x= 3 , y = 5 $ and fraction = $ { 3 \over 5} $ ;
Form the pair of linear equations in the following problems, and find their Solutions (if they exist) by the elimination method:
(ii) 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 ?
Solution :
Let the Present age of Nuri is x years
And the present age of Sonu y years
5 years ago, age of Nuri = (x – 5) years
5 years ago, age of Sonu = (y – 5) years
According to the question
$$ x - 5 = 3(y - 5) $$
$$ x - 5 = 3y - 15 $$
$$ x - 3y = -15 + 5 $$
$${ x - 3y = -10 }......... (1) $$
After 10 years from now, age of Nuri = ( x + 10 ) years and Sonu = ( y + 10) years
According to given condition, we have
$$ x + 10 = 2(y + 10) $$
$$ x + 10 = 2y + 20 $$
$$ {x - 2y = 10 }....... (2)$$
By subtracting equation (2) from equation (1), we obtain
$$ ( x - 3y) - (x - 2y )= -10 - 10 $$
$$ -y = -20 $$
$$ y = 20 $$
Substituting this value $ y = 20 $ in equation (1), we obtain
$$ { x - 3y = -10 }$$
$$ { x - 3(20) = -10 }$$
$$ { x - 60 = -10 }$$
$$ { x = 50 }$$
Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years .
Form the pair of linear equations in the following problems, and find their Solutions (if they exist) by the elimination method:
(iii) 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.
Solution :
Let digit at ten’s place = x
and Let digit at one’s place = y
Then the number = 10x + y
According to the question
$$ { x + y = 9 }......... (1)$$
By reversing the order of the digits,the number = 10y + x
According to given condition, we have
$$ 9 (10x + y) = 2 (10y + x) $$
$$ ⇒ 90x + 9y = 20y + 2x $$
$$ ⇒ 90x + 9y- 20y - 2x =0 $$
$$ ⇒ 88x - 11y =0 $$
$$ ⇒ { 8x - y =0 }......... (2)$$
Adding equation (1) and equation (2), we obtain
$$ ⇒ ( x + y ) + (8x - y )= 9 $$
$$ ⇒{ x + 8x = 9 } $$
$$ ⇒{ 9x = 9 } $$
$$ ⇒{ x = 1 } $$
Substituting this value $ x = 1 $ in equation (1), we obtain
$$ { x + y = 9 }$$
$$ ⇒{ 1 + y = 9 }$$
$$ ⇒{ y = 8 }$$
Therefore, number = 10x + y = $ { (10 × 1) + 8 = 18 }$
Form the pair of linear equations in the following problems, and find their Solutions (if they exist) by the elimination method:
(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
Solution :
Let number of Rs 100 notes = x
and let number of Rs 50 notes = y
According to the question
$$ { x + y = 25 }......... (1)$$
According to given condition, we have
$$ { 100x + 50y = 2000 }$$
$$ ⇒{ 2x + y = 40 }........ (2)$$
By subtracting equation (2) from equation (1), we obtain
$$ ( 2x + y ) - (x + y )= 40 - 25 $$
$$ ⇒{ 2x + y -x - y = 15 } $$
$$ ⇒{ x = 15 } $$
Substituting this value $ x = 15 $ in equation (1), we obtain
$$ { x + y = 25 }$$
$$ ⇒{ 15 + y = 25 }$$
$$ ⇒{ y = 10 }$$
Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10
Form the pair of linear equations in the following problems, and find their Solutions (if they exist) by the elimination method:
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution :
Let the fixed charge for the first three days = Rs. x
And additional charge for each day = Rs.y
According to the question
Saritha paid Rs.27 for a book kept for 7 days (3 fixed days + additional 4 days)
$$ { x + 4y = 27 }......... (1)$$
According to given condition, we have
Susy paid Rs.21 for a book kept for 5 days (3 fixed days + additional 2 days)
$$ { x + 2y = 21 }......... (2)$$
By subtracting equation (2) from equation (1), we obtain
$$ ( x + 4y ) - (x + 2y )= 27 - 21 $$
$$ ⇒{ x + 4y - x - 2y = 6 } $$
$$ ⇒{ 2y = 6 } $$
$$ ⇒{ y = 3 } $$
Substituting this value $ y = 3 $ in equation (1), we obtain
$$ { x + 4y = 27 }$$
$$ ⇒{ x + (4 × 3) = 27 }$$
$$ ⇒{ x + 12 = 27 }$$
$$ ⇒{ x = 15 }$$
Therefore, fixed charge for 3 days = Rs 15 And additional charge for each day = Rs. 3
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