Fast, Efficient and Scalable Solutions
Explore the comprehensive NCERT Textbook Solutions for Class X.
$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 substitution method.
(i) x + y = 14
x − y = 4.
Solution :
$$ x + y = 14 .... (1) $$
$$ x - y = 4 ..... (2) $$
By solving the equation (1) Isolate one variable in one of the equations:
$$ y = 14 - x .... (3) $$
Substitute the expression $ y = 14 - x $ into the other equation (2), we obtain
$$⇒ x - (14 - x) = 4 $$
$$⇒ x + x = 4 + 14 $$
$$⇒ 2x = 18 $$
$$⇒ x = 9 $$
Substituting this value $ x = 9 $ in equation (3), we obtain
$$⇒ y = 14 - 9 $$
$$⇒ y = 5 $$
The solution to the pair of linear equations is x = 9 and y = 5.
Solve the following pair of linear equations by the substitution method.
(ii) $$ s − t = 3 $$ $${ s \over 3} + { t \over 2} = 6 $$
Solution :
$$ s − t = 3 .... (1) $$
$$ { s \over 3} + { t \over 2} = 6 ..... (2) $$
By solving the equation (1) Isolate one variable in one of the equations:
$$ s = 3 + t.... (3) $$
Substitute the expression $ s = 3 + t $ into the other equation (2), we obtain
$$⇒ {3 + t \over 3} + { t \over 2} = 6 $$
$$⇒ {6 + 2t + 3t \over 6} = 6 $$
$$⇒ {6 + 2t + 3t } = 36 $$
$$⇒ 5t = 30 $$
$$⇒ t = 6 $$
Substituting this value $ t = 6 $ in equation (3), we obtain
$$⇒ s = 3 + 6 $$
$$⇒ s = 9 $$
The solution to the pair of linear equations is s = 9 and t = 6.
Solve the following pair of linear equations by the substitution method.
(iii)$$ 3x - y = 3 $$ $$ 9x -3y = 9 $$
Solution :
$$ 3x - y = 3 .... (1) $$
$$ 9x -3y = 9 ..... (2) $$
By solving the equation (1) Isolate one variable in one of the equations:
$$⇒ y = 3x - 3 .... (3) $$
Substitute the expression $ y = 3x - 3 $ into the other equation (2), we obtain
$$⇒ 9x -3(3x - 3) = 9 $$
$$⇒ 9x - 9x + 9 = 9 $$
$⇒$ 9 = 9 $$
This indicates that the two equations are identical.
Since the equations are the same, any pair of values for (x,y) that satisfies the first equation will also satisfy the second. Therefore, there are an infinite number of solutions.
Solve the following pair of linear equations by the substitution method.
(iv) $$ 0.2x + 0.3y = 1.3 $$ $$ 0.4x + 0.5y = 2.3 $$
Solution :
$$ 0.2x + 0.3y = 1.3 .... (1) $$
$$ 0.4x + 0.5y = 2.3 ..... (2) $$
Multiply both the equations (1) and (2) by 10, to remove the decimal number and making it easier for calculation.
$$ 2x + 3y = 13 .... (3) $$
$$ 4x + 5y = 23 ..... (4) $$
By solving the equation (3) :
$$⇒ y = {{13 - 2x } \over 3} ..... (5)$$
Substitute the expression $ y = {{13 - 2x } \over 3} $ into the other equation (4), we obtain
$$⇒ 4x + 5({{13 - 2x } \over 3}) = 23 $$
$$⇒ {12x + 65 - 10x \over 3} = 23 $$
$$⇒ {2x + 65 } = 23 × 3 $$
$$⇒ 2x = 69- 65 $$
$$⇒ x = 2 $$
Substituting this value $ x = 2 $ in equation (5), we obtain
$$ y = {{13 - 2 × 2 } \over 3} $$
$$⇒ y = { 9 \over 3}$$
$$⇒ y = 3 $$
Therefore, x = 2 and y = 3.
Solve the following pair of linear equations by the substitution method.
(v) $${ \sqrt 2x + \sqrt 3y = 0 }$$ $${ \sqrt 3x - \sqrt 8y = 0 }$$
Solution :
$${ \sqrt 2x + \sqrt 3y = 0..... (1)} $$
$${ \sqrt 3x - \sqrt 8y = 0 ..... (2)} $$
By solving the equation (1) :
$$⇒ { \sqrt 2x + \sqrt 3y = 0} $$
$$⇒{ \sqrt 2x = - \sqrt 3y} $$
$$⇒ x = {{- \sqrt 3y} \over \sqrt 2} ..... (3) $$
Substitute the expression $ x = {{- \sqrt 3y} \over \sqrt 2} $ into the other equation (2), we obtain
$$⇒{ \sqrt 3 ({{- \sqrt 3y} \over \sqrt 2}) - \sqrt 8y = 0 } $$
$$⇒{{ - 3y \over \sqrt 2} - \sqrt 8y = 0 } $$
$$⇒{{ - 3y \over \sqrt 2} - 2\sqrt 2y = 0 } $$
$$⇒{{ - 3y - \sqrt 2 × 2\sqrt 2y \over \sqrt 2 }= 0 } $$
$$⇒{{ - 3y - 4y \over \sqrt 2 }= 0 } $$
$$⇒{{ - 7y \over \sqrt 2 }= 0 } $$
$$⇒{ y = {0 \over -7}} $$
$$⇒ y = 0 $$
Substituting this value $ y = 0 $ in equation (3), we obtain
$$⇒ x = {{- \sqrt 3} × 0 \over \sqrt 2}$$
$$⇒ x = 0 $$
Therefore, x= 0 and y= 0.
Solve the following pair of linear equations by the substitution method.
(vi)$$ {{3x \over 2}- {5y \over 3} =-2 }$$ $$ {{x \over 3}+ {y \over 2} = {13 \over 6}}$$
Solution :
$$ {{3 \over 2}x- {5y \over 3} =-2 } .... (1) $$
$$ {{x \over 3}+ {y \over 2} = {13 \over 6}}..... (2) $$
By solving the equation (2) :
$$ {{x \over 3}+ {y \over 2} = {13 \over 6}} $$
$$⇒ {{x \over 3} = {13 \over 6}- {y \over 2}} $$
$$⇒ {{x \over 3} = {13 - 3y \over 6}} $$
$$⇒ {{x } = 3({13 - 3y \over 6})} $$
$$⇒ {{x } = {13 - 3y \over 2}} ..... (3)$$
Substitute the expression $ {{x } = {13 - 3y \over 2}} $ into the other equation (1), we obtain
$$ {{3x \over 2}- {5y \over 3} =-2 } $$
$$ ⇒ {{3 \over 2}({13 - 3y \over 2})- {5y \over 3} =-2 } $$
$$⇒ {{39 - 9y \over 4}- {5y \over 3} =-2 } $$
$$⇒ {{117 - 27y - 20y \over 12} =-2 } $$
$$⇒ {{117 - 47y } = -24 } $$
$$⇒ 117 + 24 = 47y $$
$$⇒ 141 = 47y $$
$$⇒ 3 = y $$
Substituting this value $ y = 3 $ in equation (3), we obtain
$$ x = {13 - 3 × 3 \over 2}$$
$$⇒ x = {4 \over 2} $$
$$⇒ x = 2 $$
Therefore, x= 2 and y= 3
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which $$ y = mx + 3 $$
Solution :
$$ {2x + 3y = 11 } .... (1) $$
$$ {2x – 4y = – 24}..... (2) $$
By solving the equation (1) :
$$ {{2x + 3y = 11 }} $$
$$ {{2x = 11 - 3y }} $$
$$ {x ={ {11 - 3y }\over 2}} ..... (3)$$
Substitute the expression $ {x ={ {11 - 3y }\over 2}} $ in equation (2), we obtain
$$ {2x – 4y = – 24} $$
$$ {2({ {11 - 3y }\over 2}) – 4y = – 24} $$
$$ { 11 -3y – 4y = – 24} $$
$$ { – 7y = – 24 -11} $$
$$ { – y = – {35\over 7}} $$
$$ { y = 5} $$
Substituting this value $ y = 5 $ in equation (3), we obtain
$$ {x ={ {11 - (3 × 5) }\over 2}}$$
$$ x = {-4 \over 2} $$
$$ x = -2 $$
Therefore, x = -2, y = 5
Now, y = mx + 3
$$ 5 = m(-2) + 3 $$
$$ 5 = – 2m + 3 $$
$$ 5 – 3 = – 2m $$
$$ 2 = – 2m $$
$$ -{2\over 2} = m $$
$$ ∴ m = - 1 $$
Therefore, x = -2, y = 5 and m = -1.
Form the pair of linear equations for the following problems and find their solution by substitution method :
(i)The difference between two numbers is 26 and one number is three times the other. Find them.
Solution :
Let the (larger) number = 'x' and another number =‘y’.
According to the question
Their difference is 26.
$$ x - y = 26 .... (1) $$
One number is three times the other,
$$ x = 3y ..... (2) $$
Substituting this value $ x = 3y $ in equation (1), we obtain
$$⇒ 3y - y = 26 $$
$$⇒ 2y = 26 $$
$$⇒ y = 13 $$
Substituting the value of ‘y’ in equation (2),
$$ x = 3y $$
$$⇒ x = 3(13) $$
$$⇒ x = 39 $$
Therefore, The two numbers are 39 and 13.
Form the pair of linear equations for the following problems and find their solution by substitution method :
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution :
In two supplementary angles, let one larger angle = 'x°' and another angle =‘y°’.
Since the angles are supplementary.
$$ x + y = 180° .... (1) $$
The larger of two supplementary angles exceeds the smaller by 18 degrees.
$$ x – y = 18° ..... (2) $$
By solving the equation (2)
$$ x = 18° + y ..... (3)$$
Substituting this value $ x = 18° + y $ in equation (1), we obtain
$$ x + y = 180° $$
$$⇒ 18° + y + y = 180° $$
$$⇒ 2y = 162° $$
$$⇒ y = 81° $$
Substitute the value of y back into Equation 3 to find x:,
$$ x = 18° + y $$
$$⇒ x = 18° + 81° $$
$$⇒ x = 99° $$
Therefore, The two supplementary angles are 99° and 81°.
Form the pair of linear equations for the following problems and find their solution by substitution method :
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
Solution :
Let the cost of each bat be Rs. x
Cost of each ball be Rs. y
According to the question
$$ 7x + 6y = 3800 .... (1) $$
$$ 3x + 5y = 1750 ..... (2) $$
By solving the equation (2)
$$ 3x + 5y = 1750 $$
$$ ⇒ 3x = 1750 - 5y $$
$$⇒ x = {{1750 - 5y} \over 3 }..... (3) $$
Substituting this value $ x = {{1750 - 5y} \over 3 } $ in equation (1), we obtain
$$ 7x + 6y = 3800 $$
$$⇒ 7({{1750 - 5y} \over 3 }) + 6y = 3800 $$
$$⇒ {{12250 - 35y} \over 3 } + 6y = 3800 $$
$$⇒ {{12250 - 35y + 18y } \over 3 } = 3800 $$
$$⇒ {12250 - 17y } = 11400 $$
$$⇒ { - 17y } = 11400 - 12250 $$
$$ - 17y = - 850 $$
$$⇒ y = {850 \over 17 } $$
$$⇒ y = 50 $$
Substitute the value of y back into Equation 3 to find x:,
$$ x = {{1750 - 5y} \over 3 } $$
$$⇒ x = {{1750 - (5 × 50)} \over 3 } $$
$$⇒ x = {1500 \over 3 } $$
$$⇒ x = 500 $$
Therefore, The cost of a bat is Rs 500 and cost of a ball is Rs 50.
Form the pair of linear equations for the following problems and find their solution by substitution method :
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution :
Let the fixed charge of the taxi be Rs. x.
And variable cost per km = Rs. y
Total cost = fixed charge + variable charge
According to the question
Charge for a distance of 10 km
$$ x + 10y = 105 .... (1) $$
Charge for a distance of 15 km
$$ x + 15y = 155 ..... (2) $$
By solving the equation (1)
$$ x + 10y = 105 $$
$$ x = 105 - 10y ..... (3)$$
Substituting this value $ x = 105 - 10y $ in equation (2), we obtain
$$ x + 15y = 155 $$
$$⇒ 105 - 10y + 15y = 155 $$
$$⇒ 5y = 50 $$
$$⇒ y = {50 \over 5 }$$
$$⇒ y = 10 $$
Substitute the value of y back into Equation 3 to find x:,
$$ x = 105 - 10y $$
$$⇒ x = 105 - (10 × 10) $$
$$⇒ x = 5 $$
Now, charge for a distance of 25 km
$$⇒ = x + 25y $$
$$⇒ = 5 + (25 × 10) $$
$$⇒ = 255 $$
∴ Fixed Charge = x= Rs. 5.
Variable Charge for each km = y = Rs. 10
Therefore, Charge for 25km = Rs. 255
Form the pair of linear equations for the following problems and find their solution by substitution method :
(v) A fraction becomes $ 9 \over 11 $. if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes $ 5 \over 6 $. Find the fraction
Solution :
Let If the Numerator is x
And denominator is y
then the Fraction is $ x \over y $
According to the question
Adding 2 to both Numerator and denominator
$$ {x + 2 \over y + 2 } = { 9 \over 11} $$
$$⇒ { 11(x + 2 )} = {9 (y + 2) } $$
$$⇒ 11x + 22 = 9y + 18 $$
$$⇒ 11x - 9y + 22 - 18 = 0 $$
$$⇒ 11x - 9y + 4 = 0 ..... (1)$$
If 3 is added to both Numerator and denominator,
$$ {x + 3 \over y + 3 } = { 5 \over 6} $$
$$⇒ { 6(x + 3 )} = {5 (y + 3) } $$
$$⇒ 6x + 18 = 5y + 15 $$
$$⇒ 6x - 5y + 18 - 15 = 0 $$
$$⇒ 6x - 5y + 3 = 0 .... (2)$$
By solving the equation (1)
$$ 11x - 9y + 4 = 0 $$
$$⇒ 11x = 9y - 4 $$
$$ x = { ( 9y - 4 )\over 11} .... (3)$$
Substituting this value $ x = { ( 9y - 4 )\over 11} $ in equation (2), we obtain :
$$ 6x - 5y + 3 = 0 $$
$$ 6 × { ( 9y - 4 )\over 11} - 5y + 3 = 0 $$
$$ { 54y - 24 \over 11} - 5y + 3 = 0 $$
$${ 54y - 24 - 55y + 33\over 11}= 0 $$
$$ -y + 9 = 0 $$
$$y = 9 $$
Substitute the value of y back into Equation 3 to find x :
$$ x = { ( 9y - 4 )\over 11} $$
$$ x = { (9 × 9) - 4 \over 11} $$
$$ x = { 77 \over 11} $$
$$ x = 7 $$
Therefore, Fraction = ${ x \over y} = { 7 \over 9} $
Form the pair of linear equations for the following problems and find their solution by substitution method :
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution :
Let the present age of Jacob be ‘x’ years
And his son’s present age is 'y' years
After 5 years, Age of Jacob will be x+5 years
And After 5 years, Age of his son will be y+5 years
According to the question
At that time Jacob’s age will be three times that of his son.
$$ {x + 5 = 3(y + 5)} $$
$$ {x + 5 = 3y + 15} $$
$$ {x - 3y = 15 - 5}$$
$$ {x - 3y = 10}.... (1)$$
5 years ago
Jacob’s age wase x-5 years
His son’s age was y-5 years
According to the question
At that time Jacob’s age was seven times that of his son.
$$ {x - 5 = 7(y - 5)} $$
$$ {x - 5 = 7y - 35} $$
$$ {x - 7y = -35 + 5}$$
$$ {x - 7y = -30}.... (2)$$
By solving the equation (2)
$$ { x - 7y = -30 } $$
$$ { x = -30 + 7y } .... (3) $$
Substituting this value $ x = -30 + 7y $ in equation (1), we obtain :
$$ {x - 3y = 10} $$
$$ { -30 + 7y - 3y = 10} $$
$$ { -30 + 4y = 10} $$
$$ { 4y = 10 + 30} $$
$$ { y = {40\over 4}} $$
$$ { y = 10} $$
Substitute the value of y back into Equation 3 to find x :
$$ { x = -30 + 7y } $$
$$ { x = -30 + (7 × 10) } $$
$$ { x = 40 } $$
∴ Present age of Jacob is 40.
Present age of Jacob’s son is 10.
Advanced courses and exam preparation.
Advanced courses and exam preparation.
Explore programming, data science, and AI.