NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Ex 3.3

Exercise 3.3

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

(i) $\mathbf{x}+\mathbf{y}=\mathbf{5}$ and $\mathbf{2x}\mathbf{3y}=\mathbf{4}$ 

Ans: Elimination method

The given equations are:

$x+y=5$          …… (i)

$2x-3y=4$   …… (ii)

Multiplying equation (ii) by $2$, we get

$2x+2y=10$   …… (iii)

Subtracting equation (ii) from equation (iii), we obtain

$5y=6$

$y=\dfrac{6}{5}$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$x=5-\dfrac{6}{5}$

$x=\dfrac{19}{5}$

Therefore, $x=\dfrac{19}{5}$ and $y=\dfrac{6}{5}$.

Substitution method:

From equation (i) we get

$x=5-y$      …… (v)

Substituting (v) in equation (ii), we get

$2\left( 5-y \right)-3y=4$

$-5y=-6$

$y=\dfrac{6}{5}$            …… (vi)

Substituting (vi) in equation (v), we obtain

$x=5-\dfrac{6}{5}$

$x=\dfrac{19}{5}$

Therefore, $x=\dfrac{19}{5}$ and $y=\dfrac{6}{5}$.

(ii) $\mathbf{3x}+\mathbf{4y}=\mathbf{10}$ and $\mathbf{2x}\mathbf{2y}=\mathbf{2}$

Ans: Elimination method

The given equations are:

$3x+4y=10$          …… (i)

$2x-2y=2$   …… (ii)

Multiplying equation (ii) by $2$, we get

$4x-4y=4$   …… (iii)

Adding equation (ii) and (iii), we obtain

$7x=14$

$x=2$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$6+4y=10$

$4y=4$

$y=1$

Therefore, $x=2$ and $y=1$.

Substitution method:

From equation (ii) we get

$x=1+y$      …… (v)

Substituting (v) in equation (i), we get

$3\left( 1+y \right)+4y=10$

$7y=7$

$y=1$            …… (vi)

Substituting (vi) in equation (v), we obtain

$x=1+1$

$x=2$

Therefore, $x=2$ and $y=1$.

(iii) $\mathbf{3x}\mathbf{5y}\mathbf{4}=\mathbf{0}$ and $\mathbf{9x}=\mathbf{2y}+\mathbf{7}$

Ans: Elimination method

The given equations are:

$3x-5y-4=0$          …… (i)

$9x=2y+7$

$9x-2y=7$   …… (ii)

Multiplying equation (i) by $3$, we get

$9x-15y-12=0$   …… (iii)

Subtracting equation (iii) from equation (ii), we obtain

$13y=-5$

$y=-\dfrac{5}{13}$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$3x+\dfrac{25}{13}-4=0$

$3x=\dfrac{27}{13}$

$x=\dfrac{9}{13}$

Therefore, $x=\dfrac{9}{13}$ and $y=-\dfrac{5}{13}$.

Substitution method:

From equation (i) we get

$x=\dfrac{5y+4}{3}$      …… (v)

Substituting (v) in equation (ii), we get

$9\left( \dfrac{5y+4}{3} \right)-2y-7=0$

$13y=-5$

$y=\dfrac{-5}{13}$            …… (vi)

Substituting (vi) in equation (v), we obtain

$x=\dfrac{5\left( \dfrac{-5}{13} \right)+4}{3}$

$x=\dfrac{9}{13}$

Therefore, $x=\dfrac{9}{13}$ and $y=\dfrac{-5}{13}$.

(iv) $\dfrac{x}{2}+\dfrac{2y}{3}=-1$ and $x-\dfrac{y}{3}=3$

Ans: Elimination method

The given equations are:

$\dfrac{x}{2}+\dfrac{2y}{3}=-1$

$3x+4y=-6$          …… (i)

$x-\dfrac{y}{3}=3$

$3x-y=9$   …… (ii)

Subtracting equation (ii) from equation (i), we obtain

$5y=-15$

$y=-3$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$3x+4\left( -3 \right)=-6$

$3x=6$

$x=2$

Therefore, $x=2$ and $y=-3$.

Substitution method:

From equation (ii) we get

$x=\dfrac{y+9}{3}$      …… (v)

Substituting (v) in equation (i), we get

$3\left( \dfrac{y+9}{3} \right)+4y=-6$

$5y=-15$

$y=-3$            …… (vi)

Substituting (vi) in equation (v), we obtain

$x=\dfrac{-3+9}{3}$

$x=2$

Therefore, $x=2$ and $y=-3$.

2. 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 $\mathbf{1}$ to the numerator and subtract $\mathbf{1}$ from the denominator, a fraction reduces to $\mathbf{1}$. It becomes $\dfrac{\mathbf{1}}{2}$ if we only add $\mathbf{1}$ to the denominator. What is the fraction?

Ans: Assuming the fraction be $\dfrac{x}{y}$.

Writing the algebraic representation using the information given in the question:

$\dfrac{x+1}{y-1}=1$

$x-y=-2$             …… (i)

$\dfrac{x}{y+1}=1$

$2x-y=1$    …… (ii)

Subtracting equation (i) from equation (ii), we obtain

$x=3$              …… (iii)

Substituting the value of (iii) in equation (i), we get

$3-y=-2$

$y=5$

Therefore, $x=2$ and $y=-3$.

Hence the fraction is $\dfrac{3}{5}$.

(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?

Ans: Assuming the present age of Nuri be $x$ and the present age of Sonu be $y$.

Writing the algebraic representation using the information given in the question:

$\left( x-5 \right)=3\left( y-5 \right)$

$x-3y=-10$             …… (i)

$\left( x+10 \right)=2\left( y+10 \right)$

$x-2y=10$    …… (ii)

Subtracting equation (i) from equation (ii), we obtain

$y=20$              …… (iii)

Substituting the value of (iii) in equation (i), we get

$x-60=-10$

$x=50$

Therefore, $x=50$ and $y=20$.

Hence Nuri’s present age is $50$ years and Sonu’s present age is $20$ years.

(iii) The sum of the digits of a two-digit number is $\mathbf{9}$. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Ans: Assuming the unit digit of the number be $x$ and the tens digit be $y$.

Therefore, the number is $10y+x$

The number after reversing the digits is $10x+y$.

Writing the algebraic representation using the information given in the question:

$x+y=9$             …… (i)

$9\left( 10y+x \right)=2\left( 10x+y \right)$

$-x+8y=0$    …… (ii)

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

$9y=9$

$y=1$              …… (iii)

Substituting the value of (iii) in equation (i), we get

$x=8$

Therefore, $x=8$ and $y=1$.

Hence the number is $10y+x=18$.

(iv) Meena went to bank to withdraw $\mathbf{Rs}\text{ }\mathbf{2000}$. She asked the cashier to give her $\mathbf{Rs}\text{ }\mathbf{50}$ and $\mathbf{Rs}\text{ }\mathbf{100}$ notes only. Meena got $\mathbf{25}$ notes in all. Find how many notes of $\mathbf{Rs}\text{ }\mathbf{50}$ and $\mathbf{Rs}\text{ }\mathbf{100}$ she received.

Ans: Assuming the number of $Rs\text{ }50$ notes be $x$ and the number of $Rs\text{ }100$ be $y$.

Writing the algebraic representation using the information given in the question:

$x+y=25$             …… (i)

$50x+100y=2000$    …… (ii)

Multiplying equation (i) by $50$, we obtain

$50x+50y=1250$     …… (iii)

Subtracting equation (iii) from equation (ii), we obtain

$50y=750$

$y=15$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$x=10$

Therefore, $x=10$ and $y=15$.

Hence Meena has $10$ notes of $Rs\text{ }50$ and $15$ notes of $Rs\text{ }100$.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid $\mathbf{Rs}\text{ }\mathbf{27}$ for a book kept for seven days, while Susy paid $\mathbf{Rs}\text{ }\mathbf{21}$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Ans: Assuming that the charge for first three days is $Rs\text{ }x$ and the charge for each day thereafter is $Rs\text{ y}$.

Writing the algebraic representation using the information given in the question:

$x+4y=27$             …… (i)

$x+2y=21$    …… (ii)

Subtracting equation (ii) from equation (i), we obtain

$2y=6$

$y=3$     …… (iii)

Subtracting equation (iii) from equation (i), we obtain

$x+12=27$

$x=15$              …… (iv)

Therefore, $x=15$ and $y=3$.

Hence, fixed charges are $Rs\text{ 15}$ and charges per day are $Rs\text{ 3}$.

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3 - Free PDF Download

Get the Class 10 Maths Free NCERT Solutions Chapter 3 Exercise 3.3 PDF. NCERT Solutions are highly helpful while doing your homework. The experienced Teachers prepare all the answers for Class 10 Maths. Detailed answers to all the questions in the NCERT TextBook Chapter 3, Maths Class 10 Pair of Linear Equations in Two Variables Exercise 3.3. 

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1. Solving pair of linear equations by Elimination method in two variables

In the Elimination method, either add or subtract the given linear equations which eliminate a variable and results in getting an equation in one variable. 

• We can subtract the equations to eliminate one variable if the coefficients of any one of the variables are the same and the sign of the coefficients is the same.

• Similarly, if the coefficients of any one of the variables are the same, but the sign of the coefficients are opposite, then we can add the linear equations to get the desired linear equation in one variable.

Steps in Elimination Method

• Step 1 - Making the same coefficients: To make the coefficients of any one of the variables (either x or y) numerically equal, multiply both the given equations by some suitable non-zero constants.

• Step 2 - Adding or Subtracting the Linear equations: After step 1, add or subtract one equation from the other in a way that one variable will be eliminated. Now, if you resulted in a linear equation in one variable, directly go to Step 3. Else;

• If we obtain a true statement that includes no variable, then the original pair of equations will have infinitely many solutions.

• If we obtain a false statement that includes no variable (2 ≠ 3), then the original pair of equations will have no solution, i.e., it is inconsistent.

• Step 3 - Simplification: Simplify the equation in one variable (x or y) to get the variable value.

• Step 4 - Substitution: Finally, Substitute the obtained variable value in any one of the given (original) equations to get the value of another unknown variable.

2. Solving pair of linear equations by Substitution method in two variables

The substitution method of solving linear equations is an algebraic method to solve simultaneous linear equations. As the name says, in the substitution method, the value of one variable from one equation is substituted in the second equation. By doing so, we get a pair of the linear equations transformed into one linear equation with only one variable, which can be simply solved.

Steps in Substitution Method

• Step 1: Firstly, Simplify the given equations by expanding the parenthesis ( ).

• Step 2: Transform one of the given linear equations for either variable x or variable y.

• Step 3: Now, Substitute the variable value obtained in step 2 in the other equation.

• Step 4: Solve the new equation obtained using basic arithmetic operations to get the variable value.

• Step 5: Substitute the obtained variable value (from step 4) in any of the given equations and find out the value of the second variable.

Difference Between Substitution Method and Elimination method

As we discussed, the substitution method is the process of substituting and solving the equation to find the variable value, where the value of a variable is substituted in the other equation. And, the elimination method is the process of eliminating the variables in the equation so that the system of the equation can be with a single variable that can be easily solved.

We can conclude that the major difference between the Substitution and the Elimination method is that the substitution method involves replacing the variable with a value, whereas the elimination method involves removing the variable from the system of linear equations.

NCERT Solutions for Class 10 Maths Chapter 3 Exercises

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