## Algebraic solution of a pair of linear equations in two variables by Variable Elimination

In this 7 part series, we have covered the topic **Pair of Linear Equations in two variables**.

Specifically this session complements **Exercise 4 of NCERT Class 10 Maths Chapter 3**, in short, Ex 3.4 Class 10.

In this session now, you will learn **algebraic solution of a pair of linear equations in two variables by elimination.**

Each part contains three components:

**First:**concepts explained in an easy to understand way,**Second:**example problems that will be used for establishing the concepts more clearly and,**Third:**the solutions to the exercises in the NCERT Chapter.

In the first and second part of the series, graphical representation and graphical solutions of linear equations have been covered. And in the third part, algebraic method by substitution has been introduced*. You may read these through the following links.*

*NCERT solutions for class 10 maths linear equations 1 graphical representation,*

*NCERT solutions for class 10 maths linear equations 2 graphical solutions, and*

*NCERT solutions for class 10 maths linear equations 3 algebraic solutions by substitution.*

**Now in this fourth of the seven part series you'll,**

*Learn the***Basic concept behind the methods**of solving a pair of linear equations in two variables**algebraically using problem examples.***Learn the***algebraic method of Variable Elimination for solving two linear equations in two variables in details, and***Understand what happens when you apply the method to solve a pair of***equivalent linear equations**or a pair of linear equations that result in a**pair of parallel lines**in their geometric representation.

**Detailed solution of problems of Exercise 3.4 will follow the concepts and methods.**

**To Skip straight to the Solutions, click here.**

### Basic concept behind algebraic methods of solving a pair of linear equations in two variables

We'll start with a problem example and **reinforce the most basic concept** of solving a pair of linear equations by reducing two equations to one equation by eliminating one variable first.

#### Example problem 1.

Can you find algebraically the solution to the following problem pairs of linear equations?

- $2x+3y=8$ and $4x+5y=7$.
- $2x-y=20$ and $3x+5y=43$, and
- $3x+2y=5$ and $2x-3y=7$.

#### Solution to example problem 1 (a)—algebraic solution of a pair of linear equations in two variables by elimination—the most basic concept of solving linear equations

The two equations are,

$2x+3y=8$ and,

$4x+5y=7$.

You know that,

If you eliminate one variable from the two equations by a specific operation getting a

single equation in the second variable, algebraically it would always be possible to find the value of the second variable from the resultant single equation, if a solution exists.

Next substituting this value in any of the equations you will get the value of the first variable and hence the solution, **if it exists**.

You know by now the **reason behind the uncertainty of getting the unique solution (which is the main objective),**

If the two equations are

equivalent with ratio of coefficients of two variables and the constants equalor if thecorresponding lines are parallel with coefficients of two variables equal to each other but not equal to ratio of constants, youcan't get a unique solution of the two equations.

Taking up solving this problem again, **first thing** you do is to **judge the degree of similarity between the two pairs of coefficients** of $x$ and $y$ and **select the one that is easier to be made equal.**

By this exercise here you find that *if you multiply the first equation by 2, the coefficients of $x$ in both equations become equal to 4. This is easier to do than making the coefficients of $y$ same. *

*In second step take the action of multiplication. *The changed first equation and unchanged second equation are then,

$4x+6y=16$, and

$4x+5y=7$,

Now in the **third step**, it is straightforward subtraction of the second from the first to **eliminate one variable $x$** and you get,

$y=9$.

In the **fourth and last step**, substitute this value of $y$ in any of the two equations, say the first (as it is simpler) to get,

$2x+27=8$,

Or, $x=-\displaystyle\frac{19}{2}$.

**Solution is** $x=-\displaystyle\frac{19}{2}$, $y=9$.

**Verify the solution** by substituting these two values in both the equations.

The result of substitution in the first equation is,

$2x+3y=8$,

Or, $-19+27=8$,

Or, $27=27$, the equation is satisfied.

Similarly substituting in the second equation you get,

$4x+5y=7$,

Or, $-38+45=7$,

Or, $7=7$.

The second equation is also satisfied. So **you are sure that the solution is correct.**

Now look back, and **consider just the result of the subtraction from which you obtained the value of $y$.** It has been,

$y=9$.

This is **a linear equation in one variable.** An algebraic truth is,

You can always find the unique value of the variable from a linear equation in one variable.

This is the * most basic mechanism which is used invariably by all the algebraic methods* that you'll learn for solving two linear equations.

*In the methods, effectively one of the two variables is eliminated first, resulting in one linear equation in one variable. Value of the variable obtained from the single linear equation is then substituted in any of the equations in two variables to reduce it to a second equation in only the second variable.*

We may generalize this algebraic truth in stating,

For a unique solution, there must be$n$ number of distinct linear equations in $n$ number of variables.

The word **'distinct'** signifies,

No two equations in the set to be equivalent (with ratios of all three parameters equal) or ratio of coefficients of two variables equal (corresponding to two lines parallel).

We have already seen examples of these two special cases in our earlier session, when you couldn't get a unique solution of two linear equations in two variables.

The **distinctiveness of the set of equations is important** because say, if two equations of the set were equivalent, effectively number of equations reduces by 1 to $(n-1)$. **And with $(n-1)$ equations in $n$ variables you can't find a unique solution.**

For example, in a two equation set in two variables, if the equations are equivalent, effectively the number of equations reduces to 1 resulting in **infinitely many solutions.**

#### Solution to example problem 1 (b)—solution of a pair linear equations algebraically by Variable Elimination

In this example, we'll solve the two equations by **variable elimination**.

The two equations are,

$2x-y=20$ and,

$3x+5y=43$.

In the **first step** of *identifying the easier coefficient to equalize*, you identify $y$ as the variable. Just multiply first equation by 5 to make coefficients of $y$ in two equations equal. The situation by this multiplication is,

$10x-5y=100$, and

$3x+5y=43$.

To eliminate $y$, instead of subtraction add the two equations getting,

$13x=143$,

Or, $x=11$.

In the **second step** substitute this value of $x$ in the **first simpler equation** to get,

$2x-y=20$,

Or, $22-y=20$,

Or, $y=2$.

**Solution is,** $x=11$, and $y=2$.

#### Solution to example problem 1 (c)—a second example of algebraic solution of two linear equations in two variables by elimination

You will understand the difficulty in solving this pair of equations graphically once you solve it algebraically.

The limitation of graphical solution of a pair of linear equations lies in the fact that,

Finding the solution of a pair of linear equations where solution is with decimals or in fractional form is perfectly acceptable in algebraic methods, but may not be possible to place on a graph accurately (when intersection point coordinates involve decimal values).

The two equations we'll solve this time are,

$3x+2y=5$ and,

$2x-3y=7$.

In the **first step of identifying the coefficient easier to equalize,** you find **difficulty of equalizing as same for both coefficients**.

So select coefficient of any of the variables, say $y$ to equalize. But this time** you have to change both the equations**—multiply the first by 3 and the second by 2 in the second step. The result of these actions is,

$9x+6y=15$, and

$4x-6y=14$.

Add the two to eliminate $y$ in the **third step**,

$13x=29$,

Or, $x=\displaystyle\frac{29}{13}$.

In the **fourth and last step** substitute this value of $x$ in the second equation to get,

$\displaystyle\frac{58}{13}-3y=7$,

Or, $3y=-7+\displaystyle\frac{58}{13}=-\displaystyle\frac{33}{13}$,

Or, $y=-\displaystyle\frac{11}{13}$.

**Solution** is, $x=\displaystyle\frac{29}{13}$, and $y=-\displaystyle\frac{11}{13}$.

It is easy to find the solution in terms of fraction values algebraically. But *as in both values a prime number 13 is the divisor*, *both values are with infinite sequence of repeating decimals that do not terminate.*

**Note:** *Being rational fractions with prime number divisor*, both decimal sequences must repeat and be non-terminating. With non-repeating non-terminating decimal digit sequence, the values **would have been irrational.**

And you can see now that you won't be able to place this solution point accurately on the graph. You may write the fraction coordinate values after finding the values algebraically, but without algebraic method, only graphically you won't be able to find the solution values.

Simply speaking, graphical method of plotting linear equations on a graph and finding solution, visually gives you an idea of the situation, and thereby makes your concept clear.

But **in general**,

Only Algebraic methods are feasible in practice for solving two linear equations in two variables.

Also if you look back to the last session where we solved the problem by substitution method, you'll realize that, elimination method is simpler than substitution.

Don't take though whatever we say as the final word—**make your own judgment.**

Now we'll take up solving two problems to show you what happens if you try to solve a pair of linear equations in two variables that are inconsistent, and that result in a pair of coincident straight lines using elimination method.

#### Example Problem 2.

Find the solution of the following pair of linear equations algebraically.

- $6x-3y+10=0$ and $2x-y+9=0$
- $9x+3y+12=0$ and $18x+6y+24=0$

#### Solution to Example Problem 2 (a)—No solution as the corresponding two lines are parallel

We'll compare the ratios of coefficients of $x$ and $y$ and the constants of the two equations,

$6x-3y+10=0$ and,

$2x-y+9=0$.

The ratios are,

$\displaystyle\frac{a_1}{a_2}=\frac{6}{2}=3$,

$\displaystyle\frac{b_1}{b_2}=\frac{-3}{-1}=3$, and

$\displaystyle\frac{c_1}{c_2}=\frac{10}{9} \neq \displaystyle\frac{a_1}{a_2}$.

The *ratios of coefficients of $x$ and $y$ are equal to each other but not equal to the ratio of constants.*

So the slopes of the two lines are equal and the lines are parallel to each other. The pair of equations will have no solution and are inconsistent.

**Let's take up solving the problem algebraically using elimination method.**

Identify that coefficient easier to equalize belong to $y$.

Multiply second equation by 3,

$2x-y+9=0$,

Or, $6x-3y+27=0$.

Subtract it from the first equation and realize that instead of getting a equation in $x$ you have eliminated $x$ also. The result is,

Or, $10-27=0$, **an invalidity**.

The **two equations cannot have any pair of (x, y) values that would satisfy both the equations.**

#### Solution to Example problem 2 (b)—a pair of coincident lines with infinitely many solutions

The two equations are,

$9x+3y+12=0$ and,

$18x+6y+24=0$.

You can easily see that the ratios of the three parameters of the two equations have the same value $\displaystyle\frac{1}{2}$. The two lines are coincident with infinitely many solutions. The equations are equivalent.

To solve the equations algebraically using elimination method, identify variable $y$ to equalize its coefficients.

Multiply first equation by 2,

$18x+6y+24=0$.

It becomes exactly same as the second equation. If you subtract result will be,

$0=0$.

Effectively, equation is only one, $3x+y+4=0$ in minimal form (cancelling out all common factors of the parameters). Infinite many solutions satisfy this single linear equation.

Now we will take up the task of solving the problems in the Exercise 3.4 of Chapter 3.

### Problems to solve as in Exercise

#### Problem 1.

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

- $x+y=5$ and $2x-3y=4$
- $3x+4y=10$ and $2x-2y=2$
- $3x-5y-4=0$ and $9x=2y+7$
- $\displaystyle\frac{x}{2} + \displaystyle\frac{2y}{3}=-1$ and $x-\displaystyle\frac{y}{3}=3$

#### Problem 2.

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

- If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\displaystyle\frac{1}{2}$ if we add 1 to the denominator. What is the fraction?
- 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?
- 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.
- 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.
- 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 7 days, while Susy paid Rs.21 for a book she kept for 5 days. Find the fixed charge and the charge for each extra day.

We'll solve now the exercise problems.

This is the last component of the session.

### NCERT Class 10 Maths Solution to the problems in exercise 3.4

#### Problem 1.i.

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

- $x+y=5$ and $2x-3y=4$

#### Solution to Problem 1.i by Elimination method

The two equations are,

$x+y=5$ and,

$2x-3y=4$.

Multiply first equation by 3 to equalize the coefficient values of $y$ in the two equations. The result is,

$3x+3y=15$.

Add it to second equation to eliminate $y$. The result of addition is,

Or, $5x=19$,

Or, $x=\displaystyle\frac{19}{5}$.

Substitute value of $x=\displaystyle\frac{19}{5}$ into the first simpler equation and get,

$\displaystyle\frac{19}{5}+y=5$,

Or, $y=5-\displaystyle\frac{19}{5}=\displaystyle\frac{6}{5}$.

**Answer:** $x=\displaystyle\frac{19}{5}$, $y=\displaystyle\frac{6}{5}$.

**Verification:** Substitute values of $x$ and $y$ in the LHSs of both the equations to get respectively,

$\displaystyle\frac{19+6}{5}=\frac{25}{5}=5$, and

$2\times{\displaystyle\frac{19}{5}}-3\times{\displaystyle\frac{6}{5}}=\displaystyle\frac{38-18}{5}=4$.

The values thus satisfy both the equations and indeed is the correct solution to the equations.

Verification takes little time, but makes you confident that your solution is the correct one.

#### Solution to Problem 1.i by Substitution method

The equations are,

$x+y=5$ and

$2x-3y=4$.

Express $y$ in terms of $x$ from the first equation,

$y=5-x$.

Substitute this expression of $y$ in the second equation,

$2x-3\times{(5-x)}=4$,

Or, $5x=4+15=19$,

Or, $x=\displaystyle\frac{19}{5}$.

Substitute value of $x$ in expression for $y$,

$y=5-x=5-\displaystyle\frac{19}{5}=\displaystyle\frac{6}{5}$.

**Answer:** $x=\displaystyle\frac{19}{5}$, $y=\displaystyle\frac{6}{5}$.

#### Problem 1.ii.

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

- $3x+4y=10$ and $2x-2y=2$

#### Solution to Problem 1.ii by elimination

Multiply second equation by 2. Result is,

$3x+4y=10$, and

$4x-4y=4$.

Add the two to eliminate $y$. Result is,

$7x=14$,

Or, $x=2$.

Substitute this into the second equation to get,

$4-2y=2$,

Or, $y=1$.

**Answer:** $x=2$, and $y=1$.

**Verification:** Substitute the values of $x$ and $y$ in the LHSs of the two equations to get respectively,

$6+4=10$, and

$4-2=2$.

The values satisfy both the equations and so is the correct solution. Verification increases your confidence on the solution.

#### Solution to Problem 1.ii by Substitution

The two equations are,

$3x+4y=10$ and,

$2x-2y=2$.

Reduce second equation by eliminating common factor 2,

$x-y=1$.

Express $y$ in terms of $x$,

$y=x-1$.

Substitute this in first equation,

$3x+4(x-1)=10$,

Or, $7x=14$,

Or, $x=2$.

Substitute in expression of $y$,

$y=x-1=2-1=1$.

**Answer:** $x=2$, $y=1$.

#### Problem 1.iii.

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

- $3x-5y-4=0$ and $9x=2y+7$

#### Solution to Problem 1.iii.

Multiply the first equation by 3 to equalize the coefficients of $x$ in the two equations to 9. The result is,

$9x-15y-12=0$, and

$9x-2y-7=0$, expressing the second equation in the same form as the first one.

Subtract the second equation from the first. Result is,

$-13y-5=0$,

Or, $y=-\displaystyle\frac{5}{13}$.

Substitute this value of $y$ in the second equation to get,

$9x=-2\times{\displaystyle\frac{5}{13}}+7=\displaystyle\frac{81}{13}$,

Or, $x=\displaystyle\frac{9}{13}$.

**Answer:** $x=\displaystyle\frac{9}{13}$ and $y=-\displaystyle\frac{5}{13}$.

**Verification:** Substitute these values of $x$ and $y$ in the two equations. The results are,

$3\times{\displaystyle\frac{9}{13}}+5\times{\displaystyle\frac{5}{13}}-4=\displaystyle\frac{52}{13}-4=4-4=0$, thus satisfying the first equation.

LHS of the second equation is,

$9\times{\displaystyle\frac{9}{13}}=\displaystyle\frac{81}{13}$.

RHS of the second equation is,

$-2\times{\displaystyle\frac{5}{13}}+7=\displaystyle\frac{81}{13}=\text{LHS}$.

Both equations being satisfied you can be sure about the correctness of the solution values.

#### Solution to Problem 1.iii by Substitution

The two equations are,

$3x-5y-4=0$ and,

$9x=2y+7$.

Express $y$ in terms of $x$ from the first equation,

$y=\displaystyle\frac{3x-4}{5}$.

Substitute this expression for $y$ in second equation,

$9x=2\times{\displaystyle\frac{3x-4}{5}}+7$,

Or, $9x-\displaystyle\frac{6x}{5}=7-\displaystyle\frac{8}{5}$,

Or, $39x=27$,

Or, $x=\displaystyle\frac{9}{13}$.

Substitute this value of $x$ in expression for $y$,

$y=\displaystyle\frac{3\times{\frac{9}{13}}-4}{5}=-\frac{5}{13}$.

**Answer:** $x=\displaystyle\frac{9}{13}$ and $y=-\displaystyle\frac{5}{13}$.

#### Problem 1.iv

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

- $\displaystyle\frac{x}{2} + \displaystyle\frac{2y}{3}=-1$ and $x-\displaystyle\frac{y}{3}=3$

#### Solution to Problem 1.iv by Elimination

First multiply first equation by 6 and second by 3 to eliminate the fraction denominators. The results are,

$3x+4y=-6$, and

$3x-y=9$.

As coefficients of $x$ are same in two equations, subtract second from first to eliminate $x$. Result is,

$5y=-15$,

Or, $y=-3$.

Substitute value of $y$ in the first equation. Result is,

$3x-12=-6$,

Or, $3x=6$,

Or, $x=2$.

**Answer:** $x=2$, $y=-3$.

**Verification:** Substitute these two values to the LHSs of the first and second equation respectively to get,

$6-12=-6$, and

$6+3=9$.

The $x$ and $y$ values of 2 and -3 satisfy both the equations and indeed is the correct solution for the two linear equations.

#### Solution to Problem 1.iv by Substitution

The two equations are,

$\displaystyle\frac{x}{2} + \displaystyle\frac{2y}{3}=-1$ and,

$x-\displaystyle\frac{y}{3}=3$.

Eliminate the fraction denominators of the coefficients of both equations by multiplying the first equation by 6, the LCM of fraction denominators 3 and 2 and the second equation by 3,

First equation becomes,

$3x+4y=-6$.

The second equation becomes,

$3x-y=9$.

Or, $y=3x-9$.

Substitute this expression in $y$ into the first equation to get,

$3x+4(3x-9)=-6$,

Or, $15x=30$,

Or, $x=2$.

Substitute in expression for $y$,

$y=3x-9=6-9=-3$.

**Answer:** $x=2$, $y=-3$.

#### Problem 2.

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

- If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\displaystyle\frac{1}{2}$ if we add 1 to the denominator. What is the fraction?
- 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?
- 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.
- 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.
- 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 7 days, while Susy paid Rs.21 for a book she kept for 5 days. Find the fixed charge and the charge for each extra day.

#### Solution to Problem 2.i.

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

- If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\displaystyle\frac{1}{2}$ if we add 1 to the denominator. What is the fraction?

Let the numerator of the fraction be $x$ and the denominator $y$.

By the first statement, the equation formed is,

$\displaystyle\frac{x+1}{y-1}=1$,

Or, $x=y-2$.

This is the first equation.

In the same way by the second statement,

$\displaystyle\frac{x}{y+1}=\frac{1}{2}$,

Or, $2x=y+1$.

This is the second equation.

Multiply the first equation by 2 and subtract the second equation from the first. The result is,

$y=5$.

Substitute it to first equation and get,

$x=5-2=3$.

**Answer:** The fraction is $\displaystyle\frac{3}{5}$.

**Verification:** Substitute $x=3$ and $y=5$ in the two equations to get,

$3=5-2=3$, and

$6=5+1=6$.

Both equations are satisfied by the solution values. The solution is correct.

#### Solution to Problem 2.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?

Assume age of Nuri as $x$ years and age of Sonu to be $y$ years.

Five years ago, by the first statement,

$x-5=3(y-5)$,

Or, $x-3y=-10$.

By the second statement, ten years later,

$x+10=2(y+10)=2y+20$,

Or, $x-2y=10$.

Subtract first equation from the second to eliminate $x$. Result is,

$y=20$.

Substitute this value of $y$ into the second equation. Result is,

$x=50$.

**Answer:** Present age of Nuri is 50 years and that of Sonu is 20 years.

Five years ago age of Nuri was 45 years old and Sonu was 15 years old, one-third the age of Nuri.

Ten years later, Nuri will be 60 and Sonu 30, half as old as Nuri. Solution is verified.

#### Solution to Problem 2.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.

Let $x$ and $y$ bet the ten's and one's digits respectively so that the number is,

$10x+y$.

By the first statement,

$x+y=9$.

By the second statement,

$9(10x+y)=2(10y+x)$,

Or, $88x=11y$.

Or, $8x-y=0$.

Add this equation to the first equation to eliminate $y$. Result is,

$9x=9$,

Or, $x=1$.

So, $y=8x=8$.

**Answer:** The number is 18.

Both the conditions are satisfied. Check for yourself.

#### Solution to Problem 2.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.

Assume number of 100 rupee notes to be $x$ and number of 50 rupees note to be $y$.

As the Meena got 25 notes in total,

$x+y=25$.

And as the total money she got was Rs.2000,

$100x+50y=2000$.

Or, $2x+y=40$.

Subtract the first equation from the second to eliminate $y$. Result is,

$x=15$.

Substitute this value in the first equation and get,

$y=25-15=10$.

**Answer:** Meena got 15 numbers of 100 rupee notes totaling Rs.1500 and 10 numbers of 50 rupee notes totaling Rs.500, together Rs.2000. Both conditions verified.

#### Solution to Problem 2.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 7 days, while Susy paid Rs.21 for a book she kept for 5 days. Find the fixed charge and the charge for each extra day.

Assume the fixed charge for first three days to be $x$ and charge for each additional day after first three days to be $y$, both in rupees.

Payment of Rs.27 by Saritha is represented by the equation,

$x+4y=27$, extra number of days being $7-3=4$.

This is the first linear equation.

And payment of Rs.21 by Susy is represented by the second equation,

$x+2y=21$, extra number of days being $5-3=2$.

Subtract the second from the first to eliminate $x$ and get,

$2y=6$,

Or, $y=3$.

With this value of $y$, get the value of $x$ from the second equation as,

$x=21-2y=21-6=15$.

**Answer: **Fixed charge is Rs.15 and extra day charge is Rs.3 per day.

In the fifth part next, we'll learn the popular cross-multiplication method of solving a pair of linear equations in two variables algebraically.

### NCERT Solutions for Class 10 Maths

#### Chapter 1: Real Numbers

**NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions**

#### Chapter 2: Polynomials

#### Chapter 3: Linear Equations

**NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection**

**NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.**

#### Chapter 4: Quadratic equations

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 3 Solution by Completing the square**

#### Chapter 6: Triangles

**NCERT solutions for class 10 maths chapter 6 Triangles 1 Similarity of Triangles and Polygons**

**Solutions to Exercise 2 Chapter 6 NCERT X Maths, Characteristics of Similar triangles**

#### Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios*

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities*

#### Chapter 8: Introduction to Trigonometry, only solutions to selected problems

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1**