NCERT Class 10 Maths Solutions: Pair of Linear Equations in Two Variables

Question:

On comparing the ratios $𝐚𝟏𝐚𝟐,𝐛𝟏𝐛𝟐$ and $𝐜𝟏𝐜𝟐$, find out whether the following pair of linear equations are consistent, or inconsistent.

3x + 2y = 5 ; 2x – 3y = 7

Concept in a Minute:

To determine if a pair of linear equations is consistent or inconsistent, we compare the ratios of the coefficients of x, y, and the constant terms.
For two linear equations in the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
1. If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the system has a unique solution and is consistent.
2. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the system has infinitely many solutions and is consistent.
3. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the system has no solution and is inconsistent.
Consistency means there is at least one solution. Inconsistency means there are no solutions.

Explanation:

The given pair of linear equations is:
Equation 1: 3x + 2y = 5
Equation 2: 2x – 3y = 7

First, let’s rewrite the equations in the standard form $ax + by + c = 0$:
Equation 1: 3x + 2y – 5 = 0
Equation 2: 2x – 3y – 7 = 0

Now, we identify the coefficients:
For Equation 1: $a_1 = 3$, $b_1 = 2$, $c_1 = -5$
For Equation 2: $a_2 = 2$, $b_2 = -3$, $c_2 = -7$

Next, we calculate the ratios of these coefficients:
Ratio of x-coefficients: $\frac{a_1}{a_2} = \frac{3}{2}$
Ratio of y-coefficients: $\frac{b_1}{b_2} = \frac{2}{-3}$
Ratio of constant terms: $\frac{c_1}{c_2} = \frac{-5}{-7} = \frac{5}{7}$

Now, we compare these ratios:
Is $\frac{a_1}{a_2} = \frac{b_1}{b_2}$?
$\frac{3}{2}$ compared to $\frac{2}{-3}$.
Clearly, $\frac{3}{2} \neq \frac{2}{-3}$.

Since the ratio of the x-coefficients is not equal to the ratio of the y-coefficients ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$), the system of linear equations has a unique solution. A system with at least one solution is considered consistent.

Therefore, the given pair of linear equations is consistent.

Question:

On comparing the ratios $𝐚𝟏𝐚𝟐,𝐛𝟏𝐛𝟐$ and $𝐜𝟏𝐜𝟐$, find out whether the following pair of linear equations are consistent, or inconsistent.

$32⁢𝑥+53⁢𝑦=7$ ; 9x – 10y = 14

Concept in a Minute:

To determine if a pair of linear equations is consistent or inconsistent, we compare the ratios of the coefficients of x, y, and the constant terms.
For two linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:
1. If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a unique point, and the system is consistent (having a unique solution).
2. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident, and the system is consistent (having infinitely many solutions).
3. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel, and the system is inconsistent (having no solution).

Explanation:

The given pair of linear equations is:
Equation 1: $32x + 53y = 7$
Equation 2: $9x – 10y = 14$

We need to compare the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$.

From Equation 1, we have $a_1 = 32$, $b_1 = 53$, and $c_1 = 7$.
From Equation 2, we have $a_2 = 9$, $b_2 = -10$, and $c_2 = 14$.

Now, let’s calculate the ratios:
Ratio of x-coefficients: $\frac{a_1}{a_2} = \frac{32}{9}$
Ratio of y-coefficients: $\frac{b_1}{b_2} = \frac{53}{-10}$
Ratio of constant terms: $\frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2}$

Comparing the first two ratios:
Is $\frac{32}{9} = \frac{53}{-10}$?
To check this, we can cross-multiply:
$32 \times (-10) = -320$
$9 \times 53 = 477$
Since $-320 \neq 477$, we have $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

According to the concept, when $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair of linear equations represents intersecting lines, which means there is a unique solution. A system with a solution (unique or infinite) is called consistent.

Therefore, the given pair of linear equations is consistent.

Question:

On comparing the ratios $𝐚𝟏𝐚𝟐,𝐛𝟏𝐛𝟐$ and $𝐜𝟏𝐜𝟐$, find out whether the following pair of linear equations are consistent, or inconsistent.

$43⁢𝑥+2⁢𝑦$ = 8; 2x + 3y = 12

Concept in a Minute:

To determine if a pair of linear equations is consistent or inconsistent, we compare the ratios of the coefficients of x, y, and the constant terms.
For two linear equations $a₁x + b₁y + c₁ = 0$ and $a₂x + b₂y + c₂ = 0$:
1. If $a₁/a₂ ≠ b₁/b₂$, the lines intersect at a unique point, meaning the system is consistent and has a unique solution.
2. If $a₁/a₂ = b₁/b₂ = c₁/c₂$, the lines are coincident, meaning the system is consistent and has infinitely many solutions.
3. If $a₁/a₂ = b₁/b₂ ≠ c₁/c₂$, the lines are parallel, meaning the system is inconsistent and has no solution.
Consistency means there is at least one solution (either unique or infinite). Inconsistency means there is no solution.

Explanation:

The given pair of linear equations are:
Equation 1: $43x + 2y = 8$
Equation 2: $2x + 3y = 12$

To compare the ratios, we first rewrite the equations in the standard form $ax + by + c = 0$:
Equation 1: $43x + 2y – 8 = 0$
Equation 2: $2x + 3y – 12 = 0$

Now, we identify the coefficients:
For Equation 1: $a₁ = 43$, $b₁ = 2$, $c₁ = -8$
For Equation 2: $a₂ = 2$, $b₂ = 3$, $c₂ = -12$

Next, we calculate the ratios:
Ratio of x coefficients: $a₁/a₂ = 43/2$
Ratio of y coefficients: $b₁/b₂ = 2/3$
Ratio of constant terms: $c₁/c₂ = -8/-12 = 8/12 = 2/3$

Now, we compare these ratios:
$a₁/a₂ = 43/2$
$b₁/b₂ = 2/3$
$c₁/c₂ = 2/3$

We observe that $a₁/a₂ ≠ b₁/b₂$ because $43/2 ≠ 2/3$.

According to the concept of comparing ratios for linear equations, if $a₁/a₂ ≠ b₁/b₂$, the pair of linear equations represents intersecting lines, which means there is a unique solution. A system with at least one solution is called consistent.

Therefore, the given pair of linear equations is consistent.

Question:

Solve the following pair of linear equation by the elimination method and the substitution method:

x + y = 5 and 2x – 3y = 4

Concept in a Minute:

The question asks to solve a system of two linear equations with two variables using two different methods: elimination and substitution. Both methods aim to find the unique values of x and y that satisfy both equations simultaneously.

Elimination method: Involves multiplying one or both equations by constants so that the coefficients of one variable are opposites. Then, adding the equations eliminates that variable, allowing you to solve for the other.

Substitution method: Involves solving one equation for one variable in terms of the other. This expression is then substituted into the second equation, reducing it to a single-variable equation that can be solved.

Explanation:

We are given the pair of linear equations:
Equation 1: x + y = 5
Equation 2: 2x – 3y = 4

Solving by Elimination Method:
1. Multiply Equation 1 by 3 to make the coefficients of y opposites:
3(x + y) = 3(5)
3x + 3y = 15 (Equation 3)

2. Add Equation 2 and Equation 3:
(2x – 3y) + (3x + 3y) = 4 + 15
2x + 3x – 3y + 3y = 19
5x = 19
x = 19/5

3. Substitute the value of x (19/5) into Equation 1:
(19/5) + y = 5
y = 5 – 19/5
y = (25 – 19)/5
y = 6/5

So, by elimination method, x = 19/5 and y = 6/5.

Solving by Substitution Method:
1. Solve Equation 1 for x:
x = 5 – y (Equation 4)

2. Substitute the expression for x from Equation 4 into Equation 2:
2(5 – y) – 3y = 4
10 – 2y – 3y = 4
10 – 5y = 4
-5y = 4 – 10
-5y = -6
y = -6 / -5
y = 6/5

3. Substitute the value of y (6/5) into Equation 4:
x = 5 – (6/5)
x = (25 – 6)/5
x = 19/5

So, by substitution method, x = 19/5 and y = 6/5.

Both methods yield the same solution: x = 19/5 and y = 6/5.

Question:

On comparing the ratios $𝐚𝟏𝐚𝟐,𝐛𝟏𝐛𝟐$ and $𝐜𝟏𝐜𝟐$, find out whether the following pair of linear equations are consistent, or inconsistent.

5x – 3y = 11 ; –10x + 6y = –22

Concept in a Minute:

To determine if a pair of linear equations is consistent or inconsistent, we compare the ratios of the coefficients of x, y, and the constant terms.
For two linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
1. If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a unique point, and the system is consistent with a unique solution.
2. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident, and the system is consistent with infinitely many solutions.
3. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel, and the system is inconsistent with no solution.

Explanation:

The given pair of linear equations are:
Equation 1: 5x – 3y = 11
Equation 2: –10x + 6y = –22

To compare the ratios, let’s rewrite the equations in the standard form $ax + by + c = 0$:
Equation 1: 5x – 3y – 11 = 0
Equation 2: –10x + 6y + 22 = 0

Now, identify the coefficients:
For Equation 1: $a_1 = 5$, $b_1 = -3$, $c_1 = -11$
For Equation 2: $a_2 = -10$, $b_2 = 6$, $c_2 = 22$

Calculate the ratios:
Ratio of x-coefficients: $\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}$
Ratio of y-coefficients: $\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}$
Ratio of constant terms: $\frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}$

Compare the ratios:
We observe that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -\frac{1}{2}$.

According to the rules for consistency:
When $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair of linear equations has infinitely many solutions, meaning the system is consistent.

Therefore, the given pair of linear equations is consistent.

Question:

Form the pair of linear equation in the following problem, and find its solutions (if they exist) by the elimination method:

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?

Concept in a Minute:

The problem involves forming a system of linear equations based on the given age-related relationships and then solving this system using the elimination method. The key is to translate the word problem into algebraic equations by assigning variables to the unknown ages. The elimination method requires manipulating the equations so that when one variable is subtracted from the other, it is eliminated, allowing for the solution of the remaining variable.

Explanation:

Let Nuri’s current age be ‘x’ years and Sonu’s current age be ‘y’ years.

Five years ago:
Nuri’s age was (x – 5) years.
Sonu’s age was (y – 5) years.
According to the problem, Nuri was thrice as old as Sonu:
x – 5 = 3(y – 5)
x – 5 = 3y – 15
x – 3y = -15 + 5
Equation 1: x – 3y = -10

Ten years later:
Nuri’s age will be (x + 10) years.
Sonu’s age will be (y + 10) years.
According to the problem, Nuri will be twice as old as Sonu:
x + 10 = 2(y + 10)
x + 10 = 2y + 20
x – 2y = 20 – 10
Equation 2: x – 2y = 10

Now, we solve these two linear equations using the elimination method.
Equation 1: x – 3y = -10
Equation 2: x – 2y = 10

Subtract Equation 2 from Equation 1:
(x – 3y) – (x – 2y) = -10 – 10
x – 3y – x + 2y = -20
-y = -20
y = 20

Now substitute the value of y = 20 into either Equation 1 or Equation 2 to find x. Let’s use Equation 2:
x – 2y = 10
x – 2(20) = 10
x – 40 = 10
x = 10 + 40
x = 50

Therefore, Nuri’s current age is 50 years and Sonu’s current age is 20 years.

Verification:
Five years ago: Nuri was 50 – 5 = 45, Sonu was 20 – 5 = 15. 45 = 3 * 15. (Correct)
Ten years later: Nuri will be 50 + 10 = 60, Sonu will be 20 + 10 = 30. 60 = 2 * 30. (Correct)

Question:

Form the pair of linear equations for the following problem 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?

Concept in a Minute:

Linear equations in two variables, representing relationships between unknown quantities. The substitution method involves solving one equation for one variable and substituting that expression into the other equation to solve for the remaining variable.

Explanation:

Let the present age of Jacob be x years and the present age of his son be y years.

According to the first condition, five years hence, Jacob’s age will be three times that of his son.
Jacob’s age after 5 years = x + 5
Son’s age after 5 years = y + 5
So, the equation is: x + 5 = 3(y + 5)
x + 5 = 3y + 15
x – 3y = 15 – 5
x – 3y = 10 (Equation 1)

According to the second condition, five years ago, Jacob’s age was seven times that of his son.
Jacob’s age 5 years ago = x – 5
Son’s age 5 years ago = y – 5
So, the equation is: x – 5 = 7(y – 5)
x – 5 = 7y – 35
x – 7y = -35 + 5
x – 7y = -30 (Equation 2)

Now, we solve these two linear equations by the substitution method.
From Equation 1, we can express x in terms of y:
x = 10 + 3y (Equation 3)

Substitute this expression for x into Equation 2:
(10 + 3y) – 7y = -30
10 – 4y = -30
-4y = -30 – 10
-4y = -40
y = (-40) / (-4)
y = 10

Now, substitute the value of y back into Equation 3 to find x:
x = 10 + 3(10)
x = 10 + 30
x = 40

Therefore, the present age of Jacob is 40 years and the present age of his son is 10 years.

We can check our answer:
Five years hence: Jacob will be 40 + 5 = 45 years old. His son will be 10 + 5 = 15 years old. 45 is indeed 3 times 15.
Five years ago: Jacob was 40 – 5 = 35 years old. His son was 10 – 5 = 5 years old. 35 is indeed 7 times 5.
The conditions are satisfied.

Question:

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

x + y = 14

x – y = 4

Concept in a Minute:

The substitution method for solving a pair of linear equations involves isolating one variable in one equation and then substituting that expression into the other equation. This results in a single equation with one variable, which can then be solved. Once one variable’s value is found, it can be substituted back into either of the original equations to find the value of the other variable.

Explanation:

We are given the following pair of linear equations:
1) x + y = 14
2) x – y = 4

Step 1: Isolate one variable in one of the equations.
Let’s take equation (2) and isolate x.
From equation (2), x – y = 4
Add y to both sides:
x = 4 + y

Step 2: Substitute this expression for x into the other equation.
Now substitute the expression for x (which is 4 + y) into equation (1):
(4 + y) + y = 14

Step 3: Solve the resulting equation for the remaining variable.
4 + 2y = 14
Subtract 4 from both sides:
2y = 14 – 4
2y = 10
Divide by 2:
y = 10 / 2
y = 5

Step 4: Substitute the value of the solved variable back into either original equation (or the isolated expression) to find the other variable.
We can substitute y = 5 into the expression we found for x:
x = 4 + y
x = 4 + 5
x = 9

Alternatively, we can substitute y = 5 into equation (1):
x + y = 14
x + 5 = 14
Subtract 5 from both sides:
x = 14 – 5
x = 9

Step 5: Check the solution.
Substitute x = 9 and y = 5 into both original equations:
For equation (1): x + y = 9 + 5 = 14 (This is correct)
For equation (2): x – y = 9 – 5 = 4 (This is correct)

Therefore, the solution to the pair of linear equations is x = 9 and y = 5.

Question:

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

3x – y = 3

9x – 3y = 9

Concept in a Minute:

The substitution method for solving a pair of linear equations involves expressing one variable in terms of the other from one equation and then substituting this expression into the other equation. This reduces the problem to a single equation with one variable, which can then be solved.

Explanation:

We are given the following pair of linear equations:
1) 3x – y = 3
2) 9x – 3y = 9

Step 1: Express one variable in terms of the other from one of the equations.
From equation (1), we can express y in terms of x:
y = 3x – 3

Step 2: Substitute this expression for y into the other equation.
Substitute y = 3x – 3 into equation (2):
9x – 3(3x – 3) = 9

Step 3: Solve the resulting equation for x.
9x – 9x + 9 = 9
9 = 9

Step 4: Interpret the result.
The equation 9 = 9 is a true statement. This indicates that the two linear equations are dependent, meaning they represent the same line. Therefore, there are infinitely many solutions to this pair of equations. Any pair of (x, y) that satisfies the first equation will also satisfy the second equation.

To express the solution set, we can say that for any real value of x, y = 3x – 3 is the corresponding value of y.

Example solutions:
If x = 1, y = 3(1) – 3 = 0. So, (1, 0) is a solution.
Check in equation (2): 9(1) – 3(0) = 9 – 0 = 9. (Correct)

If x = 2, y = 3(2) – 3 = 6 – 3 = 3. So, (2, 3) is a solution.
Check in equation (2): 9(2) – 3(3) = 18 – 9 = 9. (Correct)

Final Answer: The pair of linear equations has infinitely many solutions, and the solution set can be represented as (x, 3x – 3) for any real number x.

Question:

On comparing the ratios $𝐚𝟏𝐚𝟐,𝐛𝟏𝐛𝟐$ and $𝐜𝟏𝐜𝟐$ without drawing them, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincide.

9x + 3y + 12 = 0

18x + 6y + 24 = 0

Concept in a Minute:

For a pair of linear equations in two variables, given by $a₁x + b₁y + c₁ = 0$ and $a₂x + b₂y + c₂ = 0$, we compare the ratios of the coefficients: $\frac{a₁}{a₂}$, $\frac{b₁}{b₂}$, and $\frac{c₁}{c₂}$.
1. If $\frac{a₁}{a₂} \neq \frac{b₁}{b₂}$, the lines intersect at a unique point.
2. If $\frac{a₁}{a₂} = \frac{b₁}{b₂} \neq \frac{c₁}{c₂}$, the lines are parallel.
3. If $\frac{a₁}{a₂} = \frac{b₁}{b₂} = \frac{c₁}{c₂}$, the lines coincide.

Explanation:

We are given the pair of linear equations:
Equation 1: $9x + 3y + 12 = 0$
Equation 2: $18x + 6y + 24 = 0$

First, identify the coefficients of x, y, and the constant terms for both equations.
From Equation 1: $a₁ = 9$, $b₁ = 3$, $c₁ = 12$
From Equation 2: $a₂ = 18$, $b₂ = 6$, $c₂ = 24$

Now, calculate the ratios of the corresponding coefficients:
Ratio of x-coefficients: $\frac{a₁}{a₂} = \frac{9}{18}$
Ratio of y-coefficients: $\frac{b₁}{b₂} = \frac{3}{6}$
Ratio of constant terms: $\frac{c₁}{c₂} = \frac{12}{24}$

Simplify each ratio:
$\frac{a₁}{a₂} = \frac{9}{18} = \frac{1}{2}$
$\frac{b₁}{b₂} = \frac{3}{6} = \frac{1}{2}$
$\frac{c₁}{c₂} = \frac{12}{24} = \frac{1}{2}$

Now, compare these ratios:
We observe that $\frac{a₁}{a₂} = \frac{1}{2}$, $\frac{b₁}{b₂} = \frac{1}{2}$, and $\frac{c₁}{c₂} = \frac{1}{2}$.
Therefore, $\frac{a₁}{a₂} = \frac{b₁}{b₂} = \frac{c₁}{c₂}$.

According to the concept, when the ratios of all three coefficients are equal, the lines representing the pair of linear equations coincide. This means that both equations represent the same line, and there are infinitely many solutions.

Question:

Form the pair of linear equations for the following problem 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.

Concept in a Minute:

This question involves forming a system of linear equations from a word problem. The solution will then be found using the substitution method, where we express one variable in terms of another from one equation and substitute it into the other equation to solve for a single variable.

Explanation:

Let the cost of each bat be ₹ x and the cost of each ball be ₹ y.

From the first statement, “The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800”, we can form the first linear equation:
7x + 6y = 3800 (Equation 1)

From the second statement, “Later, she buys 3 bats and 5 balls for ₹ 1750”, we can form the second linear equation:
3x + 5y = 1750 (Equation 2)

Now, we will solve these equations using the substitution method.

Step 1: Express one variable in terms of the other from either equation.
Let’s express x from Equation 2:
3x = 1750 – 5y
x = (1750 – 5y) / 3 (Equation 3)

Step 2: Substitute this expression for x into Equation 1.
7 * [(1750 – 5y) / 3] + 6y = 3800

Step 3: Solve the resulting equation for y.
Multiply both sides by 3 to eliminate the denominator:
7 * (1750 – 5y) + 18y = 11400
12250 – 35y + 18y = 11400
12250 – 17y = 11400
-17y = 11400 – 12250
-17y = -850
y = -850 / -17
y = 50

So, the cost of each ball is ₹ 50.

Step 4: Substitute the value of y back into Equation 3 to find the value of x.
x = (1750 – 5 * 50) / 3
x = (1750 – 250) / 3
x = 1500 / 3
x = 500

So, the cost of each bat is ₹ 500.

Therefore, the cost of each bat is ₹ 500 and the cost of each ball is ₹ 50.

Question:

10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Concept in a Minute:

This problem requires setting up and solving a system of linear equations. We will define variables to represent the unknown quantities (number of boys and girls) and then translate the given information into two distinct equations. Solving these equations simultaneously will give us the values of our variables.

Explanation:

Let B be the number of boys and G be the number of girls who took part in the quiz.

From the problem statement, we have two pieces of information:
1. The total number of students is 10. This can be written as an equation:
B + G = 10 (Equation 1)

2. The number of girls is 4 more than the number of boys. This can be written as an equation:
G = B + 4 (Equation 2)

Now we have a system of two linear equations with two variables. We can solve this system using substitution.

Substitute the expression for G from Equation 2 into Equation 1:
B + (B + 4) = 10

Simplify and solve for B:
2B + 4 = 10
2B = 10 – 4
2B = 6
B = 6 / 2
B = 3

Now that we have found the number of boys (B = 3), substitute this value back into either Equation 1 or Equation 2 to find the number of girls. Using Equation 2 is simpler:
G = B + 4
G = 3 + 4
G = 7

So, the number of boys is 3 and the number of girls is 7.

To verify our answer:
Total students: 3 boys + 7 girls = 10 students (Matches the given information)
Number of girls vs boys: 7 girls is indeed 4 more than 3 boys (Matches the given information)

Therefore, there were 3 boys and 7 girls who took part in the quiz.

Question:

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

0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

Concept in a Minute:

The substitution method for solving a pair of linear equations involves solving one equation for one variable in terms of the other, and then substituting this expression into the second equation. This results in a single equation with one variable, which can then be solved. Finally, substitute the value of this variable back into the expression from the first step to find the value of the other variable.

Explanation:

Step 1: Convert the given equations into a simpler form by eliminating decimals.
Multiply the first equation by 10: (0.2x + 0.3y = 1.3) * 10 => 2x + 3y = 13 (Equation 1′)
Multiply the second equation by 10: (0.4x + 0.5y = 2.3) * 10 => 4x + 5y = 23 (Equation 2′)

Step 2: Solve Equation 1′ for one variable in terms of the other. Let’s solve for x.
From Equation 1′: 2x = 13 – 3y
x = (13 – 3y) / 2 (Equation 3)

Step 3: Substitute the expression for x from Equation 3 into Equation 2′.
4 * [(13 – 3y) / 2] + 5y = 23

Step 4: Simplify and solve the resulting equation for y.
2 * (13 – 3y) + 5y = 23
26 – 6y + 5y = 23
26 – y = 23
-y = 23 – 26
-y = -3
y = 3

Step 5: Substitute the value of y back into Equation 3 to find the value of x.
x = (13 – 3 * 3) / 2
x = (13 – 9) / 2
x = 4 / 2
x = 2

Step 6: Verify the solution by substituting the values of x and y into the original equations.
For the first equation: 0.2(2) + 0.3(3) = 0.4 + 0.9 = 1.3 (Correct)
For the second equation: 0.4(2) + 0.5(3) = 0.8 + 1.5 = 2.3 (Correct)

The solution is x = 2 and y = 3.

Question:

Form the pair of linear equation in the following problem, and find its solution (if they exist) by the elimination method:

Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

Concept in a Minute:

Linear Equations in Two Variables: Representing real-world problems using two variables and forming two linear equations.
System of Linear Equations: Solving a system of two linear equations with two variables.
Elimination Method: A method to solve a system of linear equations by eliminating one variable.

Explanation:

Let the number of ₹ 50 notes be ‘x’ and the number of ₹ 100 notes be ‘y’.

Formulate the equations based on the given information:
1. Total number of notes: The total number of notes Meena received is 25.
Equation 1: x + y = 25

2. Total amount withdrawn: The total amount withdrawn is ₹ 2000. The value of ‘x’ notes of ₹ 50 is 50x, and the value of ‘y’ notes of ₹ 100 is 100y.
Equation 2: 50x + 100y = 2000

Now, solve these two linear equations using the elimination method.

Step 1: Simplify Equation 2 by dividing by 50.
Equation 2 (simplified): x + 2y = 40

Step 2: We now have the system of equations:
Equation 1: x + y = 25
Equation 2 (simplified): x + 2y = 40

Step 3: To eliminate ‘x’, subtract Equation 1 from Equation 2 (simplified).
(x + 2y) – (x + y) = 40 – 25
x + 2y – x – y = 15
y = 15

Step 4: Substitute the value of ‘y’ (15) into Equation 1 to find ‘x’.
x + 15 = 25
x = 25 – 15
x = 10

Solution:
Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

Verification:
Total notes: 10 + 15 = 25 (Correct)
Total amount: (10 * ₹ 50) + (15 * ₹ 100) = ₹ 500 + ₹ 1500 = ₹ 2000 (Correct)

Question:

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

x – y = 8, 3x – 3y = 16

Concept in a Minute:

To determine consistency of a pair of linear equations, we compare the ratios of the coefficients of x, y and the constant terms.
For equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
1. If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair is consistent and has a unique solution.
2. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair is consistent and has infinitely many solutions.
3. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair is inconsistent and has no solution.

Explanation:

The given pair of linear equations is:
Equation 1: $x – y = 8$
Equation 2: $3x – 3y = 16$

We can rewrite these equations in the standard form $ax + by + c = 0$:
Equation 1: $x – y – 8 = 0$
Equation 2: $3x – 3y – 16 = 0$

Now, let’s identify the coefficients:
$a_1 = 1, b_1 = -1, c_1 = -8$
$a_2 = 3, b_2 = -3, c_2 = -16$

Let’s calculate the ratios:
$\frac{a_1}{a_2} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$

Comparing the ratios, we observe that $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ (since $\frac{1}{3} \neq \frac{1}{2}$).
According to the conditions for consistency, this means the pair of linear equations is inconsistent and has no solution.

Since the equations are inconsistent, we cannot obtain a solution graphically.

The final answer is $\boxed{Inconsistent}$.

Question:

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

s – t = 3

$𝑠3+𝑡2=6$

Concept in a Minute:

The substitution method for solving a pair of linear equations involves expressing one variable in terms of the other from one equation and then substituting this expression into the other equation. This results in a single equation with one variable, which can then be solved.

Explanation:

We are given the following pair of linear equations:
1) s – t = 3
2) s/3 + t/2 = 6

First, let’s simplify the second equation by finding a common denominator for the fractions:
Multiply the second equation by 6 to eliminate the denominators:
6 * (s/3 + t/2) = 6 * 6
2s + 3t = 36 (Equation 3)

Now we have a simplified system of equations:
1) s – t = 3
3) 2s + 3t = 36

From Equation 1, we can express ‘s’ in terms of ‘t’:
s = 3 + t

Now, substitute this expression for ‘s’ into Equation 3:
2 * (3 + t) + 3t = 36

Distribute the 2:
6 + 2t + 3t = 36

Combine the ‘t’ terms:
6 + 5t = 36

Subtract 6 from both sides:
5t = 36 – 6
5t = 30

Divide by 5 to solve for ‘t’:
t = 30 / 5
t = 6

Now that we have the value of ‘t’, substitute it back into the equation s = 3 + t to find ‘s’:
s = 3 + 6
s = 9

Therefore, the solution to the pair of linear equations is s = 9 and t = 6.

Verification:
Check the original equations with the found values:
Equation 1: s – t = 3 => 9 – 6 = 3 (True)
Equation 2: s/3 + t/2 = 6 => 9/3 + 6/2 = 3 + 3 = 6 (True)

Question:

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

Concept in a Minute:

Systems of linear equations, elimination method, substitution method, solving for variables, linear equation in two variables, finding the slope and y-intercept.

Explanation:

The problem asks us to solve a system of two linear equations with two variables (x and y) and then use the solution to find the value of ‘m’ in a third linear equation.

Step 1: Solve the system of linear equations.
We are given the equations:
1) 2x + 3y = 11
2) 2x – 4y = – 24

We can use the elimination method. Notice that the coefficient of ‘x’ is the same in both equations (2x).
Subtract equation (2) from equation (1):
(2x + 3y) – (2x – 4y) = 11 – (– 24)
2x + 3y – 2x + 4y = 11 + 24
7y = 35
y = 35 / 7
y = 5

Now substitute the value of y (y=5) into either equation (1) or (2) to find ‘x’. Let’s use equation (1):
2x + 3(5) = 11
2x + 15 = 11
2x = 11 – 15
2x = -4
x = -4 / 2
x = -2

So, the solution to the system of equations is x = -2 and y = 5.

Step 2: Find the value of ‘m’ using the third equation.
We are given the equation y = mx + 3.
We know that the point (x, y) = (-2, 5) must satisfy this equation because it’s the solution to the original system.
Substitute x = -2 and y = 5 into the equation y = mx + 3:
5 = m(-2) + 3
5 = -2m + 3

Now, solve for ‘m’:
5 – 3 = -2m
2 = -2m
m = 2 / -2
m = -1

Therefore, the value of ‘m’ is -1.

Question:

5x – 4y + 8 = 0,

7x + 6y – 9 = 0

Concept in a Minute:

To determine the relationship between two linear equations of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ without graphing, we compare the ratios of their corresponding coefficients: $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$.
– If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a single point.
– If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.
– If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines coincide (are the same line).

Explanation:

The given pair of linear equations are:
1) 5x – 4y + 8 = 0
2) 7x + 6y – 9 = 0

We need to compare the ratios of the coefficients.
From equation (1): $a_1 = 5$, $b_1 = -4$, $c_1 = 8$
From equation (2): $a_2 = 7$, $b_2 = 6$, $c_2 = -9$

Now, let’s find the ratios:
Ratio of x-coefficients: $\frac{a_1}{a_2} = \frac{5}{7}$
Ratio of y-coefficients: $\frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}$
Ratio of constant terms: $\frac{c_1}{c_2} = \frac{8}{-9} = -\frac{8}{9}$

Compare the ratios:
Is $\frac{a_1}{a_2} = \frac{b_1}{b_2}$?
$\frac{5}{7} \neq -\frac{2}{3}$

Since the ratio of the x-coefficients is not equal to the ratio of the y-coefficients ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$), the lines representing these two linear equations intersect at a single point.

Question:

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

The difference between two numbers is 26 and one number is three times the other. Find them.

Concept in a Minute:

Linear equations in two variables, substitution method. Substitution method involves expressing one variable in terms of another from one equation and substituting it into the other equation to solve for one variable. Then, substitute the found value back into one of the original equations to find the other variable.

Explanation:

Let the two numbers be x and y.

From the problem statement, we have two conditions:
1. The difference between two numbers is 26.
This can be written as:
x – y = 26 (Equation 1)

2. One number is three times the other.
Let’s assume x is the larger number. So,
x = 3y (Equation 2)

Now, we will solve these linear equations using the substitution method.

Step 1: Substitute the expression for x from Equation 2 into Equation 1.
Substitute x = 3y into x – y = 26:
(3y) – y = 26

Step 2: Solve for y.
2y = 26
y = 26 / 2
y = 13

Step 3: Substitute the value of y back into Equation 2 to find x.
x = 3y
x = 3 * 13
x = 39

Step 4: Verify the solution.
Check if the difference between the numbers is 26: 39 – 13 = 26 (This is correct).
Check if one number is three times the other: 39 = 3 * 13 (This is also correct).

Therefore, the two numbers are 39 and 13.

The pair of linear equations is:
x – y = 26
x = 3y

The solution by substitution method is x = 39 and y = 13.

Question:

Form the pair of linear equation in the following problem, and find its solution (if they exist) by the elimination method:

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.

Concept in a Minute:

Forming linear equations from word problems.
Solving a pair of linear equations using the elimination method.

Explanation:

Let the fixed charge for the first three days be ‘x’ rupees.
Let the additional charge for each day thereafter be ‘y’ rupees.

According to the problem:
Saritha paid Rs 27 for a book kept for seven days.
This means she paid for the first 3 days (fixed charge) and for the remaining 7 – 3 = 4 days (additional charge).
So, the equation formed is: x + 4y = 27 (Equation 1)

Susy paid Rs 21 for a book kept for five days.
This means she paid for the first 3 days (fixed charge) and for the remaining 5 – 3 = 2 days (additional charge).
So, the equation formed is: x + 2y = 21 (Equation 2)

Now, we need to solve these two linear equations using the elimination method.
We have the equations:
1) x + 4y = 27
2) x + 2y = 21

To eliminate ‘x’, we can subtract Equation 2 from Equation 1:
(x + 4y) – (x + 2y) = 27 – 21
x + 4y – x – 2y = 6
2y = 6
y = 6 / 2
y = 3

Now, substitute the value of ‘y’ (which is 3) into either Equation 1 or Equation 2 to find ‘x’. Let’s use Equation 2:
x + 2y = 21
x + 2(3) = 21
x + 6 = 21
x = 21 – 6
x = 15

Therefore, the fixed charge for the first three days is Rs 15, and the charge for each extra day is Rs 3.

Question:

6x – 3y + 10 = 0,

2x – y + 9 = 0

Concept in a Minute:

To determine the relationship between two linear equations of the form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we compare the ratios of their corresponding coefficients: a₁/a₂, b₁/b₂, and c₁/c₂.
– If a₁/a₂ ≠ b₁/b₂, the lines intersect at a single point.
– If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
– If a₁/a₂ = b₁/b₂ = c₁/c₂, the lines coincide (are the same line).

Explanation:

We are given two linear equations:
Equation 1: 6x – 3y + 10 = 0
Equation 2: 2x – y + 9 = 0

First, identify the coefficients for each equation:
For Equation 1: a₁ = 6, b₁ = -3, c₁ = 10
For Equation 2: a₂ = 2, b₂ = -1, c₂ = 9

Now, calculate the ratios of the corresponding coefficients:
Ratio of x-coefficients: a₁/a₂ = 6/2 = 3
Ratio of y-coefficients: b₁/b₂ = -3/(-1) = 3
Ratio of constant terms: c₁/c₂ = 10/9

Compare these ratios:
a₁/a₂ = 3
b₁/b₂ = 3
c₁/c₂ = 10/9

We observe that a₁/a₂ = b₁/b₂ (since 3 = 3), but a₁/a₂ ≠ c₁/c₂ (since 3 ≠ 10/9).

According to the concept in a minute, when the ratio of the x-coefficients is equal to the ratio of the y-coefficients, but not equal to the ratio of the constant terms, the lines representing the pair of linear equations are parallel.

Therefore, the lines representing the given pair of linear equations are parallel.

Question:

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Concept in a Minute:

To determine consistency of a pair of linear equations, we compare the ratios of the coefficients of x, y, and the constant terms.
For equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0:
1. Consistent (unique solution): a1/a2 ≠ b1/b2
2. Consistent (infinitely many solutions): a1/a2 = b1/b2 = c1/c2
3. Inconsistent (no solution): a1/a2 = b1/b2 ≠ c1/c2
If consistent, the solution can be obtained graphically by plotting the lines represented by the equations and finding their point of intersection.

Explanation:

We are given the pair of linear equations:
Equation 1: 2x – 2y – 2 = 0
Equation 2: 4x – 4y – 5 = 0

Let’s compare the coefficients of these equations with the general form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.
Here, a1 = 2, b1 = -2, c1 = -2
And, a2 = 4, b2 = -4, c2 = -5

Now, let’s calculate the ratios:
a1/a2 = 2/4 = 1/2
b1/b2 = -2/-4 = 1/2
c1/c2 = -2/-5 = 2/5

Comparing the ratios:
a1/a2 = 1/2
b1/b2 = 1/2
c1/c2 = 2/5

We observe that a1/a2 = b1/b2, but a1/a2 ≠ c1/c2.
This condition (a1/a2 = b1/b2 ≠ c1/c2) indicates that the given pair of linear equations is inconsistent, meaning there is no solution.

Therefore, we do not need to obtain the solution graphically as the lines are parallel and do not intersect.

Question:

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

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

Concept in a Minute:

Supplementary angles: Two angles are supplementary if their sum is 180 degrees.
Linear equations: Equations involving variables with a power of 1.
Substitution method: A method to solve a system of linear equations by substituting the expression of one variable from one equation into another equation.

Explanation:

Let the two supplementary angles be x and y.
Since the angles are supplementary, their sum is 180 degrees.
Equation 1: x + y = 180

The larger angle exceeds the smaller by 18 degrees. Let’s assume x is the larger angle and y is the smaller angle.
Equation 2: x = y + 18

Now, we will solve these two equations using the substitution method.
Substitute the expression for x from Equation 2 into Equation 1:
(y + 18) + y = 180
Combine like terms:
2y + 18 = 180
Subtract 18 from both sides:
2y = 180 – 18
2y = 162
Divide by 2:
y = 162 / 2
y = 81

Now substitute the value of y back into Equation 2 to find x:
x = y + 18
x = 81 + 18
x = 99

Therefore, the two supplementary angles are 99 degrees and 81 degrees.

To check the answer:
Sum of angles: 99 + 81 = 180 degrees (Supplementary)
Difference between angles: 99 – 81 = 18 degrees (Larger exceeds smaller by 18)

Question:

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden

Concept in a Minute:

This problem involves understanding the properties of a rectangle, specifically its perimeter and the relationship between its length and width. The concept of forming algebraic equations from given word problems is also crucial.

Explanation:

Let the width of the rectangular garden be ‘w’ meters.
According to the problem, the length of the garden is 4 m more than its width.
So, the length of the garden is ‘l’ = w + 4 meters.

The perimeter of a rectangle is given by the formula: Perimeter = 2 * (length + width).
The problem states that “half the perimeter” is 36 m.
Therefore, (1/2) * [2 * (l + w)] = 36 m.
This simplifies to l + w = 36 m.

Now, substitute the expression for ‘l’ in terms of ‘w’ into this equation:
(w + 4) + w = 36

Combine like terms:
2w + 4 = 36

Subtract 4 from both sides of the equation:
2w = 36 – 4
2w = 32

Divide by 2 to find the value of ‘w’:
w = 32 / 2
w = 16 meters

Now that we have the width, we can find the length:
l = w + 4
l = 16 + 4
l = 20 meters

So, the dimensions of the garden are:
Width = 16 meters
Length = 20 meters

Next Chapter: Polynomials

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