NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.2

Go back to  'Pair of Linear Equations in Two Variables'

Chapter 3 Ex.3.2 Question 1

Form the pair of linear equations in the following problems and find their Solutions graphically.

(i) \(10\) 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.

(ii) \(5\) pencils and \(7\) pens together cost ₹ \(50, \) whereas \(7\) pencils and \(5\) pens together cost ₹ \(46.\) Find the cost of one pencil and that of one pen.

Solution

Video Solution

(i)

What is Known?

(i) Number of students took part in Quiz \(= 10\)

(ii) Number of girls is \(4\) more than number of boys

What is Unknown?

Finding Solutions graphically for the given situation.

Reasoning:

Assuming the number of boys as \(x\) and the number of girls as \(y,\) two linear equations can be formed for the above situation.

Steps:

Total number of boys and girls is:

\[x + y = 10\]

Number of girls is \(4\) more than the number of boys,

Mathematically:

\[\begin{align}y &= x + 4\\ - x + y &= 4\end{align}\]

Algebraic representation where \(x\) and \(y\) are the number of boys and girls respectively.

\[\begin{align}x + y &= 10 \,\dots(1)\\- x + y &= 4\,\dots(2) \end{align}\]

Therefore, the algebraic representation for equation 1 is:

\[\begin{align}x + y &= 10\\y &= 10-x\end{align}\]

And, the algebraic representation is for equation 2 is:

\[\begin{align} - x + y& = 4\\y &= x + 4\end{align}\]

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in table shown below.

\(x\) \(2\) \(7\)
\(y = 10 - x\) \(8\) \(3\)

 

\(x\) \(1\) \(4\)
\(y = x + 4\) \(5\) \(8\)

The graphical representation is as follows.

Answer:

From graph solution \(\left( {x,y} \right) = \left( {3,7} \right)\)

Number of boys \(= 3\)

Number of girls\( = 7\)

(ii)

What is Known?

(i) \(5\) pencils and \(7\) pens cost ₹ \(50\) \(\left( {x,y} \right) = \left( {3,7} \right)\)

(ii) \(7\) pencils and \(5\) pens cost ₹ \(46\)

What is Unknown?

Finding Solutions graphically for the given situation.

Reasoning:

Assuming the cost of \(1\) pencil as ₹ \(x\) and the cost of \(1\) pen as ₹ \(y,\) two linear equations are to be formed for the above situation.

Steps:

Let us assume cost of \(1\) pencil be \(x\) and cost of \(1\) pen be \(y.\)

The cost of \(5\) pencils and \(7\) pens is ₹ \(50.\)

Mathematically,

\[5x + 7y = 50\]

And, the cost of \(7\) pencils and 5 pens is ₹ \(50.\)

Mathematically,

\[7x + 5y = 46\]

Algebraic representation where \(x\) and \(y\) are the cost of \(1\) pencil and \(1\) pen respectively.

\[\begin{align}5x + 7y &= 50 \,\dots (1) \\7x + 5y &= 46 \,\dots(2) \end{align}\]

Therefore, the algebraic representation for equation \(1\) is:

\[\begin{align}5x+ 7y &= 50\\7y &= 50-5x\\y &= \frac{{50 - 5x}}{7}\end{align}\]

And, the algebraic representation for equation \(2\) is:

\[\begin{align}7x + 5y &= 46\\5y &= 46-7x\\y &= \frac{{46 - 7x}}{5}\end{align}\]

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in table shown below.

\(x\) \(3\) \( - 4\)
\(y = \frac{{50 - 5x}}{7}\) \(5\) \(10\)

 

\(x\) \(3\) \(8\)
 \(y = \frac{{46 - 7x}}{5}\) \(5\) \( - 2\)

From graph Solution \(\left( {x,y} \right) = \left( {3,5} \right)\)

Cost of one pencil \(=\)\(3\)

Cost of one pen \(=\)\(5\)

Chapter 3 Ex.3.2 Question 2

On comparing the ratios \(\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}, \end{align}\) find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 

\(\begin{align}\quad 5x-4y + 8 &= 0\\7x + 6y-9 &= 0\end{align}\)

(ii)

\(\begin{align}\quad 9x + 3y + 12& = 0\\18x + 6y + 24 &= 0\end{align}\)

(iii)

\(\begin{align}\quad 6x-3y + 10 &= 0\\2x-y + 9 &= 0\end{align}\)

Solution

Video Solution

(i) What is Known?

\(\begin{align}5x-4y + 8 &= 0\\7x + 6y-9 &= 0\end{align}\)

What is Unknown?

Whether the lines are

(i) Intersecting

(ii) Parallel

(iii) Coincident

Reasoning:

For any pair of linear equation

\(a_1 x + b_1 y + c_1 = 0\\ a_2 x + b_2 y + c_2 = 0\)

a) \(\begin{align}\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\end{align}\) (Intersecting Lines)

b)  \(\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align}\) (Coincident Lines)

c)  \(\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\end{align}\) (Parallel Lines)

Steps:

\[\begin{align}{a_1} &= 5 \;\,\qquad{b_1} = - 4 \qquad {c_1} = 8\\{a_2} &= 7 \qquad \;\,{b_2} = 6 \;\;\;\qquad {c_2} = - 9\end{align}\]

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{5}{7} \qquad \;\;\; \dots(1)\\
\frac{{{b_1}}}{{{b_2}}} = \frac{{ - 4}}{6} &= \frac{{ - 2}}{3} \qquad \dots (2)\end{align}\]

From (i) and (ii)

\[\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\]

Therefore, they are intersecting lines at a point

(ii) What is Known?

\(\begin{align}9x + 3y + 12& = 0\\18x + 6y + 24 &= 0\end{align}\)

What is Unknown?

Whether the lines are

(i) Intersecting

(ii) Parallel

(iii) Coincident

Reasoning:

For any pair of linear equation

\[\begin{align}{a_1}x + {b_1}y + {c_1} &= 0\\{a_2}x + {b_2}y + {c_2} &= 0\end{align}\]

a) \(\begin{align}\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\end{align}\)(Intersecting Lines)

b) \(\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align}\)(Coincident Lines)

c) \(\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\end{align}\)(Parallel Lines)

Steps:

\[\begin{align}{a_1} &= 9, \quad {b_1} = 3 \quad {c_1} = 12\\{a_2}& = 18\quad {b_2} = 6\quad {c_2} = 24\end{align}\]

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{9}{{18}} = \frac{1}{2} \quad \dots(1)\\
\frac{{{b_1}}}{{{b_2}}} &= \frac{3}{6} = \frac{1}{2} \quad \;\;\dots(2)\\
\frac{{{c_1}}}{{{c_2}}} &= \frac{{12}}{{24}} = \frac{1}{2} \quad \dots(3)
\end{align}\]

From (1), (2) and (3)

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} = \frac{1}{2}\]

Therefore, they are coincident lines

(iii) what is Known?

\(\begin{align}6x-3y + 10 &= 0\\2x-y + 9 &= 0\end{align}\)

What is Unknown?

Whether the lines

(i) Intersecting

(ii) Parallel

(iii) Coincident

Reasoning:

For any pair of linear equation

\[\begin{align}{a_1}x + {b_1}y + {c_1} = 0\\{a_2}x + {b_2}y + {c_2} = 0\end{align}\]

a) \(\begin{align}\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\end{align}\) (Intersecting Lines)

b) \(\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align}\) (Coincident Lines)

c) \(\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\end{align}\) (Parallel Lines)

Steps:

\[\begin{align}{a_1} &= 6,\quad {b_1} = - 3\quad{c_1} = 10\\{a_2} &= 2 \quad \; {b_2} = - 1 \quad \,{c_2} = 9\end{align}\]

\[\begin{align}\frac{{{a_1}}}{{{a_2}}}& = \frac{6}{2} = 3 \qquad \quad \dots(1)\\\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 3}}{{ - 1}} = 3 \qquad \dots(2)\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{10}}{9} \qquad  \qquad \;\dots(3)\end{align}\]

From \((1), (2)\) and \((3)\)

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\]

Therefore, they are parallel lines.

Chapter 3 Ex.3.2 Question 3

On comparing the ratios \(\begin{align}\frac{{{a_1}}}{{{a_2}}},\frac{{{b_1}}}{{{b_2}}},\frac{{{c_1}}}{{{c_2}}}\end{align}\), find out whether the following pair of linear equations are consistent, or inconsistent. 

(i)

\(3x + 2y = 5;\;\;2x-3y = 7\)

(ii)

\(2x-3y = 8;\;\;4x-6y = 9\)

(iii)

\(\begin{align} \frac{3}{2}x + \frac{5}{3}y = 7;\;\;9x-10y = 14 \end{align}\)

(iv)

\(5x-3y = 11;\;\;-10x + 6y = -22\)

(v)

\(\begin{align} \frac{4}{3}x + 2y = 8;\;\;2x + 3y = 12 \end{align}\)

Solution

Video Solution

What is Unknown?

To find out whether the linear equations are consistent or inconsistent.

Reasoning:

For any pair of linear equation

\[\begin{align}{a_1}x + {b_1}y + {c_1} &= 0\\{a_2}x + {b_2}y + {c_2} &= 0\end{align}\]

Consistent means pair of linear equations have one solution or infinitely many solutions.

\[\begin{align}&\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\\& \begin{bmatrix} \text{Intersecting lines/one} \\ \text{Solution} \end{bmatrix} \\&\,\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\\&\left[ \begin{array} \ {\text{Coincident Lines/ Infinitely}}\\{\text{ many Solutions}} \end{array}\right]\end{align}\]

Inconsistent means, the lines may be parallel and do not have any Solution)

\(\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \,\,\frac{{{c_1}}}{{{c_2}}}\end{align}\) (Parallel lines/No Solution)

(i) What is Known?

\(\begin{align}3x + 2y - 5 &= 0{\rm{ }}\\2x-3y - 7 &= 0\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{3}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{2}{{ - 3}}\\
\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 5}}{{ - 7}} \\&= \frac{5}{7}\end{align}\]

From above

\[\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\]

Therefore, lines are intersecting and have one solution,

Hence, the pair of equations are consistent.

(ii) What is Known:

\(\begin{align}2x-3y - 8 &= 0\\4x-6y - 9 &= 0\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{2}{4} = \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 3}}{{ - 6}} = \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 8}}{{ - 9}} = \frac{8}{9}\end{align}\]

From above

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \,\,\frac{{{c_1}}}{{{c_2}}}\,\]

Therefore, lines are parallel and have no solution,

Hence, the pair of equations are inconsistent.

(iii) What is Known?

\(\begin{align}\frac{3}{2}x + \frac{5}{3}y &= 7{\rm{ }}\\9x - 10y &= 14\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{{\frac{3}{2}}}{9} \\&= \frac{3}{2} \times \frac{1}{9}\\& = \frac{1}{6}\\
\frac{{{b_1}}}{{{b_2}}} &= \frac{{\frac{5}{3}}}{{ - 10}} \\&= \frac{5}{3} \times \frac{1}{{ - 10}}\\& = \frac{1}{{ - 6}}\\
\frac{{{c_1}}}{{{c_2}}} &= \frac{7}{{14}} \\&= \frac{1}{2}\end{align}\]

From above

\[\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\,\]

Therefore, lines are intersecting and have one solution.

Hence, they are consistent.

(iv) What is Known?

\(\begin{align}5x-3y - 11 &= 0{\rm{ }}\\-10x + 6y + 22 &= 0\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{5}{{ - 10}} \\&= \frac{{ - 1}}{2}\\
\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 3}}{6} \\&= \frac{{ - 1}}{2}\\\frac{{{c_1}}}{{{c_2}}}& = \frac{{ - 11}}{{22}} \\&= \frac{{ - 1}}{2}\end{align}\]

From above

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \,\,\frac{{{c_1}}}{{{c_2}}}\,\]

Therefore, lines are coincident and have infinitely many solutions.

Hence, they are consistent.

(v) What is Known?

\(\begin{align}\frac{4}{3}x + 2y &= 8\\2x + 3y &= 12\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}}& = \frac{{\frac{4}{3}}}{2} = \frac{4}{3} \times \frac{1}{2} \\&= \frac{2}{3}\\
\frac{{{b_1}}}{{{b_2}}} &= \frac{2}{3}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 8}}{{ - 12}}\\& = \frac{2}{3}\end{align}\]

From above

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\,\]

Therefore, lines are coincident and have infinitely many solutions.

Hence, they are consistent.

Chapter 3 Ex.3.2 Question 4

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

(i)

\(x + y = 5, \;2x + 2y = 10\)

(ii)

\(x-y = 8, \;3x-3y = 16\)

(iii)

\(2x + y-6 = 0, \;4x-2y-4 = 0\)

(iv)

\(2x-2y-2 = 0, \;4x-4y-5 = 0\)

Solution

Video Solution

What is Unknown?

Whether the linear equations are consistent or inconsistent and graphical solution, if consistent.

Reasoning:

Consistent means pair of linear equations have one solution or infinitely many solutions.

\[\begin{align}{a_1}x + {b_1}y + {c_1} &= 0\\{a_2}x + {b_2}y + {c_2} &= 0\end{align}\]

\[\begin{align}&\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\\& \begin{bmatrix} {\text{Intersecting lines/one}}\\ {\text{Solution}} \end{bmatrix} \\&\,\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\\&\left[ \begin{array} \ {\text{Coincident Lines/Infinitely}}\\{\text{ many Solutions}} \end{array}\right]\end{align}\]

Inconsistent means, the lines may be parallel and do not have any Solution)

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \,\,\frac{{{c_1}}}{{{c_2}}}\) (Parallel lines/ No Solution)

 (i) What is Known?

\(\begin{align}x + y-5 &= 0\\2x + 2y-10 &= 0\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 5}}{{ - 10}} \\&= \frac{1}{2}\end{align}\]

From above

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \,\,\frac{{{c_1}}}{{{c_2}}}\]

Therefore, lines are coincident and have infinitely many solutions.

Hence, they are consistent.

\[\begin{align}x + y-5 &= 0\\y &= - x + 5\\y& = 5-x\end{align}\]

\(x\)

\(1\)

\(2\)

\(y = 5-x\)

\(4\)

\(3\)

\[\begin{align}2x + 2y-10 &= 0\\2y &= 10-2x\\y &= 5-x\end{align}\]

\(x\) \(3\) \(4\)
\(y = 5-x\) \(2\) \(1\)

All the points on coincident line are solutions for the given pair of equations.

(ii) What is Known?

\(\begin{align}x-y-8 &= 0\\3x-3y-16 &= 0\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{1}{3}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 1}}{{ - 3}}\\&= \frac{1}{3}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 8}}{{ - 16}} \\&= \frac{1}{2}\end{align}\]

From above

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \,\,\frac{{{c_1}}}{{{c_2}}}\,\]

Therefore, lines are parallel and have no solution,

Hence, the pair of equations are inconsistent.

\[\begin{align}x - y - 8 &= 0\\y &= x - 8\end{align}\]

\(x\) \(8\) \(6\)
\(y{\rm{ }} = {\rm{ }}x{\rm{ }}-{\rm{ }}8\) \(0\) \( - 2\)

\[\begin{align}3x - 3y - 16 &= 0\\3y &= 3x - 16\\y &= \frac{{3x - 16}}{3}\end{align}\]

\(x\) \(2\) \(4\)
\(y = \frac{{3x - 16}}{3}\) \( - 3.3\) \( - 1.3\)

(iii) What is Known?

\(\begin{align}2x + y-6 &= 0\\4x-2y-4 &= 0\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{2}{4}\\& = \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{1}{{ - 2}} \\&= - \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 6}}{{ - 4}} \\&= \frac{3}{2}\end{align}\]

From above:

\[\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\,\]

Therefore, lines are intersecting and have one solution.

Hence, they are consistent.

\[\begin{align}2x + y-6 &= 0\\y &= 6-2x\end{align}\]

\(x\) \(0\) \(2\)
\(y = 6-2x\) \(6\) \(2\)

\[\begin{align}4x - 2y - 4 &= 0\\2y &= 4x - 4\\y& = 2x - 2\end{align}\]

\(x\) \(2\) \(3\)
\(y = 2x - 2\) \(2\) \(4\)

\(x=2\) and \(y=2\) are solutions for the given pair of equations.

(iv) What is Known?

\(\begin{align}2x-2y-2 &= 0\\4x-4y-5& = 0 \\\end{align} \)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{2}{4} \\&= \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 2}}{{ - 4}}\\&= \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 2}}{{ - 5}}\\& = \frac{2}{5}\end{align}\]

From above:

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \,\,\frac{{{c_1}}}{{{c_2}}}\,\]

Therefore, lines are parallel and have no solution,

Hence, the pair of equations are inconsistent.

\[\begin{align}2x - 2y - 2 &= 0\\2y &= 2x - 2\\y &= x - 1\end{align}\]

\(x\) \(1\) \(3\)
\(y = x-1\) \(0\) \(2\)

\[\begin{align}4x - 4y - 5 &= 0\\4y &= 4x - 5\\y& = \frac{{4x - 5}}{4}\end{align}\]

\(x\) \(4\) \(3\)
\(y = \frac{{4x - 5}}{4}\) \(2.8\) \(1.8\)

Chapter 3 Ex.3.2 Question 5

Half the perimeter of a rectangular garden, whose length is \(4 \,\rm{m}\) more than its width, is \(36\,\rm{ m.}\) Find the dimensions of the garden.

Solution

Video Solution

What is Known?

(i) Half the perimeter of rectangular garden \( = 36{\rm{\,m }}\)

(ii) Length is \(4{\rm{ \,m}}\) more than width

What is Unknown?

Dimensions of the garden

Reasoning:

Assuming length of the garden as \(x\) and width of the garden as \(y,\) two linear equations can be formed for the known data.

Perimeter of rectangle \(=\) 2(Length \(+\) Breadth)

Steps:

Let the length of the garden be \(x\) and breadth be \(y\)

Then,

\[\begin{align}x& = y + 4\\ x - y &= 4\\y &= x - 4\end{align}\]

\(x\) \(8\) \(16\)
\(y = x - 4\) \(4\) \(12\)

Half perimeter of the rectangle be \(x + y = 36\)

\(y = 36 - x\)

\(x\) \(16\) \(26\)
\(y = x - 4\) \(20\) \(10\)

The answer is 

Length \(x = 20{\rm{\; m}}\)

Breadth \(y = 16{\rm{\; m}}\)

Chapter 3 Ex.3.2 Question 6

Given the linear equation \(2x + 3y-8 = 0,\) write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

Solution

Video Solution

What is Known?

One linear equation \(2x + 3y-8 = 0\)

What is Unknown?

Another linear equation such that given is satisfied.

Reasoning:

Same as Exercise 3.2 (2)

(i) Intersecting lines

Condition: \(\begin{align} \frac{{{a_1}}}{{{a_2}}} \ne \,\frac{{{b_1}}}{{{b_2}}}\, \end{align}\)

\[\begin{align}2x + 3y-8 &= 0\\{a_1}& = 2\\{b_1} &= 3\end{align}\]

So, considering \({a_2} = 3\) and \({b_2} = 2\) will satisfy the condition for intersecting lines \({c_2}\) can be any value.

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{2}{3} \quad \frac{{{b_1}}}{{{b_2}}} = \frac{3}{2}\\\frac{2}{3} &\ne \frac{3}{2}\,\end{align}\]

\(\therefore\) Another linear equation is

\(3x + 2y-6 = 0\)

(ii) Parallel Lines

Condition: \(\begin{align} \frac{{{a_1}}}{{{a_2}}} = \,\frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}} \end{align}\)

\[\begin{align}2x + 3y-8 &= 0\\{a_1} &= 2\\{b_1} &= 3\\{c_1} &=  - 8\end{align}\]

So, considering \({a_2} = 4,\;\;{b_2} = 6,\;\;{c_2} = 9\) will satisfy the condition for parallel lines.

\[\begin{align}\frac{{{a_1}}}{{{a_2}}}&= \frac{2}{4} = \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{3}{6} = \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}}& = \frac{{ - 8}}{9}\end{align}\]

From above:

\[\frac{{{a_1}}}{{{a_2}}} = \,\frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\]

Therefore, another linear equation is

\( = 4x + 6y + 9 = 0\)

(iii) Coincident lines:

Condition: \(\begin{align} \frac{{{a_1}}}{{{a_2}}} = \,\frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} \end{align}\)

\[\begin{align}2x + 3y-8 &= 0\\{a_1} &= 2\\{b_1} &= 3\\{c_1}& =  - 8\end{align}\]

So, considering \({a_2} = 4,\;\;{b_2} = 6,\;\;{c_2} = - 16\) will satisfy the condition for parallel lines.

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{2}{4} = \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{3}{6} = \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 8}}{{ - 16}} = \frac{1}{2}\end{align}\]

From above:

\[\frac{{{a_1}}}{{{a_2}}} = \,\frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\]

Therefore, linear equation is

\(4x + 6y-16 = 0\)

Chapter 3 Ex.3.2 Question 7

Draw the graphs of the equations \(x-y + 1 = 0\) and \(3x + 2y-12 = 0.\) Determine the coordinates of the vertices of the triangle formed by these lines and the \(x-\)axis and shade the triangular region.

Solution

Video Solution

What is Known?

Linear equation

\[\begin{align}x - y + 1 &= 0\\3x + 2y-12 &= 0\end{align}\]

What is Unknown?

Coordinates of the vertices of the triangle formed by intersecting lines and the \(x\)-axis

Reasoning:

From graph of two linear equations and \(x\)-axis, triangle can be shaded, and vertices can be located.

Steps:

\[\begin{align}x - y + 1 &= 0\\y &= x + 1\end{align}\]

\(x\) \(0\) \(1\)
\(y = x + 1\) \(1\) \(2\)

\[\begin{align}3x + 2y-12 &= 0\\2y &= 12 - 3x\\y &= \frac{{12 - 3x}}{2}\end{align}\]

\(x\) \(0\) \(2\)
\(y = \frac{{12 - 3x}}{2}\) \(6\) \(3\)

From graph, Vertices are \((-1, \,0),\; (4,\, 0),\) and \((2, \,3)\)

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