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. Solution
Let number of boys be =$x$
Let numbers of girls be =$y$
Given: 10 students of Class X took part in a Mathematics quiz i.e $x+y=10$
and number of girls is 4 more than the number of boys i.e $y=x+4$
Plotting the equations graphically.
Ans: Total number of girls = 7 and total number of boys=4
(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
Let the cost of 1 pencils be ₹ $x$
Let the cost of 1 pen be ₹ $y$
Given :5 pencils and 7 pens together cost ₹ 50
i.e $5x+7y = 50$ ...................(1)
7 pencils and 5 pens together cost ₹ 46
i.e $7x+5y=46$ .....................(2)
Plotting the equations graphically.
Cost of 1 pencil is ₹ 3;
Cost of 1 pen is ₹ 5;
2. On comparing the ratios ${a_1/a_2}$ , ${b_1/b_2}$ and ${c_1/c_2}$ , find out whether the lines representing the
following pairs of linear equations intersect at a point, are parallel or coincident:
(i) ${5x – 4y + 8 = 0}$ ; ${7x + 6y – 9 = 0}$
Solution
(i) Given: ${5x – 4y + 8 = 0}$ i.e $a_1=5 ; b_1= {-4} ; c_1 = 8$
; and ${7x + 6y – 9 = 0}$ i.e $a_2=7; b_2 = 6 ;c_2 = -9 $
Therefore $a_1/a_2 = 5/7$
$b_1/b_2 = {-4}/6 = -2/3$
$c_1/c_2 = 8/{-9} = -8/9$
Ans: Since $a_1/a_2 ≠ b_1/b_2$, therefore the pair of lines are intersecting.
Important Note: As you know that a linear equation in 1 degree represent by the equation $y=mx+c$ is always a
straight line passing through a point $c$ on y-axis and making an angle of $tan^{-1}{(m)}$ degree with the horizontal axis.
So, if a pair of lines are given by the equations,
$a_1x+b_1y + c_1 = 0$
$a_2x+b_2y + c_2 = 0$
So, for the given pair of lines to be parallel, their slopes must be equal, i.e $m_1=m_2$
i.e $a_1/b_1$=$a_2/b_2$ i.e $a_1/a_2 = b_1/b_2 $
Also, for the pair of lines to coincides, their intercept of y-axis must also be same,
i,e $c_1/b_1=c_2/b_2$
i.e $c_1/c_2=b_1/b_2$
Therefore for the pair of lines to be parallel and coincides,
$a_1/a_2 = b_1/b_2= c_1/c_2$
Things to Remember: Kindly arrange both the polynomials in standard form of $ax+by+c=0$ or
$ax+by=c$ before solving otherwise it may show wrong result.
(iii) ${6x – 3y + 10 = 0}$ ; ${2x – y + 9 = 0}$
Solution:
Given: ${6x – 3y + 10 = 0}$ i.e $a_1=6 ; b_1=-3 ; c_1 = 10$
and ${2x – y + 9 = 0}$ i.e $a_2=2 ; b_2=-1 ; c_2 = 9$
Therefore: $a_1/a_2 = 6/2 = 3 $
$b_1/b_2 = {-3}/{-1} = 3 $
$c_1/c_2 = 10/9 = 10/9 $
Since $a_1/a_2$ = $b_1/b_2$ , therfore the lines are parallel,
but $b_1/b_2$ ≠ $c_1/c_2$ , therefore the lines are not coinciding or equivalent.
3. On comparing the ratios ${a_1/a_2}$ , ${b_1/b_2}$ and ${c_1/c_2}$ find out whether the following pair of linear
equations are consistent, or inconsistent.
(i) ${3x + 2y = 5 }$; ${2x – 3y = 7}$
Solution:
GIven: ${3x + 2y = 5 }$ i.e $ a_1=3 ; b_1=2 ; c_1 = 5 $
and ${2x – 3y = 7}$ i.e $a_2=2 ; b_2=-3 ; c_2 = 7$
Therefore: $a_1/a_2 = 3/2 $
$b_1/b_2 = 2/{-3} = -2/3 $
$c_1/c_2 = 5/7 = 10/9 $
Since $a_1/a_2$≠ $b_1/b_2$, the pair of linear equation is consistent.
Remember: If the pair of linear equations has a solution, then they are consistent. i.e
they must be intersecting or must be coinciding.
(ii) ${2x – 3y = 8 }$; ${4x – 6y = 9}$
Solution
Given: ${2x – 3y = 8 }$ i.e $ a_1=2 ; b_1={-3} ; c_1 = 8 $
and ${4x – 6y = 9}$ i.e $a_2=4 ; b_2={-6} ; c_2 = 9$
Therefore: $a_1/a_2 = 2/4 = 1/2 $
$b_1/b_2 = {-3}/{-6} = 1/2 $
$c_1/c_2 = 5/7 = 10/9 $
Since, $a_1/a_2 = b_1/b_2 ≠ c_1/c_2$, therefore the pair of linear equations are inconsistent.
(i.e pair of linear equation are parallel and has no solutions.)
Remember: If the pair of linear equations has no solution, then they are inconsistent. i.e
they are parallel to each other.
(iii) ${3/2 x + 5/3 y = 7}$ ; ${9x – 10y = 14}$
Solution
Given: ${3/2 x + 5/3 y = 7}$ i.e $a_1 = 3/2 ; b_1 = 5/3 ; c_1 = 7$
${9x – 10y = 14}$ i.e $a_2 = 9 ; b_1 = -10 ; c_1 = 14$
Therefore: $a_1/a_2 = {3/2}/9 = 1/6 $
$b_1/b_2 = {5/3}/{-10} = -1/6 $
$c_1/c_2 = 7/14 = 1/2 $
Since, $a_1/a_2$ ≠ $b_1/b_2$, therefore the lines are intersecting and has solution.
Therefore they are consistent.
(iv) ${5x – 3y = 11}$ ; ${– 10x + 6y = –22}$
Solution
Given: ${5x – 3y = 11}$ i.e $a_1=5 ; b_1={-3} ; c_1=11$
${– 10x + 6y = –22}$ i.e $a_2= {-10} ; b_2={6} ; c_2={-22}$
Therefore: $a_1/a_2 = 5/{-10} = -1/2 $
$b_1/b_2 = {-3}/6 = -1/2$
$c_1/c_2 = 11/{-22} = -1/2 $
Since $a_1/a_2=b_1/b_2=c_1/c_2$, the pair of lines are coinciding and has many solutions.
Therefore they consistent.
(v) ${4/3 x + 2y = 8}$ ; ${ 2x + 3y = 12 }$
Solution
Given: ${4/3 x + 2y = 8}$ i.e $a_1=4/3 ; b_1=2 ; c_1= 8 $
${ 2x + 3y = 12 }$ i.e $a_2 = 2 ; b_2 = 3 ; c_2= 12 $
Therefore: $a_1/a_2 = {4/3}/2 = 2/3 $
$b_1/b_2 = 2/3 = 2/3$
$c_1/c_2 = 8/12 = 2/3 $
Since $a_1/a_2=b_1/b_2=c_1/c_2$, the pair of lines are coinciding and has many solutions.
Therefore they consistent.
4. Which of the following pairs of linear equations are consistent/inconsistent?
(i) ${x + y = 5}$, ${2x + 2y = 10}$
Solution
$a_1/a_2$ = $1/2$
$b_1/b_2$ = $1/2$
$c_1/c_2$=$5/10$=$1/2$
Since $a_1/a_2=b_1/b_2=c_1/c_2$, the pair of lines are coinciding and has many solutions.
Therefore they consistent
(ii) ${x – y = 8}$, ${3x – 3y = 16}$
Solution
$a_1/a_2$ = $1/3$
$b_1/b_2$ = $1/3$
$c_1/c_2$ = $8/16$ = $1/2$
Since, ${a_1/a_2=b_1/b_2≠c_1/c_2}$, the pair of lines are parallel. Hence inconsistent.
(iii) ${2x + y – 6 = 0}$, ${4x – 2y – 4 = 0}$
Solution
$a_1/a_2$ = $2/4$ = $1/2$
$b_1/b_2$ = $1/(-2)$ = $-1/2$
$c_1/c_2$ = $8/16$ = $(-6)/(-4)$ = $3/2$
Since $a_1/a_2≠ b_1/b_2$, so the pair of lines are intersecting and has solution.
Hence they are consistent.
(iv) ${2x – 2y – 2 = 0}$, ${4x – 4y – 5 = 0}$
solution
$a_1/a_2$ = $2/4$ = $1/2$
$b_1/b_2$ = $(-2)/(-4)$ = $1/2$
$c_1/c_2$ =$(-2/(-5))$ = $2/5$
Since, ${a_1/a_2=b_1/b_2≠c_1/c_2}$, the pair of lines are parallel. Hence inconsistent.
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m.
Find the area of the garden.
Solution
Let the width be $x$ $m$
Hence length will be $(x+4)$ $m$
Therefore perimeter = 2(l+b) = 2(x+4+x) = 2(2x+4)
Given Half the perimeter = 36 $m$
${2(2x+4)}/2 = 36$
${(2x+4)} = 36$
${2x} = 36-4$
${2x} = 32$
$x=16$ $m$
Therefore width of rectangular field = $16$ $m$
and length = $20$ $m$
Therefore area = length × breadth = 20×16
= 320 $m^2$
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:
(i) intersecting lines.
For intersecting lines $a_1/a_2≠b_1/b_2$
given $a_1=2$ $b_1=3$
Therefore $2/a_2≠3/b_2$
i.e $a_2/b_2≠2/3$
Therefore for all real values of $a_2$ and $b_2$ where $a_2/b_2≠2/3$ will form a linear equation.
(ii)parallel lines
For parallel lines $a_1/a_2=b_1/b_2≠c_1/c_2$
Since $a_1=2$ $b_1=3$ , $c_1=-8$
therefore for all real values of $a_2$,$b_2$ and $c_2$ where the ratio of
$a_1/a_2=b_1/b_2≠c_1/c_2$ will form a linear equation.
i.e $4x+6y-10=0$
(iii) coincident lines
For coincident lines $a_1/a_2=b_1/b_2=c_1/c_2$
Since $a_1=2$ $b_1=3$ , $c_1=-8$
therefore for all real values of $a_2$,$b_2$ and $c_2$ where the ratio of
$a_1/a_2=b_1/b_2=c_1/c_2$ are equal will form a linear equation.
i.e $4x+6y-16=0$
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
Consider the equation ${x – y + 1 = 0}$
therefore $y=x+1$
(i) Area enclosed by the pair of lines and x-axis will the the region enclosed by triangle ABC as
shown in the graph.
Therefore Base (BC) = 1+4=5
and Height (AD) = 3
Area of triangle = $1/2$× base× height
= $1/2$× 5× 3
=7.5 sq.unit
(ii)Area enclosed by the pair of lines and x-axis will the the region enclosed by triangle AEF as
shown in the graph.
Therefore base (EF) =(6-1) = 5 unit
and height (AG) = 2 unit
Area of triangle = $1/2$× base× height
= $1/2$×5×2
= 5 sq.unit
Ans: Total number of girls = 7 and total number of boys=4 (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.
Cost of 1 pencil is ₹ 3; Cost of 1 pen is ₹ 5;
(i) Area enclosed by the pair of lines and x-axis will the the region enclosed by triangle ABC as
shown in the graph.
Therefore Base (BC) = 1+4=5
and Height (AD) = 3
Area of triangle = $1/2$× base× height
= $1/2$× 5× 3
=7.5 sq.unit
(ii)Area enclosed by the pair of lines and x-axis will the the region enclosed by triangle AEF as
shown in the graph.
Therefore base (EF) =(6-1) = 5 unit
and height (AG) = 2 unit
Area of triangle = $1/2$× base× height
= $1/2$×5×2
= 5 sq.unit