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.
Let the no. of boys taking part be x
Let the no. of girls taking part be y
Since, total students taking part in quiz are 10.
So,
x + y = 10
y = 10 – x. —- (i)
A/Q,
y = x + 4. —–(ii)
Table for eq (i)
| x | 4 | 5 | 6 |
| y | 6 | 5 | 4 |
Table for eq (ii)
| x | 0 | 1 | -1 |
| y | 4 | 5 | 3 |
The point of intersection of both lines is (3,7)
So,
No. of boys taking part is x = 3 boys
No. of girls taking part is y = 7 girls
(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
let the cost of one pencil be ₹x
cost of one pen = ₹y
A/Q,
5x + 7y = 50
5x = 50 – 7y
x = \frac{50 -7y}{5}
and
7x + 5y =46
5y = 46 -7y
y = \frac{46 – 7x}{5}
Table for eq (i)
| x | 3 | 10 |
| y | 5 | 0 |
Table for eq (ii)
| x | 8 | 3 |
| y | -2 | 5 |
The point of intersection of both lines is (3,5)
So,
Cost of one pencil = x = 3
Cost of one pen = y = 5
On comparing the ratios \frac{a_{1}}{a_{2}}
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
\frac{a_{1}}{a_{2}} = \frac{5}{7}
\frac{b_{1}}{b_{2}} = \frac{-4}{6}
\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}
So, the line are intersecting at a point
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
\frac{a_{1}}{a_{2}} = \frac{9}{18} = \frac{1}{2}
\frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}
\frac{c_{1}}{c_{2}} = \frac{12}{24} = \frac{1}{2}
\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}
So, the line are coincident
(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
\frac{a_{1}}{a_{2}} = \frac{6}{2} = 3
\frac{b_{1}}{b_{2}} = \frac{-3}{-1} = 3
\frac{c_{1}}{c_{2}} = \frac{10}{9}
\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}
So, the line are parallel
On comparing the ratios \frac{a_{1}}{a_{2}}
(i) 3x + 2y = 5 ; 2x – 3y = 7
\frac{a_{1}}{a_{2}} = \frac{3}{2}
\frac{b_{1}}{b_{2}} = \frac{2}{-3}
\frac{c_{1}}{c_{2}} = \frac{5}{7}
\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}
So, the line are intersecting at a point and has a solution. The lines are consistent.
(ii) 2x – 3y = 8 ; 4x – 6y = 9
\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}
\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}
So, the line are parallel and has no solution
\therefore
(iii) \frac{3}{2}x
\frac{a_{1}}{a_{2}} = \frac{\frac{3}{2}}{9} = \frac{3}{2 \times 9} = \frac{1}{2\times3} = \frac{1}{6}
\frac{b_{1}}{b_{2}} = \frac{\frac{5}{3}}{-10} = \frac{5}{3 \times (-10)} = \frac{-1}{3 \times 2} = \frac{-1}{6}
\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}
So, the line are intersecting at a point and has solution
\therefore
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
\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}
\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}
So, the line are coincident and has infinitely many solutions
\therefore
(v) \frac{4}{3}x
\frac{a_{1}}{a_{2}} = \frac{\frac{4}{3}}{2} = \frac{4}{3 \times 2} = = \frac{2}{3}
\frac{b_{1}}{b_{2}} = \frac{2}{3}
\frac{c_{1}}{c_{2}} = \frac{8}{12} = \frac{2}{3}
\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}
So, the line are coincident
\therefore
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
\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}
\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}
So, the line are coincident and has infinitely many solutions
\therefore
y = 5 – x —-(i)
Table for eq(i)
| x | 2 | 3 |
| y | 3 | 2 |
y = \frac{10 – 2x}{2}
Table for eq(ii)
| x | 4 | 5 |
| y | 1 | -0 |
(ii) x – y = 8, 3x – 3y = 16
\frac{a_{1}}{a_{2}} = \frac{1}{3} = \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}
\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}
So, the line are parallel and has no solution
\therefore
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
\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}
\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}
So, the line are intersecting and has solution
\therefore
y = 6 – 2x
Table for eq(i)
| x | 2 | 1 |
| y | 2 | 4 |
2y = 4x – 4
y = \frac{4x – 4}{2}
Table for eq(ii)
| x | 1 | 0 |
| y | 0 | -2 |
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
\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}
\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}
So, the line are parallel and has no solution
\therefore
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.
let the length of garden be x m
the width of garden = y m
A/Q,
length is 4 more than its width
x = 4 + y
x – y = 4 ——(i)
and
\frac{2(x + y)}{2}
x + y = 36 ——(ii)
Adding eq (i) & (ii)
| x – y = 4 |
| x + y = 36 |
| 2x = 40 |
x = 20 m
Putting value of x in eq (i)
20 – y = 4
y = 16
So, length = x = 20 m
width = y= 16 m
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
Intersecting lines :- for having intersecting lines
\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}
\therefore
(ii) parallel lines
Parallel lines :- for having parallel lines
\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}
\therefore
(iii) coincident lines
Coincident lines :- for having coincident lines
\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}
\therefore
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.
x – y + 1 = 0
x = y – 1 —-(i)
Table for eq(i)
| x | 0 | 1 |
| y | 1 | 2 |
3x + 2y – 12 = 0
2y = 12 – 3x
y = \frac{12 – 3x}{2}
Table for eq(ii)
| x | 4 | 2 |
| y | 0 | 3 |
Coordinates of triangle formed are (-1,0), (2,3) and (4,0)