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NCERT Class 10 Mathematics Chapter 3 Pair of Linear Equations in two Variables
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Pair of Linear Equations in two Variables
Chapter – 3
| Exercise 3.1 |
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.
Ans: Let number of boys be x and the number of girls be y.
According to the given conditions
x + y = 10 and y = x + 4
We get the required pair of linear equations as
x + y-10 = 0, x – y + 4 = 0
Graphical solution
x + y – 10 = 0 … (1)
| x | 2 | 5 |
| Y = 10- x | 8 | 5 |
x – y + = …..(2)
| x | 2 | 4 |
| Y = x + 4 | 6 | 8 |
From the graph, we have x = 3, y = 7
Common solution of the two linear equations.
Hence, the number of boys = 3
and the number of girls = 7
(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Ans: Let the cost of one pencil be Rs. x and the cost of one pes be Rs. y
According to the given condition,
Cost of 5 pencils and 7 pens = Rs. 50
Cost of 7 pencils and 5 pens = Rs. 46
5x + 7y = 50 and 7x+5y=46
Hence, the required linear equations are
5x + 7y=50, 7x+5y = 46
Graphical solution 5 × + y = 50 …(1)
| × | 2 | 3 | 6 |
| Y= 50 – 5× /7 | 5.7 | 5 | 2.8 |
7× + 5y = 46 …(2)
| × | 2 | 3 | 4 |
| Y = 46 – 7 × /5 | 6.4 | 5 | 3.6 |
From the graph, we have the common solution x = 3, y = 5 Hence, the cost of one pencil = Rs. 3 and the cost of one pen = Rs.
2. One compares the ratios a and find out whether the lines C2 representing the following pairs of linear equations intersect at a point, are parallel or coincident.
(i) 5x-4y+8=0… (1)
7x+6y-9=0… (2)
Ans: 5x-4y+8=0
7x+6y-9=0
a₁ /b₂ 5/7, b₁/b₂ = -4/6 = -4/6 = -⅔
⇒ a₁/b₂ ≠ b₁/b₂
⇒ Lines represented by (1) and (2) intersect at a point
(ii) 9x+3y+12=0 … (1)
18x+6y+ 24 = 0… (2)
Ans: 9 × + 3y + 12 = 0
18 × + 6 y + 24 = 0
a₁/a₂ = 8/19 = ½ , b₁/b₂= 3/6 = ½ ⇒a₁a₂ /b₁/b₂ = C₁ /C ₂
⇒ Liner represented by (i) ans (ii) are coincident
(iii) 6x-3y+10=0… (1)
2x-y+9=0… (2)
Ans: 6x-3y+10=0
2x-y+9=0
a₁/a₂ = 6/2 = 3, b₁/b₂ = -3/-1 = 3 ,C₁ /C ₂ = 10/9
a₁/a₂ = b₁/b₂ ≠ C₁ /C
⇒ lines represented by (1) and (2) are parallel.
3. On the comparing the rations a₁/a₂ = b₁/b₂ and C₁ /C find out whether the following pairs of lines questions are consistent for inconsistent.
(i) 3x+2y=5; 2x-3y=7
Ans: 3× + 2y – 5 = 0
2 × – 3y – 7 = 0
a₁/a₂ = 3/2 ; b₁/b₂ = 2/-3 = -⅔ ⇒ a₁/a₂≠C₁ /C₂
⇒ The equations have a unique Hence, consistent
(ii) 2x-3y = 8; 4x-6y=9
Ans: 2× – 3y – 8 = 0
4× – 6y – 9 = 0
a₁/a₂ = 2/4 = ½ ; b₁b₂ = -3 /-6 = ½; C₁ /C₂
= -8/-9 = 8/9 ⇒ a₁/a₂ = b₁b₂ ⇒ C₁ /C₂
⇒ The equation have no solution.
Hence, inconsistent.
(iii) 3/2 + ⅗y = 7; 9x-10y = 14
Ans: 3/2 + ⅗y – 7 = 0
9x-10y – 14 = 0
a₁/a₂ = (3/2)/9 = ⅙ ; ≠ = (5/3)/-10 = ⅙ ⇒
a₁/a₂ ≠ b₁b₂
⇒ The equation have no solution. Hence, inconsistent.
(iv) 5× – 3y = 11; – 10× + 6y = -22
Ans: 5x – 3y – 11 ,= 0
10x – 6y – 22 = 0
→ Equations have infinitely many solutions. Hence, consistent.
(v) 4/3× 2y = 8; +3y = 12
Ans:
⇒ The equation have no solution. Hence, inconsistent.
4. Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5, 2x + = 10
Ans: x + y – 5 = 0 ……(i)
2x + ½ , b₁b₂ = ½, C₁ /C₂ = -5/-10 = ½
i.e a₁/a₂ = b₁b₂ = C₁ /C₂
Hence, the pair of linear equation is consistent (1) and (2) are same equations and hence the graph is a consistent straight Liner.
| X | 1 | 3 |
| Y = 5 – x | 4 | 2 |
Graph of liner equations (1) and (2) is the coincident line passing through the points (1,4) and (3,2) . every point on the line given a solution.
(ii) X- y = 8, 3X- 3y = 16
Ans: x-y -8= 0 (i)
3x- 3y -16 = 0…(ii)
a₁/a₂ = ⅓, b₁b₂= -1/-3, ⅓ C₁/C₂ = -8/-16,= ½ ; a₁/a₂ = b₁b₂ ≠ C₁/C₂
The pair of equations is inconsistent and the graph of two equations is a pair of parallel straight lines.
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
Ans: 2x + y – 6 = 0 (1)
4x – 2y – 4 = 0… (2)
a₁/a₂ = 2/4 = ½, b₁b₂ = 1/-2 = -½