1.Pair of Linear Equations in Two Variables 10th

Pair of Linear Equations in Two Variables 10th

In this article,”Pair of Linear Equations in Two Variables 10th” we will learn to find out solution of equation,whether equations are consistent or inconsistent.

Also Read This Article:- Linear equation

2.Pair of Linear Equations in Two Variables 10th Illustrations

Illustration: 1. Form the pair of linear equations in the following problems, and find their solutions
Illustration: 1 (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 the number of boys = x
number of girls= y
x + y = 10 \quad \cdots{(1)} \\ y - x = 4 \quad \cdots{(2)}
Adding (1) and (2), we get
2y = 14 \Rightarrow y = \frac{14}{2} = 7
Putting above value in (1)
x + 7 = 10 \Rightarrow x = 3
Illustration: 1 (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 one pencil and pen are x and y respectively.
5x + 7y = 50 \quad \cdots{\textcircled{1}} \\ 7x + 5y = 46 \quad \cdots{\textcircled{2}} \\ \text{Multiply \textcircled{1} by 5 and \textcircled{2} by 7, we get} \\ 25x + 35y = 250 \quad \cdots{\textcircled{3}} \\ 49x + 35y = 322 \quad \cdots{\textcircled{4}} \\ \underline{\hspace{5cm}} \quad \text{subtract} \\ -24x = -72 \Rightarrow x = \frac{-72}{-24} = 3 \\ \text{Substitute in equation \textcircled{1}, we get} \\ 5 \times 3 + 7y = 50 \\ \Rightarrow 7y = 50 - 15 \\ \Rightarrow y = \frac{35}{7} = 5 \\ x = 3, y = 5
Illustration: 2. On comparing the ratios \frac{a_1}{a_2}, \frac{b_1}{b_2} \text{ and } \frac{c_1}{c_2}, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or consistent: 
Illustration: 2 (i). \begin{aligned} 5x - 4y + 8 &= 0 \\ 7x + 6y - 9 &= 0 \end{aligned}
\text{Solution: } 5x - 4y + 8 = 0 \quad \cdots{\textcircled{1}} \\ 7x + 6y - 9 = 0 \quad \cdots{\textcircled{2}} \\ a_1 = 5, b_1 = -4, c_1 = 8 \\ a_2 = 7, b_2 = 6, c_2 = -9 \\ \frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3} \\ \frac{a_1}{a_2} \neq \frac{b_1}{b_2}
Lines intersecting each other at a point
Illustration: 2 (ii). 9x + 3y + 12 = 0 ; 18x + 6y + 24 = 0
Solution: 9x + 3y + 12 = 0 ; 18x + 6y + 24 = 0
a_1 = 9, b_1 = 3, c_1 = 12 \\ a_2 = 18, b_2 = 6, c_2 = 24 \\ \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2} \\ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
Lines are coincident
Illustration: 2 (iii). 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution: 6x – 3y + 10 = 0
2x – y + 9 = 0
a_1 = 6, b_1 = -3, c_1 = 10 \\ a_2 = 2, b_2 = -1, c_2 = 9 \\ \frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9} \\ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}
Lines are Parallel

Pair of Linear Equations in Two Variables 10th

Illustration: 3. On comparing the ratios \frac{a_1}{a_2}, \frac{b_1}{b_2} \text{ and } \frac{c_1}{c_2} find out whether the following pair of linear equations are consistent, or inconsistent.
Illustration: 3 (i). 3x + 2y = 5; 2x – 3y = 7
\text{Solution: } 3x + 2y - 5 = 0 \\ 2x - 3y - 7 = 0 \\ a_1 = 3, b_1 = 2, c_1 = -5 \\ a_2 = 2, b_2 = -3, c_2 = -7 \\ \frac{a_1}{a_2} = \frac{3}{2}, \quad \frac{b_1}{b_2} = \frac{2}{-3}, \quad \frac{c_1}{c_2} = \frac{-5}{-7} = \frac{5}{7} \\ \frac{a_1}{a_2} \neq \frac{b_1}{b_2}
Consistent
Illustration: 3 (ii). 2x – 3y = 8; 4x – 6y = 9
Solution: 2x - 3y = 8; \quad 4x - 6y = 9 \\ 2x - 3y - 8 = 0, \quad 4x - 6y - 9 = 0 \\ a_1 = 2, b_1 = -3, c_1 = -8 \\ a_2 = 4, b_2 = -6, c_2 = -9 \\ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9} \\ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}
Inconsistent
Illustration: 3 (iii). \frac{3}{2}x + \frac{5}{3}y = 7; 9x - 10y = 14
Solution: \frac{3}{2}x + \frac{5}{3}y - 7 = 0 \\ 9x - 10y - 14 = 0 \\ a_1 = \frac{3}{2}, \quad b_1 = \frac{5}{3}, \quad c_1 = -7 \\ a_2 = 9, \quad b_2 = -10, \quad c_2 = -14 \\ \frac{a_1}{a_2} = \frac{\frac{3}{2}}{9} = \frac{3}{2 \times 9} = \frac{1}{6} \\ \frac{b_1}{b_2} = \frac{\frac{5}{3}}{-10} = -\frac{5}{3 \times 10} = -\frac{1}{6} \\ \frac{c_1}{c_2} = \frac{-7}{-14} = \frac{1}{2}
\frac{a_1}{a_2} \neq \frac{b_1}{b_2}
Inconsistent
Illustration: 3 (iv). 5x – 3y = 11;-10x + 6y = -22
Solution:5x – 3y – 11 = 0; -10x + 6y + 22 = 0
a_1 = 5, b_1 = -3, c_1 = -11\\ a_2 = -10, b_2 = 6, c_2 = 22\\ \frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}
\frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
Consistent
Illustration: 3 (v). \frac{4}{3}x + 2y = 8; 2x + 3y = 12
Solution: \frac{4}{3}x + 2y - 8 = 0, \quad 2x + 3y - 12 = 0\\ a_1 = \frac{4}{3}, b_1 = 2, c_1 = -8\\ a_2 = 2, b_2 = 3, c_2 = -12\\ \frac{a_1}{a_2} = \frac{\frac{4}{3}}{2} = \frac{4}{2 \times 3} = \frac{2}{3} \\ \frac{b_1}{b_2} = \frac{2}{3}, \quad \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3} \\ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
Consistent
Illustration: 5. 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.
Solution: Let width of garden = y
and length of garden = x
x - y = 4 \quad \cdots{(1)}
x + y = 36 \quad \cdots{(2)}
Adding (1) and (2), we get
2x = 40 \Rightarrow x = 20
Putting the value of x in (2), we get
20 + y = 36 \Rightarrow y = 36 - 20 \Rightarrow y = 16
Illustration: 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) 2x + 3y – 8 = 0
a_1 = 2, b_1 = 3, c_1 = -8
Intersecting lines
\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \quad \text{hence another line}
5x + 2y - 15 = 0
(ii) Parallel lines
\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}
\text{hence } 6x + 9y + 9 = 0
(iii) Coincident lines
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
6x + 9y - 24 = 0
By solving the above illustrations,one can understand the Pair of Linear Equations in Two Variables 10th.

Also Read This Article:- Revisiting Irrational Numbers Class 10

3.Practice Questions of Pair of Linear Equations in Two Variables 10th for Students

Pair of Linear Equations in Two Variables 10th

On comparing the ratios and,find out whether the following pair of linear equations are consistent or inconsistent
(1.)3x-y=2,6x-2y=4
(2.)2x-2y=2,4x-4y=5
Answers:(1.)consistent (2.)inconsistent
By solving the above questions,you can understand the Pair of Linear Equations in Two Variables 10th well because the concept is well
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Also Read This Article:- Fundamental Theorem of Arithmetic 10th

4.Frequently Asked Questions Related to Pair of Linear Equations in Two Variables 10th

5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

Pair of Linear Equations in Two Variables 10th

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
यदि 2x^3 + 4x^2 + 2ax + b, पूरी तरह x^2 - 1 से विभाजित हो जाए, तो a और b के मान क्या होंगे?
(If 2x^3 + 4x^2 + 2ax + b, is completely divisible by x^2 - 1 , what will be the values of a and b?)
*पिछली प्रश्नोत्तरी का हल*
माना संख्याएँ 7x और 2x हैं
7x \times 2x = 126 \\ 14x^2 = 126 \\ \Rightarrow x^2 = 9 \\ \Rightarrow x = 3
अतः संख्याएँ 21 और 6 हैं
संख्याओं के वर्गों का अन्तर
= (21)^2 - 6^2 \\ = 441 - 36 \\ = 405
*Previous Quiz Solution*
Let the numbers be 7x and 2x.
7x \times 2x = 126 \\ \Rightarrow 14x^2 = 126 \\ \Rightarrow x^2 = 9 \\ \Rightarrow x = 3
Hence, the numbers are 21 and 6
Difference of squares of numbers
= (21)^2 - 6^2 \\ = 441 - 36 \\ = 405

*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*