दोस्तों अब हम 👉NCERT SOLUTIONS FOR CLASS 10 MATHS EXERCISE 3.3 CHAPTER 3
👉 EX. 3.3 CLASS 10
👉 MATHS NCERT EX 3.3 CLASS 10 SOLUTIONS
👉EX. 3.3 CLASS 10 QUESTION 1
👉 CLASS 10 MATHS CHAPTER 3 EXERCISE 3.3 QUESTION 1 के सारे प्रश्न हल करेंगे।
प्रश्न - (1) निम्न रैखीक समीकरण युग्मो को प्रतिस्थापन विधि से हल कीजिए ?
(I) X + Y =14 ; X - Y = 4 (II ) s - t = 3 ; s / 3 + t / 2 = 6
(iii) 3x - y = 3 ; 9x - 3y = 9 (iv) 0. 2 x + 0 . 3 y =1 . 3 ; 0. 4 x + 0 . 5 y = 2 . 3
(v) \/2 x + \/3 y = 0 ; \/3 x - \/8 y =0 (vi) 3x / 2 - 5y /3 = -2 ; x / 3 + y / 2 = 13 / 6
हल : (i) दिया गया समीकरण , X + Y =14 --------> ( 1 )
X - Y = 4 --------> (2)
समीकरण (1) से ,
X + Y =14
x = 14 - y --------> ( 3 )
समीकरण (3 ) से x का मान समीकरण ( 2 ) में रखने पर ,
X - Y = 4
=> 14 -y -y = 4
=> 14 -2y = 4
=> 14 -4 = 2y
=> 10 = 2y
=> 10 /2 = y
=> 5 = y
=> y = 5
y का यह मान समीकरण (3 ) में रखने पर ,
=> x = 14 -y
=> x = 14 - 5
=> x = 9
उत्तर की जाँच: x + y
= 9 +5 = 14
= r . h . s
(II ) s - t = 3 ; s / 3 + t / 2 = 6
हल : दिया गया रेखिक समीकरण ,
s - t = 3 --------> (1)
s / 3 + t / 2 = 6 ----------> (2)
समीकरण (1 ) से , s = 3 + t --------> ( 3 )
s का यह मान समीकरण ( 2 ) में रखने पर ,
s / 3 + t / 2 = 6
(3 + t) / 3 + t / 2 = 6
=> [ 2 (3 + t ) + 3t ] / 6 = 6
=> 6 + 2t + 3t = 36
=> 6 + 5t = 36
=> 5t = 36 -6
=> 5t = 30
=> t = 30 / 5
=> t = 6
t का यह मान समीकरण ( 3 ) में रखने पर ,
s = 3 + t
=> s = 3 + 6
=> s = 9
उत्तर की जाँच : समीकरण ( 1 ) में s और t का मान रखने पर ,
s - t = 3
L.H.S = s - t
= 9 - 6
= 3 = R .H .S
(iii) 3x - y = 3 ; 9x - 3y = 9
हल : 3x - y = 3 -------> ( 1 )
9x - 3y = 9 -------> ( 2 )
समीकरण ( 1 ) से ,
3x - y = 3
3X - 3 = Y
या
Y = 3X - 3 समीकरण ( 2 ) में रखने पर ,
=> 9x - 3y = 9
=> 9x - 3 (3X - 3 ) = 9
=> 9x - 9x + 9 = 9
=> 9 = 9 जो की x से रहित समीकरण हे ,
इसलिए x के सभी मानो के लिए सत्य हे। यहाँ समीकरण के अन्नंत हल होंगे।
यहाँ पर 9 को समीकरण के रूप में प्रकार लिख सकते हे।
9 = 9
जैसे , 0 * x + 0 * y = 9 और 0 * x + 0 * y = 9
यहाँ पर दोनों ही समीकरण एक हि हे। इसका अर्थ यह हुआ के दोनों रेखा एक ही हे। जिससे समीकरण निकाय के अनंत अनेक हल होंगे।
(iv) 0. 2 x + 0 . 3 y =1 . 3 ; 0. 4 x + 0 . 5 y = 2 . 3
हल : दिया गया रेखिक समीकरण ,
0. 2 x + 0 . 3 y =1 . 3 --------> ( A )
0. 4 x + 0 . 5 y = 2 . 3 --------> (B )
समीकरण ( A ) व समीकरण ( B ) को सरल करने के लिए , इसमें 10 गुणा करने पर ,
2 x + 3y = 13 ------> ( 1 )
4x + 5y = 23 -----> ( 2 )
समीकरण ( 1 ) से , 2x = 13 - 3y
x = (13 - 3y) / 2 -------> (3)
x का यह मान समीकरण ( 2 ) में रखने पर ,
=> 4x + 5y = 23
=> 4 * (13 - 3y) / 2 + 5y = 23
=> 2 * ( 13 - 3y ) + 5y = 23
=> 26 - 6y + 5y = 23
=> -y = -26 + 23
=> -y = -3
=> y = 3
y का यह मान समीकरण ( 3 ) में रखने पर ,
=> x = (13 - 3y) / 2
=> x = 13 -3 * 3 / 2
=> x = ( 13 - 9 ) / 2
=> x = 4 / 2
=> x = 2
उत्तर : x = 2 , y = 3
(v) \/2 x + \/3 y = 0 ; \/3 x - \/8 y =0
हल : दिया गया रेखिक समीकरण ,
\/2 x + \/3 y = 0 -------> ( 1 )
\/3 x - \/8 y = 0 --------> ( 2 )
समीकरण ( 1 ) से ,
Friends, now we are going to NCERT SOLUTIONS FOR CLASS 10 MATHS EXERCISE 3.3 CHAPTER 3
EX. 3.3 CLASS 10
Maths NCERT EX 3.3 CLASS 10 SOLUTIONS
EX. 3.3 CLASS 10 QUESTION 1
CLASS 10 MATHS CHAPTER 3 EXERCISE 3.3 QUESTION 1 will solve all the questions.
Question - (1) Solve the following pair of linear equations by substitution method?
(I) X + Y =14 ; X - Y = 4 (II) s - t = 3 ; s / 3 + t / 2 = 6
(iii) 3x – y = 3 ; 9x - 3y = 9 (iv) 0. 2 x + 0 . 3 y =1 . 3 ; 0. 4 x + 0 . 5 y = 2 . 3
(v) \/2 x + \/3 y = 0 ; \/3 x - \/8 y =0 (vi) 3x / 2 - 5y /3 = -2 ; x / 3 + y / 2 = 13 / 6
Solution : (i) Given equation, X + Y =14 --------> ( 1 )
X - Y = 4 --------> (2)
From equation (1),
X + Y =14
x = 14 - y --------> ( 3 )
Substituting the value of x from Equation (3) in Equation (2),
X - Y = 4
14 -y -y = 4
14 -2y = 4
14 -4 = 2y
10 = 2y
10 /2 = y
5 = y
y = 5
Substituting this value of y in Equation (3),
=> x = 14 -y
=> x = 14 - 5
=> x = 9
Answer check: L.H.S = x + y
= 9 +5
= 14
= R.H.S
(II) s - t = 3 ; s / 3 + t / 2 = 6
Solution : Given linear equation,
s - t = 3 --------> (1)
s / 3 + t / 2 = 6 ----------> (2)
From Equation (1) , s = 3 + t --------> ( 3 )
Substituting this value of s in Equation (2),
s / 3 + t / 2 = 6
(3 + t) / 3 + t / 2 = 6
=> [ 2 (3 + t ) + 3t ] / 6 = 6
=> 6 + 2t + 3t = 36
=> 6 + 5t = 36
=> 5t = 36 -6
=> 5t = 30
=> t = 30 / 5
=> t = 6
Substituting this value of t in Equation ( 3 ) ,
s = 3 + t
=> s = 3 + 6
=> s = 9
Check the answer : Substituting the values of s and t in Eqn. ( 1 ) ,
s - t = 3
L.H.S = s - t
= 9 - 6
= 3 = R .H .S
Solution : 3x - y = 3 -------> ( 1 )
9x - 3y = 9 -------> ( 2 )
From equation (1),
3x - y = 3
3X - 3 = Y
Or
Putting Y = 3X - 3 in equation ( 2 ) ,
=> 9x - 3y = 9
=> 9x - 3 (3X - 3 ) = 9
=> 9x - 9x + 9 = 9
=> 9 = 9 which is an equation without x,
Hence is true for all values of x. Here the equation will have infinite solutions.
Here 9 can be written in the form of an equation.
9 = 9
For example, 0 * x + 0 * y = 9 and 0 * x + 0 * y = 9
Here both the equations are same. This means that both the lines are the same. Thus the system of equations will have infinitely many solutions.
(iv) 0. 2 x + 0 . 3 y =1 . 3 ; 0. 4 x + 0 . 5 y = 2 . 3
Solution : Given linear equation,
0. 2 x + 0 . 3 y =1 . 3--------> ( A )
0. 4 x + 0 . 5 y = 2 . 3 --------> (B )
To simplify equation (A) and equation (B), multiplying by 10,
2 x + 3y = 13 ------> ( 1 )
4x + 5y = 23 -----> ( 2 )
From equation ( 1 ), 2x = 13 - 3y
x = (13 - 3y) / 2 -------> (3)
Substituting this value of x in Equation (2),
=> 4x + 5y = 23
=> 4 * (13 - 3y) / 2 + 5y = 23
=> 2 * ( 13 - 3y ) + 5y = 23
=> 26 - 6y + 5y = 23
=> -y = -26 + 23
=> -y = -3
=> y = 3
Substituting this value of y in equation ( 3 ) ,
=> x = (13 - 3y) / 2
=> x = 13 -3 * 3 / 2
=> x = ( 13 - 9 ) / 2
=> x = 4 / 2
=> x = 2
Answer: x = 2, y = 3
(v) \/2 x + \/3 y = 0 ; \/3 x - \/8 y =0
Solution : Given linear equation,
\/2 x + \/3 y = 0 -------> ( 1 )
\/3 x - \/8 y = 0 --------> ( 2 )
From equation (1),
Substituting the value of x from Equation (3) in Equation (2),
Substituting this value of y in equation ( 3 ) ,
Answer : x = 0, y = 0
(vi) 3x / 2 - 5y /3 = -2
x / 3 + y / 2 = 13 / 6
Solution : Given equation,
In equation (1) and (2) multiplying by 6,
9x - 10y = -12 ------- > ( 3 )
2x + 3y = 13 --------- > ( 4 )
From equation (4),
Substituting this value of x in equation ( 3 ) ,
117 - 27 y - 20 y = -24
117 + 24 = 27 y + 20 y
141 = 47y
Substituting this value of y in the equation for x,
x = 2 , y = 3 answer