Question 1. Solve the following pair of linear equations by the substitution method.r / Chapter 3: Pair of Linear Equations in Two Variable Maths Class 10 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. Question 1. Solve the following pair of linear equations by the 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) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3 (v)√2 x+ √3y = 0 ; √3 x - √8y = 0 (vi)3/2 x - 5/3y = -2 ; x/3 + y/2 = 13/6 is solved by our expert teachers. You can get ncert solutions and notes for class 10 chapter 3 absolutely free. NCERT Solutions for class 10 Maths Chapter 3: Pair of Linear Equations in Two Variable is very essencial for getting good marks in CBSE Board examinations
Question 1. Solve the following pair of linear equations by the 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) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3
(v)√2 x+ √3y = 0 ; √3 x - √8y = 0 (vi)3/2 x - 5/3y = -2 ; x/3 + y/2 = 13/6
solve first equation
x + y = 14
x = 14 - y …………..(1)
plug this value in equation second we get
x – y = 4
14 – y - y = 4
Add 14 both side we get
- 2 y = 4 - 14
- 2 y = - 10
Y = -10/-2
Y = 5
Plug y = 5 in equation first we get
X = 14 – y
X = 14 – 5
X = 9
Answer x = 9 , y = 5
Solve first equation
s – t = 3
s = 3+t …………(1)
Plug this value in equation second we get
s/3 + t/2 = 6
Multiply by 6 to remove all denominator we get
6 + 2t + 3t = 36
5t = 36 – 6
5t = 30
Divide by 2 we get
t = 30/5
t =6
Plug the value in equation first we get
s = 3 +6
s = 9
Answer s = 9 , t = 6
solve first equation
3x – y = 3
3x =3 +y
x = (3 + y)/3 …………..(1)
Plug this value in equation second we get
9 is always equal to 9
So it is a true condition and in such condition
Equations will have many or infinity solutions
0.4x + 0.5y = 2.3 ...(2)
0.2x + 0.3y = 1.3
Solve first equation we get
0.2x = 1.3 – 0.3y
Divide by 0.2 we get
X = 1.3/0.2 - 0.3/0.2
X = 6.5 – 1.5 y …(3)
Plug the value in equation second we get
0.4x + 0.5y = 2.3
(6.5 – 1.5y)*0.4x + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
-0.1 y = 2.3 – 2.6
Y = - 0.3/-0.1
Y = 3
Plug this value in equation 3 we get
X = 6.5 – 1.5 y
X = 6.5 – 1.5(3)
X =6.5 - 4.5
X =2
Answer x = 2, y = 3
√3 x - √8y = 0 ... (2)
√2 x+ √3y = 0
Subtract √3y both side we get
√2 x = -√3y
Divide by √2
Plug the value of x in equation first we get
√3 x - √8y = 0
Plug this value in equation (3) we get
Our answer x = 0 and y = 0
x/3 + y/2 = 13/6 .......(2)
Solve first equation
3/2 x - 5/3y = -2
Multiply by 6 to remove all denominator we get
9 x – 10 y = -12
9x = -12 + 10y
Divide by 9 we get
x = (-12 + 10 y)/9 …..(3)
plug this value in equation (2) we get
Multiply by LCM of denominator = 27* 2 = 54 we get
2(-12 +10 y ) + 27 y= 117
-24 + 20 y + 27 y = 117
47 y = 117 + 24
Y = 141/47 = 3
Plug this value in equation (3) we get
x = (-12 + 10*3)/9
x=18/9
x = 2
Answer x = 2 and y = 3
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