Solution:
We will solve the pair of linear equations given by substitution method.
(i) x + y = 14 ...(1)
x - y = 4 ...(2)
By solving the equation (1)
y = 14 - x...(3)
Substitute y = 14 - x in equation (2), we get
x - (14 - x) = 4
2x - 14 = 4
2x = 4 + 14
2x = 18
x = 9
Substituting x = 9 in equation (3), we get
y = 14 - 9
y = 5
Thus, x = 9, y = 5
(ii) s - t = 3...(1)
s/3 + t/2 = 6...(2)
By solving equation (1)
s - t = 3
s = 3 + t...(3)
Substitute s = 3 + t in equation (2), we get
(3 + t) / 3 + t/2 = 6
(6 + 2t + 3t) / 6 = 6
6 + 5t = 6 × 6
5t = 36 - 6
t = 30/5
t = 6
Substituting t = 6 in equation (3), we get
s = 3 + 6
s = 9
Thus, s = 9, t = 6
(iii) 3x - y = 3 ...(1)
9x - 3y = 9 ...(2)
By solving the equation (1)
3x - y = 3
y = 3x - 3 ...(3)
Substitute y = 3x - 3 in equation (2), we get
9x - 3(3x - 3) = 9
9x - 9x + 9 = 9
9 = 9
This shows that the lines are coincident having infinitely many solutions.
x can take any value. i.e. Infinitely many Solutions.
(iv) 0.2x + 0.3y = 1.3 ...(1)
0.4x + 0.5 y = 2.3 ...(2)
Multiply both the equations (1) and (2) by 10, to remove the decimal number and making it easier for calculation.
[0.2x + 0.3y = 1.3] × (10)
⇒ 2x + 3y = 13 ...(3)
[0.4x + 0.5y = 2.3] × (10)
⇒ 4x + 5y = 23 ...(4)
By solving the equation (3)
2x + 3y = 13
3y = 13 - 2x
y = (13 - 2x) / 3 ...(5)
Substitute y = (13 - 2x)/3 in equation (4), we get
4x + 5 [(13 - 2x) / 3] = 23
(12x + 65 - 10x) / 3 = 23
2x + 65 = 23 × 3
2x = 69 - 65
x = 4/2 = 2
Substituting x = 2 in equation (5), we get
y = (13 - 2 × 2) / 3
y = 9/3 = 3
Thus, x = 2, y = 3
(v) √2x + √3y = 0...(1)
√3x - √8y = 0...(2)
By solving equation (1)
√2x + √3y = 0
√3y = - √2x
y = - (√2x/3) ...(3)
Substitute y = - √2x/3 in equation (2), we get
√3x - √8(- √2x/3) = 0
√3x + (√16x/3) = 0
(3√3x + 4x)/3 = 0
x(3√3 + 4) = 0
x = 0
Substituting x = 0 in equation (3), we get
y = - (√2 × 0) / 3
y = 0
Thus, x = 0, y = 0
(vi) (3x/2) - (5y/3) = - 2 ...(1)
x/3 + y/2 = 13/6 ...(2)
Multiply both the equations (1) and (2) by 6, to remove the denominatores and making it easier for calculation.
[3x/2 - 5y/3 = - 2] × 6
9x - 10y = - 12 ...(3)
[x/3 + y/2 = 13/6] × 6
2x + 3y = 13 ...(4)
By solving equation (3)
9x - 10y = - 12
10y = 9x + 12
y = (9x + 12) / 10 ...(5)
Substituting y = (9x + 12) / 10 in equation (4), we get
2x + 3 [(9x + 12) / 10] = 13
(20x + 27x + 36) / 10 = 13
47x = 130 - 36
x = 94/47 = 2
Substituting x = 2 in equation (5), we get
y = (9 × 2 + 12) / 10
y = 30/10
y = 3
Thus, x = 2, y = 3
☛ Check: NCERT Solutions for Class 10 Maths Chapter 3
Video Solution:
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 + 0.5 y = 2.3 (v) √2x + √3y = 0; √3x - √8y = 0 (vi) (3x/2) - (5y/3) = -2; x/3 + y/2 = 13/6
NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.3 Question 1
Summary:
On solving the pair of linear equations by the substitution method we get the variables as: (i) x + y = 14; x - y = 4 where, x = 9, y = 5, (ii) s - t = 3; s/3 + t/2 = 6 where, s = 9, t = 6, (iii) 3x - y = 3; 9x - 3y = 9 where, x can have infinitely many solutions, (iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5 y = 2.3 where, x = 2, y = 3, (v) √2x + √3y = 0; √3x - √8y = 0 where, x = 0, y = 0, (vi) (3x/2) - (5y/3) = -2; x/3 + y/2 = 13/6 where, x = 2, y = 3.
☛ Related Questions:
Consider #Equation(1)#
Add #color(red)(y)# to both sides
#" "color(green)(3x-ycolor(red)(+y)" "=" "4 color(red)(+y))#
But #-y+y=0# giving:
#" "3x+0=4+y#
#" "3x=4+y#
Subtract #color(red)(4)# from both sides
#" "color(green)(3xcolor(red)(-4)" "=" "4color(red)(-4)+y)#
#" "3x-4=y#
#" "y=3x-4" "......Equation(1_a)#
As we have just used equation(1) we now need to use equation(2)
Using #Equation(1_a)# substitute for #y# in #Equation(2)#
#color(green)(2x-3color(red)(y)=-9" "->" "2x-3(color(red)(3x-4))=-9)#
#" "2x-9x+12=-9#
#" "-7x+12=-9#
Subtract 12 from both sides
#" "color(green)(-7x+12color(red)(-12)" "=" "-9color(red)(-12))#
#" "-7x+0=-21#
Divide both sides by -7
#" "color(green)((-7)/(color(red)(-7))color(white)(.) x=(-21)/(color(red)(-7)))#
#" "+1xx x=+3#
#" "x=3# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of y")#
I chose equation 1 as it is the most strait forward one to use.
Substitute for x in equation 1 giving:
#3x-y=4" "->" "3(3)-y=4#
#y=5#
We think you wrote:
This solution deals with linear equations with two unknowns.
[1] 2x + 3y = 17 [2] 3x - y = 9 3y + 2x = 17 y + 3x = 9
// Solve equation [2] for the variable y
// Plug this in for variable y in equation [1]
// Solve equation [1] for the variable x
[1] 11x = 44 [1] x = 4// By now we know this much :
x = 4 y = 3x-9// Use the x value to solve for y
Solution :
{x,y} = {4,3}
Page 2
We think you wrote:
This solution deals with linear equations with two unknowns.
[1] 2x + 3y = 17 [2] 3x - y = 9 3y + 2x = 17 y + 3x = 9
// Solve equation [2] for the variable y
// Plug this in for variable y in equation [1]
// Solve equation [1] for the variable x
[1] 11x = 44 [1] x = 4// By now we know this much :
x = 4 y = 3x-9// Use the x value to solve for y
Solution :
{x,y} = {4,3}