3. Pair of Linear Equation in two variables

Class X
Exercise 3.3

Q.No. 1. Solve the following pair of linear equation 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)
0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)
√2x + √3y = 0
√3x - √8y = 0
(vi)
3x/2 + 5y/3 = -2
x/3 + y/2 = 13/6

Solution:

(i)
From equation no. (i)
x = 14 - y ------------- (iii)
Now, substitute the value of x in equation no. (ii)
14 - y - y = 4
or, 14 - 2y = 4
or, 2y = 14 - 4 = 10
∴ y  = 10/2 = 5
Now, putting the value of y in equation no. (iii)
 x = 14 - 5 = 9
Hence, the value of x = 5 and y = 9 Ans.

(ii)
From equation no. (i)
s = 3 + t ---------- (iii)
Now, from equation no. (ii)
(3 + t)/3 + t/2 = 6
or, 2(3 + t) + 3t = 6x6
or, 6 + 2t + 3t = 36
or, 6 + 5t = 36
or, 5t = 36 - 6 
or, t = 30/5 = 6
Now putting the value of t in equation no. (iii)
s = 3 + 6 = 9
Hence, the value of s = 9 ant t = 6 Ans.

(iii)
From equation no. (i)
x = (3 + y)/3 ---------------- (iii)
Now, substitute the value of x in equation no. (ii)
9(3 + y)/3 - 3y = 9
9 + 3y - 3y = 9
9 = 9
This is independent of y.
Hence, y has infinite many solution.
Therefore x has infinite many solution.

(iv)

From equation no. (i)
x = (1.3 - 0.3y)/0.2 ----------- (iii)
Now, substitute the value of x in equation no. (ii)
0.4(1.3 - 0.3y)/0.2 + 0.5y = 2.3
or, 2.6 - 0.6y + 0.5y = 2.3
or, 0.1y = 0.3
or, y = 3
Now, putting the value of y in equation no. (iii)
x = (1.3 - 0.9)/0.2
or, x = 2
Hence, x = 2 and y = 4.

(v)
From equation no. (i)
x = -√3y/√2 ------------ (iii)
Now substitute the value of x in equation no. (ii)
-√3x√3y/√2 - √8y = 0
or, 3y - 4y = 0
or, y = 0
Now, putting the value of y in equation no. (iii)
x = 0
Hence, x = 0 and y = 0. Ans.

(vi)
From equation no. (i)
x = (-12 - 10y)/9 -------------- (iii)
Now, substitute the value of x in equation no.(ii)
(-12 - 10y)/27 + y/3 = 13/6
or, -12 - 10y + 9y = 13x27/6
or, -12 - y = 39/2
y = (-24 - 39)/2
y = -63/2
Now, putting the value of y in equation no. (iii)
x = (-12 - 165)/9
x = - 177/9
x = -59/3
x = -59/3 and y = -63/2 Ans.

Exercise 3.4

Q.No.1. Solve the following pair of linear equations by the elimination methods and substitution method:

(i) x + y = 5 and 2x - 3y = 4
(ii) 3x + 4y = 10 and 2x - 3y = 4
(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y = -1 and x - y/3 = 3

(i) Solution:
Elimination Method
Multiply equation no. (i) by 3 and (ii) by 1
3x + 3y = 15 ------------ (iii)
2x - 3y = 4 --------------- (iv)
Now, add equation no. (iii) and (iv)
5x = 19
x = 19/5
Now, putting the value of x in equation no. (iii)
3x19/5 + 3y = 15
3y = 15 - 57/5
3y = (75 - 57)/5
y = 18/15 = 6/5
Required answer 
x = 19/5 and y = 6/5

(ii)
Solution:
Elimination Method
Multiply the equation no. (i) by 3 and (ii) by 4
9x + 12y = 30 ---------- (iii)
8x - 12y = 16 ----------- (iv)
Now, add the equations
17x = 46
x = 46/17
Now, putting the value of x in equation no.(ii)
92/17 - 3y = 4
3y = 92/17 - 4
3y = (92 - 68)/17
y = 24/51
Required answer 
x = 46/17 and y = 24/51 Ans.

(iii)
Solution:
Elimination Method
3x - 5y = 4 ------------ (i)
9x - 2y = 7 ----------- (ii)
Multiply equation no. (i) by 2 and (ii) by 5
6x - 10y = 8 ----------- (iii)
45x - 10y = 35 -------- (iv)
Now, subtract equation no.(iii) from (iv)
39x = 27
x = 27/39
x = 9/13
Now, substitute the value of x in equation no.(i)
27/13 - 5y = 4
5y = 27/13 - 4
5y = (27 - 52)/13
y = -25/65
y = -5/13
Required Answer
x = 9/13 and y = -5/13 Ans.

(iv)
Solution:
x/2 + 2y = -1 ------- (i)
x - y/3 = 3
Multiply equation no.(i) by 1/3 and (ii) by 2
x/6 + 2y/3 = -1/3 ---------- (iii)
2x - 2y/3 = 6 ---------- (iv)
Now, add equation no. (iii) and (iv)
x/6 + 2x = -1/3 + 6
13x/6 = 17/3
x = 34/13
Now, substitute the value of x in equation no.(ii)
34/13 - y/3 = 3
y/3 = 34/13 - 3
y/3 = (34 - 39)/13
y = -5/39
Required Answer
x = 34/13 and y = -5/39 Ans.
Continue ............................. 

2. Polynomials

Class X

Exercise 2.1

Q.No.1. The graph of y = p(x) are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:

The numbers of zeroes in Fig. (i) = 0

The number of zero in Fig. (ii) = 1

The numbers of zeroes in Fig. (iii) = 3

The numbers of zeroes in Fig (iv) = 2

The numbers of zeroes in Fig. (v) = 4

The numbers of zeroes in Fig. (vi) = 3

Exercise 2.2

Q.No.1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x² - 2x - 8 (ii) 4s² - 4s + 1 (iii) 6x² - 3 - 7x (iv) 4u² + 8u (v) t² - 15 (vi) 3x² - x - 4

Solution:

(i)

x² - 2x - 8 = x² - 4x + 2x - 8 {By the method of splitting the middle term}

= x(x - 4) + 2(x - 4) = (x + 2)(x - 4)

So, the value of the polynomial is zero when 

x + 2 = 0  or, x - 4 = 0

x = -2 or x = 4.

So, the zeroes of the given polynomial are -2 and 4.

Sum of zeroes = -2 + 4 = 2

Product of zeroes = -2x4 = -8

Verification

Sum of zeroes = -(Coefficient of x)/ Coefficient of x²

or, Sum of zeroes = -(-2)/1 = 2

Product of zeroes = Constant term/ Coefficient of x²

or, Product of zeroes = -8/1 = -8

Hence, the relationship between the zeroes and coefficients are verified.

(ii)

4s² - 4s + 1 = 4s² - 2s - 2s + 1 {By the method of splitting the middle term}

= 2s(2s - 1) - 1(2s - 1) = (2s - 1)(2s - 1)

So, the value of the polynomial is zero when
2s - 1 = 0 or, 2s - 1 = 0
or, s = 1/2
So, the zeroes of the given polynomial are 1/2 or 1/2
Sum of the zeroes = 1/2 + 1/2 = 1
Product of zeroes = 1/2x1/2 = 1/4


Verification
Sum of zeroes = -(Coefficient of s)/ Coefficient of s²
Hence, the relationship between the zeroes and coefficients are verified.
(iii)
(iv)(v)(vi)Q.No.2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.Solution:Therefore, (ii)Therefore, (iii)Therefore, (iv)Therefore, (v)Therefore, (vi)Therefore, 






or, Sum of zeroes = -(-4)/4 = 1
Product of zeroes = Product of zeroes = Constant term/ Coefficient of s²
or, Product of zeroes = 1/2x1/2 = 1/4


6x² - 3 - 7x = 6x² - 9x + 2x - 3 {By the method of splitting the middle term}
= 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3)
So, the value of the polynomial is zero when
x = -1/3 or x = 3/2
So, the zeroes of the polynomial are 1/3 or 3/2
Sum of the zeroes = -1/3 + 3/2 = (-2 + 9)/6 = 7/6
Product of the zeroes = -1/3x3/2 = -3/6 = -1/2

Verification

Sum of zeroes = -(Coefficient of x)/ Coefficient of x²

or, Sum of zeroes = -(-7)/6 = 7/6

Product of zeroes = Constant term/ Coefficient of x²

or, Product of zeroes = -3/6 = -1/2

Hence, the relationship between the zeroes and coefficients are verified.

4u² + 8u = 4u(u + 2)
So, the value of the polynomial is zero when
u = 0 or, u = -2
So, the zeroes of the polynomial are 0 and -2.
Sum of the zeroes = 0 + (-2) = -2
Product of the zeroes = 0x-2 = 0

Verification

Sum of zeroes = -(Coefficient of u)/ Coefficient of u²

or, Sum of zeroes = -8/4 = -2

Product of zeroes = Constant term/ Coefficient of u²

or, Product of zeroes = 0/4 = 0

Hence, the relationship between the zeroes and coefficients are verified.

t² - 15 = (t + √15)(t - √15)
So, the value of the polynomial is zero when
t = √15 or, t = -√15
So, the zeroes of the polynomial are √15 and -√15.
Sum of the zeroes = -√15 + √15 = 0
Product of the zeroes = -√15x√15 = 15

Verification

Sum of zeroes = -(Coefficient of t)/ Coefficient of t²

or, Sum of zeroes = 0/1 = 0

Product of zeroes = Constant term/ Coefficient of t²

or, Product of zeroes = 15/1 = 15

Hence, the relationship between the zeroes and coefficients are verified.

3x² - x - 4 = 3x² - 4x + 3x - 4 {By the method of splitting the middle term}
= x(3x - 4) + 1(3x - 4) = (x + 1)(3x - 4)
So, the value of the polynomial is zero when
x = -1 or 4/3.
So, zeroes of the polynomial are -1 and 4/3.
Sum of the zeroes = -1 + 4/3 = 1/3
Product of the zeroes = -1x4/3 = -4/3

Verification

Sum of zeroes = -(Coefficient of x)/ Coefficient of x²

or, Sum of zeroes = -(-1)/3 = 1/3

Product of zeroes = Constant term/ Coefficient of x²

or, Product of zeroes = -4/3 = -4/3

Hence, the relationship between the zeroes and coefficients are verified.

(i) 1/4 , -1 (ii) √2, 1/3 (iii) 0, √5 (iv) 1, 1 (v) -1/4, 1/4 (vi) 4, 1

(i)
Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = 1/4 
Product of zeroes = -1

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  4, b = -1 and c = -4
Hence, the required quadratic polynomial
4x² - x - 4

Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = √2
Product of zeroes = 1/3

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  3, b = -3√2 and c = 1
Hence, the required quadratic polynomial

3x² - 3√2x + 1

Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = 0
Product of zeroes = √5

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  1, b = 0 and c = √5 
Hence, the required quadratic polynomial


x²  + √5

Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = 1
Product of zeroes = 1

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  1, b = -1 and c = 1
Hence, the required quadratic polynomial


x²  - x + 1

Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = -1/4
Product of zeroes = 1/4

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  4, b = -1 and c = 1
Hence, the required quadratic polynomial


4x²  + x + 1


Let ax² + bx + c be the required quadratic polynomial.
Sum of zeroes = 4
Product of zeroes = 1

ax² + bx + c =k(x² - sum of zeroes + product of zeroes)
a =  1, b = -4 and c = 1
Hence, the required quadratic polynomial

x²  - 4x + 1

Exercise 2.3

Q.No. 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x³ - 3x² + 5x - 3, g(x) = x² - 2

(ii) p(x) = x⁴ -3x² + 4x +5, g(x) = x² + 1 - x

(iii) p(x) = x⁴ - 5x + 6, g(x) = 2 - x²

Solution:

Degree of p(x) = 3

Degree of g(x) = 2

Polynomial for first step = x³ - 3x² + 5x - 3

Division for first step = (x³ - 3x² + 5x - 3) ÷ (x² - 2)

Therefore, first term of quotient = x

Polynomial for second step =x³ - 3x² + 5x - 3 - x³ + 2x = -3x² + 7x - 3

Division for second step = (-3x² + 7x - 3)÷ (x² - 2)

Second term of quotient = -3

Polynomial for third step =-3x² + 7x - 3 + 3x² - 6 = 7x - 9

Now degree became less than divisor.

So,

Quotient = x - 3

Remainder = 7x - 9

Continue ................