MANISH KUMAR
MATHEMATICS
POLYNOMIALS
POLYNOMIAL : An algebraic expression of the form p(x) = anxn + an-1x
n-1
+ an-2x
n-2
1
+ …… + a1x + a0x
0
where a0 0 and a0, a1, a2 ....an are real numbers and each power of x is a positive integer is called a polynomial. n
n-1
…….. x and anxn, an-1x ,an-2x
n
n-1
+ …. a1 x + a0 is a polynomial in variable x.
n
n-1
+ …. a1 y + a0 is a polynomial in variable y.
n
n-1
+ …. a1 z + a0 is a polynomial in variable z.
Hence, an, an-1, an-2, are coefficient of x , x Example -
(i) (ii) (iii)
p(x) = anx + an-1 x
p(y) = any + an-1 y p(z) = anz + an-1 z
0
n-1
n-2
are terms of the polynomial.
above all are called polynomials if
Example (i) p(u) = (ii)
1 3 u - 3u2 + 2y - 4 is a polynomial in variable u. 2
1 3 1 u , -3u2, 2u, -4 are known as terms of polynomial and , -3, 2, -4 are their coefficients. 2 2
TYPES OF POLYNOMIALS Generally we divided the polynomials in two categories. Types of polynomials
Bases no number of terms
(A)
Based on degree
BASED ON NUMBER OF TERMS :- These are follows : (a) Monomial :
Polynomials having only one term are called monomials. (‘Mono’ means ‘one’) 2x, 5x2, -5x2, y, u4
(b) Binomial
:
A polynomial of two terms is called binomial. p(x) = x+ 1, q(x) = x2, r(q) = y30 + 1
(c) Trinomial :
A polynomial of three terms is called a trinomial. p(x) = n + x2 + q(x) =
(B)
2 + x + x2
BASED ON DEGREE :Degree of Polynomials : The highest power of variable in a polynomial is known as it degree. For example : (a) p(y) = 2y2 - 3y + 7 is a polynomial in the variable y of degree 2.
MANISH KUMAR (b) q(x) =
MATHEMATICS 2 x + 13x4 + 5x6 is a polynomial in variable x of degree 6.
MANISH KUMAR
MATHEMATICS
(i) Linear Polynomial : A polynomial of degree one is called a linear polynomial. Ex.
p(x) = 4x + 5 q(y) = 2y r(t) = t +
2
s(u) = 3 - u (ii) Quadratic Polynomial : A polynomial of degree two is called a quadratic polynomial. Ex.
2x2 + 5 5x2 + 3x + x2 +
2 x 5
(iii) Cubic Polynomial : A polynomial of degree three is called a cubic polynomial. Ex.
4x2 . 2x3 + 1, 5x3 + x2 2x3 + 4x2 + 6x + 7
(vi) Biquadrate polynomial : A polynomial of fourth degree is called a biquadrate polynomial. Ex.
4x4 + 2x3 + 5x2 + x + 1
Type of Polynoials
Degree
Expression
(i)
Constant polynomials
Zero
F(x) = c, {(x) = 0, is zero polynomial [Degree not defined]
(ii)
Linear polynomials
1
F(x) = ax + b, a 0
(iii)
Quadritic polynomials
2
F(x) = ax + bx + c, a 0
(iv)
Cubic polynomials
3
F(x) = ax + bx + cx + d a 0
(v)
Bi-quadratic polynomials
4
F(x) = ax + bx + cx + dx + e, a 0
2 3 4
2 3
2
VALUE OF POLYNOMIALS : If p(x) is a polynomial in variable and is any real number, then the value obtained by replacing x by in p(x) is called value of p(x) at = and is denoted by p( ). For example : Find the value of p(x) = x3 - 6 x2 + 11x - 6 at x = -2
p(-2) = (-2)3 - 6 (-2)2 + 11(-2) -6 = -8 -24 - 22 - 6 p(-2) = - 60
ZERO OF POLYNOMIALS : A real number is a zero of the polynomial p(x) if p( ) = 0
Ex:
p(x) = x3 - 6x2 + 11x - 6 p(1) = (1)3 - 6(1)2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 p(2) = (2)3 - 6(2)2 + 11(2) - 6 = 8 - 24 + 22 - 6 = 0
MANISH KUMAR
MATHEMATICS
p(3) = (3)3 - 6(3)2 + 11(3) - 6 = 27 - 54 + 33 - 6 = 0 Thus, 1,2 and 3 are called the zeros of polynomial p(x).
REMAINDER THEOREM : If p(x) is any polynomial of degree greater than or equal to 1 and be any real number. If p(x) is divided by x-a. then the remainder is equal to p(a). Dividend = Divisor × quotient + Remainder
Ex.1
Find the remainder when the polynomial f (x) = 2x4 - 6x3 + 2x2 - x + 2 is divided by x + 2.
Sol.
We have , (x + 2) = 0 x = - 2. So, by remainder theorem, when f(x) is divided by (x + 2) = 0 x = - 2 the remainder is equal to f (-2).
f(x) = 2x4 - 6x3 + 2x2 - x + 2
f(-2) = 2(-2)4 - 6(-2)3 + 2(-2)2 - (-2) + 2
f(-2) = 2 × 16 - 6 × - 8 + 2 × 4 + 2 + 2 = 32 + 48 + 8 + 2 + 2 = 92 Required remainder = 92
Ex.2
Divide the polynomial 3x4 - 4x3 - 3x - 1 by (x - 1)
Sol.
By long division
3x 3 x 2 x 4 x 1 3x 4 4 x 3 3x 1 3x 4 3x 3 4 x 3 3x 1 x3 x2 x 2 3x 1 x2 x
4x 1 4x 4 1 -5 Here, the remainder is -5. Now, the zero of (x - 1) is 1. So, putting x = 1 is p(x), we see that p(1) = 3(1) 4 4(1)3 -3(1) - 1 =3-4-3-1 = - 5, which is the remainder.
FACTOR THEOREM : If p(x) is polynomial of degree n 1 and is any real number, then (i) x - a is a factor of p(x), if p(a) = 0 and (ii) p(a) = 0 if x - is a factor of p(x).
Proof : By the remainder theorem, p(x) = (x - a) (x) + p(a).
MANISH KUMAR
MATHEMATICS
(i) If p(a) = 0, then p(x) = (x - a) q(x), which shows that x - a is a factor of p(x). (ii) Since x - a is a factor of p(x), p(x) = (x - a) g(x) for same polynomial g(x). In this case, p(a) = (a - a) g(a) = 0. (i) (x + a) is a factor of p(x) p(-a) = 0
Note : -
b (ii) (ax - b) is a factor of p(x) = p = 0 a b (iii) (ax + b) is a factor of p(x) = p 0 a Ex-1
Using factor theorem, show that (x - 1) is a factor of 2x 4 + 9x3 + 6x2 - 11x - 6
Sol.
Let f(x) = 2x4 + 9x3 + 6x2 - 11x -6 By factor theorem, (x - 1), x - 1 = 0 x = 1 will be factor of f(x), if f(1) = 0. Now, f(x) = 2x4 + 9x3 + 6x2 - 11x - 6
f(1) = 2 × 14 + 9 × 13 + 6 × 12 - 11 × 1 - 6 = 2 + 9 6 - 11 - 6 = 0 (x - 1) is factor of f(x). Hence, (x - 1) is a factor of the given polynomial.
Ex.2
Determine the value of a for which the polynomial 2x4 - ax3 - 4x2 + 2x + 1 is divisible by 1 - 2x.
Sol.
Let p(x) = 2x4 - ax3 - ax2 + 2x + 1. If the polynomial z p(x) is divisible by (1 - 2x), then (1 - 2x) is a factor of
p(x).
1 p 0 2
1 2
4
1 2
3
1 2
2
1 2
2 a 4 2 1 0
2 a 4 2 1 a 1 0 111 0 16 8 4 2 8 8
25 a a 25 8 8
Hence, the given polynomial will be divisible by 1 - 2x, if a = 25.
TYPE OF FACTORIZATION :-
(i)
Factorization by taking out the common factors
Ex.
ab(a2 + b2 - c2 ) + bc (a2 + b2 - c2) - ca(a2 + b2 - c2)
Sol.
We have ab(a2 + b2 - c2) + bc (a2 + b2 - c2) - ca(a2 + b2 - c2) =
(ii)
(a2 + b2 - c2) (ab + bc - ca)
Factorization by grouping the terms
MANISH KUMAR Ex.
(x2 + 3x)2 - 5(x2 + 3x) - y(x2 + 3x) + 5y
Sol.
We have (x2 + 3x)2 - 5(x2 + 3x) - y(x2 + 3x) + 5y =
(x2 + 3x) {(x2 + 3x) -5} - y{(x2 + 3x) -5} = (x2 + 3x - 5) (x2 + 3x - y)
MATHEMATICS
MANISH KUMAR
MATHEMATICS
(iii)
Factorization by making a perfect square
Ex.
a2 + b2 - 2(ab - ac + bc)
Sol.
We have a2 + b2 - 2(ab - ac + bc) =
a2 + b2 - 2ab + 2ac - 2bc
=
(a - b)2 + 2c (a - b)
=
(a - b) {(a - b) + 2c} = (a - b) (a - b + 2c)
(iv)
Factorization the difference of two squares
Ex.
x8 - y8
Sol.
We have x8 - y8 = {(x4)2 - (y4)2} = (x4 - y4) (x4 + y4) =
{(x2)2 - (y2)2 (x4 + y4) = (x2 - y2) (x2 + y2) (x4 + y4)
=
(x - y) (x + y) (x2 + y2) (x4 + y4)
=
(x - y) (x + y) (x2 + y2) {( x2)2 + (y2)2 + 2x2y2 - 2x2y2)
=
9x - y) (x + y) (x2 + y2) {(x2 + y2)2 - ( 2 xy)2}
=
9x - y) (x + y) (x2 + y2) (x2 + y2 -
2 xy) (x2 + y2 +
2 xy)
(v)
Factorization of quadratic polynomials by splitting the middle term
Ex.
x2 + 3 3 x - 30
Sol.
In order to factorize x2 + 3 3 x - 30, we have to find two numbers p and q such that p + q = 3 3 and pq = 30. Clearly, 5 3 + (-2 3 ) = 3 3 and 5 3 × -2 3 = - 30 So. we write the middle term 3 3x as 5 3x -2 3x
x2 + 3 3x - 30
=
x2 + 5 3x - 2 3x - 30
=
(x2 + 5 3x ) -( 2 3x + 30)
=
(x2 + 5 3x ) ( 2 3x + 10 3 ×
=
x(x + 5 3 ) -2 3 (x + 5 3 ) = (x + 5 3 ) (x - 2 3 )
3)
Algebraic Identities : (i) (a + b)2 = a2 + 23ab + b2 (ii) (a - b)2 = a2 - 2ab + b2 (iii) a2 - b2 = (a + b) (a - b) (iv) (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab (a + b)
MANISH KUMAR
MATHEMATICS
(v) (a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - b3 - 3ab (a - b)
MANISH KUMAR
MATHEMATICS
(vi) a3 - b3 = (a - b) (a2 + ab + b2) (vii) a3 + b3 = (a + b) (a2 -ab + b2) (viii)
a3 + b3 + c3 - 3 abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)
(ix) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Solve the following : Ex.
(i) (3x + 4y)2 (ii) (2x + 3y) (2x -3y) (iii) (9x + 2y + z)2 (iv) (2x + 3y)3 (v) (x + 3y) (x2 - 3xy + 9x2) (vi) (x + y + 2z) (x2 + y2 + 4z2 - xy - 2yz - 2zx) (vii) If x2 +
Sol.
1 x
2
= 83. Find the value of x3 -
1 x3
.
(i) We have, (3x + 4y)2 = (3x)2 + 2 × 3x × 4y + (4y)2 = 9x2 + 24xy + 16y2
(ii) We have, (2x + 3y) (2x - 3y) =
(2x)2 - (3y)2
=
4a2 - 9y2
[Using (a + b) (a - b) = a2 - b2)]
[using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
(iii) We have, (9x + 2y + z)2 =
(9x)2 + (2y)2 + z2 + 2 × 9x × 2y + 2 × 2y xz + 2 × 9x × z
=
81x2 + 4y2 + z2 + 36xy + 4yz + 18xy 3
(iv) We have,
3
3
[using (a + b) = a + b + 3ab (a + b)] (2x + 3y)3
=
(2x)3 + (3y)3 + 3 × 2x × 3y × (2x + 3y)
(2x + 3y)3 = 8x3 + 27y3 + 18xy × 2x + 18xy × 3y
(2x + 3y)3 = 8x3 + 27y3 + 36x2y + 54xy2
MANISH KUMAR (v)
MATHEMATICS
We have, (x + 3y) (x2 - 3xy + 9y2) =
(x + 3y) {(x2 - x × 3y + (3y)2]
=
(a + b) (a2 - ab + b2), where, a = x and b = 3y
=
a 3 + b3
=
x3 + (3y)3 = x3 + 27y3
[ (a + b) (a2 - ab + b2) = a3 + b3]
(vi) We have, (x + y + 2x) (x2 + y2 + 4z2 - xy - 2yz - 2zx) =
(x + y + 2z) (x2 + y2 + (2z)2 - x × y - y × 2x - 2z × x]
=
(a + b + c) (a2 + b2 + c2 - ab - bc - ca), where a = x, b = y, c = 2z
=
a3 + b3 + c3 - 3abc
=
x3 + y3 + (2z)3 - x × xy × 2z
=
x3 + y3 + 8z3 - 6xyz
(vii) We have, 2
1 1 x x2 2 2 x x
2
=
1 x 83 2 x
=
1 x 81 x
=
1 x 92 x
=
x
=
1 x 93 x
=
x3
[Putting x2 +
1 x2
= 83]
2
1 9 x
[Taking square root of both sides]
3
1 3 x 729 x x 1
3
[Cubing both sides]
MANISH KUMAR =
MATHEMATICS x3
1 x
3
3 9 729 x 3
1 x
3
729 27 x 3
1 x3
756
MANISH KUMAR
MATHEMATICS LINEAER EQUATIONS IN TWO VARIABLES
EQUATIONS : A statement of equality of two algebraic expressions which involve one or more unknown quantity is known as an equation. Example : 2x - 5 = 23. An equation has two parts. The part which is on the left side to the quality sign is known as left hand side (L.H.S.) and the part which is not the right side to the equality sign is known as right hand side (R.H.S.) Equality Consider an equation
2x – 5 = 23 L.H.S.
R.H.S.
VARIABLE : The unknown quantities used in any equation are known as variables. Generally, they are denoted by the last English alphabets x, y, z etc.
LINEAR EQUATION : An equation in which the maximum power of variable is one is called a linear equation.
LINEAR EQUATION IN ONE VARIABLE :- An equation of the form ax + b = 0. where x is variable, a.b. are real numbers and a 0 is called Linear equation in one variable. SOLUTION : Any value of the variable that satisfies the given equation is called a solution or roots of the equation.
Ex.
Solve : 3x + 2 = 11
Sol.
3x + 2 = 11
3x = 11 - 2 3x = 9 x =
9 =3 3
LINEAR EQUATIONS IN TWO VARIABLES : DEFINIATION : An equation of the form ax + by + c = 0, where a,b,c are real number, a 0, b 0 and x. SOLUTION : Let ax + by + c = 0, where a,b,c are real number such that a 0 & b 0 then any pair of values of x & y which satisfies the equation ax + by + c = 0 is called a solution of it.
Notes : (i) Solution of a linear equation is written as ordered pair (, ) (ii) Any equation in two variable has at least two solution. (iii) A linear equation in to variable at least two solution. Ex.
Show that (x = 1, y = 1) is a solution of 4x - y - 3 = 0.
Sol.
If we put x = 1 and y = 1 in the given equation. we have L.H.S. = 4 × 1 - 1 - 3 = 0 = R.H.S. so, x = 1, y = 1 is a solution of 4x - y - 3 = 0
Ex.
Which of the following equations have a unique solution or infinitely many solutions. (i) x + 1 = 16
Sol.
(ii) y = 3x + 2
(iii)
y =2 x
(i) The given equation is x + 1 = 16
x=6-1x=5 x + 0.y - 5 =0,
....(i)
MANISH KUMAR which is linear equation is one variable. Hence, equation (i) has unique solution
MATHEMATICS
MANISH KUMAR
MATHEMATICS
(ii) The given equation is y = 3x + 2
3x + (-1)y + 2 = 0 which is linear equation in to variable.
....(ii)
Hence, equation (ii) has infinitely many solution (iii) The given equation is
y 2 x
2x - y = 0
2x + (-1) y + 0 = 0
….(iii)
which is linear equation in two variable. Hence, equation (iii) has infinitely many solution
(A)
GRAPH OF A LINEAR EQUATION IN ONE VARIABLES :
In order to draw the graph of linear equation in one variable we may follow the following algorithm. Step-I : Obtain the linear equation. Step-II :
b b If the equation is of the form ax = b, a 0 then plot the point ,0 and one more point , , where a a
is any real number on the graph paper. If the equation is of the form ay = b, a 0, then plot the point
b b 0, and , , where is any real number on the graph paper. a a Step-III:
Join the points potted in step-II to obtained the required line:-
Ex.
Remark I : If in the linear equation ax = b, a 0 and we have b = 0, then its graph is x - axis. Draw the graph of the following linear equation x - 2 = 0.
Sol.
We have x - 2 = 0 x = 2 Clearly, it does not contain y, so its graph is a line parallel of y axis passing through the point (2, 0). In fact it passes through every point whose x-co-ordinate is 2. Thus we have the following table exhibiting the coordinates of points on the line represented by x = 0. x
2
2
2
y
0
4
-1
Plotting any two point say (2, 0), (2, 4) and (2, -1) given by the above table on the graph paper and joining them we obtain the straight as shown fig.
MANISH KUMAR (B)
MATHEMATICS
GRAPH OF A LINEAR EQUATION IN TWO VARIABLE, WHERE a 0, b 0: In order to draw the graph of a linear equation ax + by + c = 0, a 0, b 0, we may follow the following algorithm. Step-I :
Obtain the linear equation let the equation be ax + by + c = 0.
Step-II :
ax c Express y in terms of x to obtain y = - . b
Step-III :
Put any two or three values of a x and calculate the corresponding value of y from the
expression in step-II to obtain two solution say 1, 1 and ( 2 , 2 ) if possible take value of x as integers in such a manner that the corresponding value of y are also integers. Step-IV :
Plot points (1, 1 ) and ( 2 , 2 ) .
Step-V :
Join the points marked in step IV to obtain a line. The line obtained is the graph of the equation ax + by + c = 0.
Ex.
Draw the graph of the following equation y - x = 2.
Sol.
We have, y-x=2 y=x+2 when
x = 1, y = 1 + 2 = 3
when
x = 2, y = 2 + 2 = 4
when
x = 3, y = 3 + 2 = 5
Thus we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation. x
1
2
3
y
3
4
5
Plotting the points (1, 3), (2, 4) and (3, 5) on the graph paper and drawing the line joining them we obtain the graph of the line represented by the given equation as shown fig.
MANISH KUMAR DIFFERENT FORMS OF A LINE :-
MATHEMATICS
Slope of a line : If la line makes an angle with positive direction of x-axis then tangent of this angle is called the slope of a line, it is denoted by m i.e. m = tan . The equation of a line passing through origin is y = mx. Here c = 0 then the line passes always from origin m is called the slope of the line.
MANISH KUMAR
MATHEMATICS
EXERCISE - 1 1.
POLYNOMIAL
One of the factor of the expression x4 + 8x is (A) x + 2
2.
(B) 3a2 - b 3
(C) 2b
(D) (a + b) (a - b)
2
If (p(x) = 2x + 5x - 3x - 2 is divided by (x - 1) then remainder is (A) 2
4.
(D) x2 + 2x + 2
One of the factor of the expression (a + b)3 - (a - b)3 is (A) a
3.
(C) x 2 + 8
(B) x - 2
(B) -1 4
(C) 0 3
(D) -2
2
Zero of the polynomials 2x + 9x + 11x + 4x - 6 is (A) 1
(B) -1
(C) 3
(D) -3
3
5.
If x + (x + m leaves the remainder 7, when divided by (x - 1) or (x + 1) find and m -
6.
Write the coefficient of x2 in each of the following (i) 2 + x2 + x
7.
(ii) 3 - x2 + x3
(iii)
2 x x 2
(
iv)
2x 3
Verify whether the following are zeros (roots) of the polynomials, indicated against them (i) p(x) = 3x + 1, x =
1 3
(ii) p(y) = 5y -
4 , 5
4 is root of the polynomial p(x) = 6x3 - 11x2 + 9x - 20, find the value of a. 3
8.
If x =
9.
Find the remainder, when x3 - ax2 + 6x - a is divided by (x - a). Also find the remainder using division method.
10.
Using factor theorem, show that (x - 1) is a factor of 2x 4 + 9x3 + 6x2 - 11x - 6.
11.
Using factor theorem show that (x + 2 ) is a factor of ( 2 2 x2 + 5x + 2 )
12.
Find the value of k for which the polynomial x4 - x3 - 4x2 - x + k is divided by (x + 3).
13.
Factories : x3 - 3x2 - 9x -5
14.
Factories :
15.
Factories : 2x3 - x2 - 13x - 6.
3 2 4 x x 2 3
POLYNOMIAL 1. A 7. (i)
2. C
1 , 3
EXERCISE -1 3. A
(ii) ( 4 ) 0
13. (x - 5) (x + 1)2
4 3x 14. x 2 3
4. A
5. 0 = 1, m = 7
8. 19
9.59
10. 0
, (iv) 0 2
6. (i) 1 (ii) -1
(iii)
11. 0
12. -12
15. (x + 2) (x - 3) (2x + 1)
MANISH KUMAR
MATHEMATICS
MANISH KUMAR
MATHEMATICS
EXERCISE - 2
LINEAR EQUATION IN TWO VARIABLES.
VERY SHORT ANSWER TYPE QUESTIONS : 1.
In each of the following verify whether the given value of the x is a solution or not. (i)
x x 8, x = 12 3 4
(ii) 3x + 2 = 11,x = 3 (iii) 2x + 6 = x - 8, x = - 14 (iv) 4x - y - 3 = 0, x = -, y = 1 (v) 3x + 5y - 2 = 0 x = 2, y = -1 2.
Solution of the equation x - 2y = 2 is/are (A) x = 4. y = 1
3.
(B) x = 2, y = 0
(D) All of these
4 (C) ,0 5
5 (D) ,0 4
The graph of line 5x + 3y = 4, cuts y axis at the point -
4 (A) 0, 3 4.
(C) x = 6, y = 2
3 (B) 0, 4
If x = 1, y = 1 is a solution of equation 9ax + 12ay = 63 then the value of a is (A) -3
(B) 3
(C) 7
(D) 5
5.
Find three solution of equation : 2x + y = 7.
6.
Find the value of k if x = 2. y = 1 is solution of the equation 2x + 3y = k
7.
(2p - 1, p) is solution of equation 10 x - 9y = 15, find the value of p.
8.
A linear equation of the form y = mx, always pass through the -
9.
If the co-ordinates of a point p are (-2, 5) then abscissa & ordinates are -
10.
The graph of a linear equation in two variables is a ..........
11.
Slop of the line ax + by + c = 0 is.......
12.
Equation of x-axis is ....... and equation of y-axis is ............
13.
The graph of the line x = a is parallel to ..............
14.
The graph of the line y = b is parallel to ...........
LINEAR EQUATIONS IN TWO VARIABLES 1. (i) No. 6. k = 7
(ii) Yes, 7. p
EXERCISE - 2
(iii) Yes,
(iv) Yes,
(v) No 2. A
3. B
4. B
25 11
8. Origin,
9. -2, 5
10. Straight line
5. (0, 7), (1, 5), (2, 3) 11.
a b
MANISH KUMAR 12. y = 0, x= 0 13. y-axis
MATHEMATICS 14. x-axis
MANISH KUMAR
MATHEMATICS
EXERCISE - 3 1.
Find out which of the following equation have x = 2, y = 1 as a solution. (i)
2.
LINEAR EQUATION IN TWO VARIABLES
2x + 5y = 9
(ii) x+ y + 4 = 0
(iii) 2x - 3y + 7 = 8
(iv) 2x - 3y = 1
Find the value of ‘a’ so that each of the following equation may have x= 1, y= 1 as a solution. (i)
3x + ay = 6
(ii) x - y = a
(iii) 5x + 2ay = 3a
(iv) 9ax + 12ay = 63
3.
Find the value of x & y given equation : 2x - 3y - 3 = 0,
4.
Draw the graph of each of the following equations (i)
2(x + 3) -3(y + 1 ) = 0
(iii) x - 2y = 4 5.
2x 1 4y 0 3 2
(ii) (x - 4) - y + 4 = 0 (iv) 5x + 3y = 4
Draw the graph of the line 4x - y = 5 & 5y - 4x = 7 on the same graph pager & find the co-ordinate of their point of intersection.
6.
Draw the graph of x - y + 1 = 0 & 3x + 2y - 12 = 0 on + 4 same graph. Calculate the area bounded by these line & x-axis.
7.
The solution of 3x - 2y = 5 is -
8.
Draw the graph of the equation 3x + 7y = 10. Find whether x = 1 and y = 1 is the solution of these equation.
9.
Draw the graph y = - 2x. Show that the point (2, - 5) is not on the graph.
10.
If 2x - 8 = 8, them find the value of x 2 + x - 70.
LINEAR EQUATIONS IN TWO VARIABLES 1. (i), (iii), (iv)
2. (i) a = 3,
7. x = 1, y = -1
10. 2
(ii) a = 0,
EXERCISE - 3 (iii) a = 5,
(iv) a = 3.
3. x
21 3 . y 10 10
