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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 1.
Linear Equation An equation of the form ax + by + c = 0 or ax + by = c , where a, b, c, are real numbers, x, y are variables, is called linear equation in two variables.
2.
Simultaneous pair of linear equations in two variables A pair of linear equations in two variables is said to form a pair of linear equations. a 1 x + b1 y = c1 ; a2 x + b2 y = c 2 form a pair of linear equations in two variables. The following cases occur: (i)
if
a1 b1 , it has a unique a 2 b2
solution. The graph of lines intersect at one point. The system is consistent
(x, y)
For example: To find whether the given pair of linear equations are consistent 2x + 5y + 1 = 0 3 x + 4y − 7 = 0 , a1 3 = , a2 2
As
b1 4 = b2 5
a1 b1 a 2 b2
It has a unique solution. The system is consistent.
(ii)
if
a1 b1 c1 .It has infinite many = = a2 b2 c 2
solutions. Every solution of one equation is a solution of other also. The graph of both equations is coincident lines. The system is dependent consistent.
For Example: To find the values of a and b so that the following system of linear equations has infinite number of solutions.
2x − 3y = 7 and (a + b)x − (a + b − 3)y = 4a + b Here
b c a1 3 7 2 , 1 = , 1 = = a2 a + b b2 a + b − 3 c 2 4a + b
For infinite many solutions a1 b1 c1 2 3 7 = = or = = a + b a + b − 3 4a + b a 2 b2 c 2
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https://www.entrancei.com/ Taking first two terms 2 3 = a+b a+b−3 3a + 3b = 2a + 2b − 6 a + b = −6 a = −6 − b
Again, taking second two terms 3 7 = a + b − 3 4a + b 12a + 3b = 7a + 7b − 21 5a − 4b = −21
5( −6 − b) − 4b = −21 − 9b = 30 − 21 = 9 b = −1
(iii)
If
a = −6 + 1 = –5 a = −5 , b = – 1
a1 b1 c1 . It has no solution. = a 2 b2 c 2
The graph of both line are parallel to each other. The system is inconsistent
For Example: To find the value of k for which system has no solution 3x − 4y + 7 = 0 , kx + 3y − 5 = 0 . a1 3 = ; a2 k
b1 4 =− ; b2 3
Condition for no solution is
c1 7 =− c2 5 3 4 =− k 3
k=−
9 4
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https://www.entrancei.com/ Table for consistency/Inconsistency conditions Pair of linear equations a1 x + b1y + c1 = 0 a2 x + b2 y + c 2 = 0
Algebraic conditions
Graphical representation
(i) Consistent
a1 b1 a2 b2
Intersecting lines
(Independent)
(ii) Consistent (Dependent)
a1 b1 c1 = = a2 b2 c 2
(iii) Inconsistent
a1 b1 c1 = a2 b2 c 2
Algebraic interpretation
Exactly one solution (unique solution) Coincident lines Infinitely many solution Pair of parallel No solution lines
A pair of value of x and y satisfying each one of the equations in x and y is called a solution of the system. Methods of solving simultaneous pair of linear equations in two variables: (i) Graphical Method (ii) Algebraic Method Let the system of linear equation in two variables be a 1 x + b1 y = c1 and a2 x + b2 y = c 2 3.
4.
. Graphical method: • Method of solving simultaneous linear equations in two variables: Step 1: Get three solutions of each of given linear equations. Step 2: Plot these points on graph in order to draw the lines representing these equations. Step 3: Get the solution of equations. Algebraic Method: (i) Substitution method Step 1: From one equation find the value of one variable, (say y) in terms of other variable, i.e., x. Step 2: Substitute the value of variable obtained in step 1, in the other equation to get an equation in one variable. Step 3: Solve the equation obtained in step 2 to get the value of one variable. Step 4: Substitute the value of variable so obtained in any given equation to find the value of other variable. For Example: Solve for x and y using Substitution method; ax + y = m and bx − y = n ax + y = m
…(i)
bx − y = n
…(ii)
Step 1: From (i), we have ax + y = m y = m − ax
…(iii)
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https://www.entrancei.com/ Step 2: Substituting value of y in equation (ii) bx − (m − ax ) = n bx − m + ax = n bx + ax = n + m x (a + b ) = m + n m+n x= a+b Step 3: Substituting value of x in equation (iii), we get m + n y = m − a a+b y=
ma + mb − am − an a+b
y=
bm − an a+b
Solution is x =
m+n bm − an , y= a+b a+b
(ii) Elimination method Step 1: Obtain the two equations. Step 2: Multiply the equations so as to make the coefficients of one of the variables equal, to be eliminated. Step 3: (a) If the coefficients of the variable to be eliminated are having same sign, then subtract the equation obtained in step 2. (b) If the coefficients of the variable to be eliminated are having opposite sign, then add the equation obtained in step 2. Step 4: Solve the equation obtained in step 3 to obtain the value of one variable. Step 5: Substitute the value of variable so obtained in any of the given equation to find the value of other variable. For Example: Solve for x and y using elimination method: 3x + 4y = 10 and 2x − 2y = 2 …(i) 3 x + 4y = 10 …(ii) 2 x − 2y = 2 Step 1: Multiplying (i) by 2 and (ii) by 3 to make the coefficients of x equal. …(iii) 6 x + 8y = 20 …(iv) 6 x − 6y = 6 Step 2: Subtract equation (iv) from (iii) to eliminate x, because the coefficients of x are the same. 6 x + 8y = 20 6 x − 6y = 6 − + − 14y = 14 14 y= =1 14
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https://www.entrancei.com/ Step 3: Substituting the value of y in (i), we get 3 x + 4 1 = 10 3 x = 10 − 4 = 6
x=
6 =2 3
Solution is x = 2, y = 1 (iii) Comparison Method Step 1: From each equation find the value of one variable in terms of other. Step 2: Equate them to get an equation in one variable and solve. For Example: Solve for x and y by Comparison Method: 2x + 3y = 8 ; x − 5y = −9
2x + 3y = 8 x − 5y = −9
…(i) …(ii)
For equation (i), 2x = 8 − 3y or
x=
8 − 3y 2
…(iii)
From equation (ii),
x = −9 + 5y
…(iv)
From equation (iii) and (iv), we have 8 − 3y = − 9 + 5y 2
8 − 3y = −18 + 10 y 8 + 18 = 10 y + 3y 26 = 13 y
26 =y 13
2=y Putting value of y in equation (iv), we have x=
8 −32 2
x =1
Solution x = 1, y = 2 (iv) Cross Multiplication Method For system of linear equations a1x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0 when, a1 b1 b c − b2c1 a c −c a the unique solution is given by x = 1 2 and y = 2 1 2 1 . a1b2 − a2b1 a1b2 − a2b1 a2 b2 For Example: Solve the following system of equations in x and y by Cross Multiplication Method
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https://www.entrancei.com/ ax + by − a + b = 0 , bx − ay − a − b = 0 By cross multiplication, we have x y
+1
b
–a+b
a
b
–a
–a–b
b
–a
y x 1 = = (b )( −a − b ) − ( −a )( −a + b ) ( −a + b)(b) − ( −a − b )(a ) (a )( −a ) − b(b )
or or or or 5.
x y 1 = = 2 2 2 2 ( −ab − b ) − (a − ab ) ( −ab + b ) − ( −a − ab ) − a − b 2 2
x −a −b 2
2
=
y a +b 2
2
=
1 − a − b2 2
a2 + b2 − a2 − b2 y = and − (a 2 + b 2 ) − a2 − b2 x = 1 and y = –1
x=
Equations reducible to system of linear equations (i) ax + by = cxy ; dx + ey = fxy We have two types of solutions: (a) Trivial solution, when x = 0 and y = 0 when x 0 and y 0
(b) Non-trivial solutions, Steps for non-trivial solution:
Step 1: Divide both sides by xy to obtain Step 2: Substitute
a b d e + = c, + = f y x y x
1 1 = u, = v to get av + bu = c, dv + eu = f x y
Step 3: Solve these equations to get values of u and v . Step 4: Use
1 = u and = v to get values of x and y . x y
For Example: Solve for u and v : 2u + v = 2u + v =
7 uv 3
11 7 uv ; u + 3v = uv 3 3
…(i)
11 uv …(ii) 3 Dividing equation (i) and (ii) by uv, we have 2 1 7 + = v u 3 1 3 11 + = v u 3 u + 3v =
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https://www.entrancei.com/ 1 1 = x and = y ; we have v u 7 2x + y = …(iii) 3 11 x + 3y = …(iv) 3 Multiplying equation (iv) by 2 and subtracting (iv) from (iii) to eliminate x, we get 2x + y = 7 / 3 2x + 6y = 22 / 3 – – –
Putting
15 3 15 =1 y= 35 Putting y = 1 in equation (iii), we get 7 4 2x = − 1 = 3 3 2 x= 3
–5y = −
Now,
1 =x v
or
1 = 1 or u
1 2 = v 3
v=
3 2
u=1
Solution is u = 0, v = 0 and u = 1, v =
3 2
p q r s + = e, + =f ax + by cx + dy ax + by cx + dy 1 1 Step 1: Substitute =v = u and ax + by cx + dy Step 2: Solve the equations obtained, for u and v . Step 3: Solve for x and y by substituting the values of u and v in Step 1.
(ii)
For Example: Solve for x and y :
10 2 15 5 + =4; − = −2 ; x y, x – y x+y x−y x+y x−y
10 2 + =4 x+y x−y 15 5 − = −2 x+y x−y
Put
…(i) …(ii)
1 1 = u and =v x+y x−y
10u + 2v = 4
…(iii) 15u − 5v = −2 …(iv) Multiplying equation (iii) by 5 and equation (iv) by 2 to eliminate v 50u + 10v = 20
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https://www.entrancei.com/ 30u − 10v = −4
80u
= 16 16 1 u= = 80 5 Putting value of u in equation (iii), we get 1 10 + 2v = 4 5 2 + 2v = 4 2v = 2 v=1 1 1 1 = ; =1 x+y 5 x−y or
x+y =5
…(v)
x −y =1
…(vi)
Adding equation (v) and (vi), we get 2x = 6 x=3 Putting value of x in equation (v), we get 3+y =5 y=2 Solution is x = 3; y = 2 (iii) ax + by = c, bx + ay = d Step 1: Add both the equations and reduce it to similar form by dividing by a + b . Step 2: Subtract one equation from other and reduce it to simpler form by dividing throughout by a − b. Step 3: Solve equations obtained by step 1 and 2. For Example: Solve for x and y : 217 x + 131y = 913 and 113 x + 217 y = 827 We have
217 x + 131y = 913
…(i)
131x + 217 y = 827
…(ii)
Adding equation (i) and (ii), we have 348 x + 348 y = 1740 or
x+y =5
Subtracting equation (ii) from (i), we have 86 x − 86 y = 86 or x −y =1 From equation (iii) and (iv), we get x+y =5 x −y =1
…(iii)
…(iv)
2x = 6 x =3
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https://www.entrancei.com/ Putting x = 3 in equation (iii), we have 3+y =5
y =2 Solution is x = 3, y = 2 6.
To solve the Word Problems (i)
Read the problem carefully and identify the unknown quantities. Give these quantities a variable name like x, y, u, v etc.
(ii) Identify the variables to be determined. (iii) Read the problem carefully and formulate the equations in terms of the variables to be determined. (iv) Solve the equations obtained in step (iii) using any one of the methods used earlier. For Example: A train covered a certain distance at a uniform speed. If the train would have been 6 km/hr faster it would have taken 8 hrs less than the scheduled time. And if the train were slower by 6 km/hr, it would have taken 12 hrs more than the scheduled time. Find the distance of the journey. OR Let the uniform speed of train = x km/hr and scheduled time of train = y hrs
Distance (length) of journey = xy km
[ Distance = speed × time]
Case 1: Increased speed = (x + 6) km/hr and changed time = (y – 8)hrs According to given condition (x + 6)(y – 8) = xy or
xy – 8x + 6y – 48 = xy
or
– 8x + 6y = 48
..... (i)
Case 2: Changed speed = (x – 6) km/hr Changed time = (y + 12) hrs
According to given condition (x – 6)(y + 12) = xy xy + 12x – 6y – 72 = xy 12x – 6y = 72
..... (ii)
From (i) and (ii), we have – 8x + 6y = 48 12x – 6y = 72 Adding (i) and (ii), we have 4x = 120 or
x = 30 km/hr
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https://www.entrancei.com/ Putting value of x in (i), we have – 8(30) + 6y = 48 6y = 48 + 240 y=
288 = 48 hrs 6
Distance of the journey = xy = 30 × 48 = 1440 km
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