(Simple and simultaneous equations)( p q )

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IndexMeaning Root of an EquationRoot of an EquationSimple Equation

Solving of simple equationSimultaneous Equationq

Substitution methodEli i ti th d

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Elimination method

What is an Algebraic Equation?An Algebraic equation is a mathematical g qstatement which equates two algebraic expressionsexpressions.Examples :

5.

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( )( )6. (9+x)(9-x)=17

7. s

8. .

are some of the examples of algebraic equations with one variable 'x'

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equations with one variable x .

Root of an EquationTh l f th i bl hi h ti fiThe value of the variable which satisfies the equation is called the ROOT (or solution) of the equation.

Example 1 : Consider the equation 4x+1=94x+1=9

when x=2, 4x2+1=9

Vikasana - CET 20129=9

Therefore x=2 is the value of 'x' which satisfies the equation 4x+1=9satisfies the equation 4x+1=9.Hence, x=2 is the 'ROOT' of 4x+1=9.

Example 2 : Consider the equationx2-5x+6=0x 5x 6 0

When x=2, 4-10+6 =0i.e., 0=0i.e., 0 0

Also when x=3, 9-15+6=0 0=0x=2 and x =3 are the roots of

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x 2, and x 3 are the roots of x2-5x+6=0.

Note:N th l f ' ' ti f th iNo other values of 'x' satisfy the given

equation, other than its roots.Process of finding the roots is known

as solving of the equation.g qThe degree of an equation determines

the number of roots of the equationthe number of roots of the equation.

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Simple EquationsAn equation, which when reduced to a simple form involves only one unknown quantity (variable) of degree one, is called Simple Equation.

Examples:1. 4x+2=42. 5x-7x+8x=12-5+7+103 3(2 x)=4(5x 6) 9(3x 4)+6

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3. 3(2-x)=4(5x-6)-9(3x-4)+6

2 24. x2 + 4x -5 = (x +1)2

5. (x+1) (2x+1)=(x+3) (2x+3)-14i.e. which reduces to 3x+1=9x+9-14 These are some of the examples ofThese are some of the examples of

Simple Equations.Note:Note:

For any given Simple equation, there exists exactly one value for the unknown whichexactly one value for the unknown, which satisfies the equation. This value of the unknown is called the root

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This value of the unknown is called the root of the equation

Illustration:4x-1=7 has exactly one value of x,y ,i.e. x=2 which satisfies the equation

4 2 1 74 x 2 - 1 = 77 = 7

Therefore x = 2 is the root of 4x-1=7

Note:The process of finding the root is

Vikasana - CET 2012known as solving the equation

Examples on Solving the Equations:

1 Solve the equation 3x 2=71. Solve the equation 3x-2=7Soln: Given equation: 3x-2=7

By adding '2' on both sides.y g3x-2+2=7+2

3x=93x=9Now divide both sides by 3

Vikasana - CET 2012x=3 is the solution.

2. Solve the equation 15(x-1)+4(x+3)=2(x+7)15(x 1)+4(x+3) 2(x+7)Soln: Given equation:Given equation:

15(x-1)+4(x+3)=2(x+7)15 15 4 12 2 1415x-15+4x+12=2x+1419x-3=2x+1419x-2x=14+3

17x=17

Vikasana - CET 2012so, x=1

3. Solve the equation (x+1)/3+(x+3)/4=16+(x+4)/5Soln: LCM of 3,4,5(denominators) is 60., , ( )Multiply the give equation by 60

20(x+1)+15(x+3)=60(16) +12(x+4)20(x+1)+15(x+3)=60(16) +12(x+4)20x+20+15x+45=960+12x+48

b i lifi ti tby simplification we get,23x=943

Vikasana - CET 2012So, x=41

4. Solve the equation 2(x+1)(x+3)+8=(2x+1)(x+5)( )( ) ( )( )Soln:

2(x2+4x+3)+8 = 2x2+11x+52(x2+4x+3)+8 = 2x2+11x+58x+14 = 11x+5

-3x = -9x = 3x = 3.

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5. Solve the equation (x+1)2+2(x+3)2=3x(x+2)-21( ) ( ) ( )Soln:

x2+2x+1+2(x2+6x+9) = 3x2+6x 21x2+2x+1+2(x2+6x+9) = 3x2+6x-21x2+2x+1+2x2+12x+18= 3x2+6x-21

14x+19 = 6x-218x = -408x = -40

x = -5

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6 Fi d b h th t i th f it6. Find a number such that six-sevenths of it shall exceed four fifths of it by 2.Soln:

let x be the required number.qgiven that (6/7)x = (4/5)x+2LCM of 7 and 5 is 35LCM of 7 and 5 is 35Multiply both sides by 35

35(6/7)x = 35(4/5)x+35(2)5(6x)= 7(4x)+70

Vikasana - CET 2012We get, 2x = 70 x = 35

Simultaneous EquationsConsider the equation 2x+y=5 which

contains two unknowns 'x' and 'y‘ each of degree of oneof degree of one.

We can rewrite this equation as 5 2 5 2y=5-2x .y=5-2x

For each value of 'x' there exists a unique value of 'y'value of y

x 0 1 2 3 . . . . . . .5 3 1 1

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y 5 3 1 -1 . . . . . . .

Thus we can find infinite number of pair of values for 'x' and 'y' as we pleaseof values for x and y as we please, which satisfy the equation. But if we have

d i f ki d 9a second equation of same kind 5x-y=9Same may be rewritten asyy=5x-9 .............. (2)

If we now seek the values of 'x' and 'y' which satisfy both equations

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which satisfy both equations.

Th l f ' ' i i 1 d 2 bThe values of 'y' in equations 1 and 2 must be identical.

5 9 5 25x-9 = 5-2x7x = 14

2x = 2Substituting this value of 'x' in the first equation (or in second equation).

We get y=1Thus if both equations are to be satisfied by the same values of 'x' and 'y', there is

Vikasana - CET 2012only one solution possible.

D fi iti Wh T tiDefinition : When Two or more equations having two or more unknown quantities

hi h i fi d b h lwhich are satisfied by the same values of the unknown quantities, are called simultaneous equations.

N tNote:In the present chapter we shall confine

our attention to the simultaneous equations having two unknowns each of

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q gdegree one.

oExamples :

1. 3x+7y=272 165x+2y=16

2. 4x-y=0 y2x+y=18

In these equations unknown quantitiesIn these equations unknown quantities are 'x' and 'y' each of degree one.

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Methods of solving Simultaneous EquationsEquations

Equations with two unknowns each of degree one: a1x+b1y=c1

a2x+b2y=c2

Can be solved using:o Substitution Methodo Elimination Method

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Substitution MethodI thi th d h t ' ' iIn this method we have to express 'y' interms of 'x' (or vice versa) using one of the given equations and substitute this value in the other equation. This reduces to a Simple equation with one variable, whichcan be solved. This value is to be substituted in one of the given simultaneous equations to get the value of

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q gthe other unknown.

Example 1: Solve by substitution method

4x+7y=29 ............. 1x+3y = 11 ............ 2

Soln:From (2) x=11-3y( ) ySubstituting this value of 'x' in........(1)

4(11-3y)+7y = 29( y) y44-12y+7y = 29

-5y = -15

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yy = 3

Substituting this value y=3 in one of theSubstituting this value y 3 in one of the given equations we get x=2.

Therefore x=2 y=3 is the solutionTherefore x=2, y=3 is the solution.

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Example 2 Solve :5x-7y=0 ......(1)7x+5y=74......(2)

By substitution method.Soln:From ....(1) 5x-7y = 0

-7y = -5xy = 5x/7

Substituting in (2) 7x+5(5x/7)=74

Vikasana - CET 2012(49x+25x)/7=74

Solving we get x=7Put x=7 in y=5x/7we get y=5we get y=5Therefore x=7, y=5 is the solution.

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Elimination methodSolution of Simultaneous equations with two variables each of degree one by g yelimination method, means elimination of one of the variables x (or y). To doof one of the variables x (or y). To do this we have to make the coefficients of one of the variables (x or y) same inone of the variables (x or y) same in both the equations.

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Then by subtracting one equation from the other, we get a Simple equation in one variable that can be solved.one variable that can be solved.Finally, by substituting this value in one of the given equations we can find the valuethe given equations we can find the value of the other unknown.

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Example 1: Solve :3x+7y=27......... (1)5x+2y=16 ........ (2)

by elimination methodby elimination method.Soln: Multiply 1st equation by 5

15x+35y=135 (3)15x+35y=135 ........ (3)Multiply 2nd equation by 3

15 6 48 (4)15x+6y=48............ (4)So that the coefficient of 'x' is same in both

Vikasana - CET 2012the equations.

subtracting equation (4: 15x+6y=48) from equation (3: 15x+35y=135)

we get, 29y=87 therefore, y=3

S b tit ti 3 i (2 5 2 16)Substituting y=3 in (2: 5x+2y=16) we get, 5x+2 x 3 = 16

5x+6=165x=105x 10x=2

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Therefore x=2, y=3 is the solution.

Example 2:Solve the Simultaneous equations :4x+7y=29 .......... (1)x+3y=11 (2)x+3y=11 ........... (2) by elimination methodS l E i (2) 4 4 12 44Soln: Equation (2) x 4, 4x+12y=44

From (1) 4x+7y=29Subtracting we get,

5y=15

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5y=15y=3

Substituting this value of 'y' in (2: x+3y=11)in .... (2: x+3y=11)

x+3y=11x+3x3=11x=11-9x=11-9x=2

Solution is x=2, y=3.

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Example 3: one third of sum of two numbers is 14 and one half of their difference is 4. Find the numbersnumbers.

Soln: Let the two numbers be x and yGivenGiven ,

I e x + y = 42 (1)I.e.., x + y = 42 …………..(1)x – y = 8 ……………(2)

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Adding equations (1: x + y = 42 ) and (2 8 )(2: x – y = 8 )

We get,2x=502x=50

x=25Form (1) 25+y=42Form (1) 25+y=42

y=42-25y=17y=17

Therefore x=25 and y=17 is the solution.

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Note: Simultaneous equations in 'x' and 'y' q y

(each of degree one) are of the forma x+b y=ca1x+b1y=c1

a2x+b2y=c2

Will have unique solution if a1/a2≠b1/b2

In this case graph of the aboveIn this case graph of the above equations are intersecting lines.

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There are infinitely many solutionif a /a =b /b =c /cif a1/a2=b1/b2=c1/c2

In this case graph of the above equations are coincident lines.

There is no solution if a1/a2=b1/b2≠c1/c2

In this case graph of the aboveIn this case graph of the above equations are parallel lines.

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SummarySimple Equations involves one variable of degree one. gSimultaneous Equations involves two or more variablesmore variables.Both Simple and Simultaneous

i l i l iequations plays very important role in the filed of Science and Engineering.

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