Equations

SIMPLE EQUATIONS & SIMULTANEOUS EQUATIONS

Simple equations and Simultaneous equations in 2 variables.

A simple equation is an equation in a single variable, whose value must be determined.

(1) ax + b = 0 is called a simple equation in one unknown.(2) ax + by + c = 0 is the general form of a linear equation in two variables.(3) ax + by + c = 0 is a single linear equation in two variables which admits of

infinite number of solutions.(4) a1x + b1y + c1 = 0 ----- (1)

a2x + b2y + c2 = 0 ----- (2)For the above equations (1) and (2) in two variables x and y, the solution is

and

(i) The above system of equations has a unique solution if . Such a

system is a consistent system. The graph consists of two intersecting lines.

(ii) If , then there is no solution to the system. It is called inconsistent.

The graph consists of two parallel lines.

If , the system has infinite solutions. The graph consists of two coincident

lines.

1. Solve Solution:

2. Find a number such that the difference between nine times the number and four times the number is 55.Solution:

Let the number be ‘x’ say.9x – 4x = 55 5x = 55 x = 11

3. A father is now three times as old as his son. Five years ago. he was four times as old as his son. Find their present ages.Solution:

Let the present age of the son be ‘x’ years, say.Father’s present age = 3x yearsFive years ago, age of the son = (x – 5) yearsFathers age = (3x – 5) yearsGiven that, 5 years ago. Father’s age = 4 son’s age

3x – 5 = 4(x – 5) = 4x – 20 x = 15

Son’s present age = 15 yearsFather’s present age = 45 years

4. Find four consecutive odd numbers whose sum is 56.Solution:

Let the first odd number be assumed as ‘x’.The other three consecutive odd numbers in ascending order will be x + 2, x

+ 4, x + 6.Given x + (x + 2) + (x + 4) + (x + 6) = 56

4x + 12 = 564x = 44

Hence the four consecutive odd numbers are 11, 13, 15, 17.

5. ‘A’ is 7 years older than ‘B’. 15 years ago, ‘B’ is age was of A’s age. Find their present ages.Solution:

Let the present ages of B and A be ‘x’ and (x + 7) respectively.15 years ago, ‘B’s age was (x – 15) years15 years ago, ‘A’ is age was [(x + 7) – 15] = (x – 8) yearsGiven B’s age = (A’s age) 15 years ago

x – 15 = (x – 8)4x – 60 = 3x – 24x = 60 – 24 = 36 = B’s present ageA’s present age = x + 7 = 36 + 7 = 43 years

6. If ‘A’ gives B Rs.4. B will have twice as much as A. If B gives A Rs.15, A will have 10 times as much as B. How much each has originally?Solution:

Let the amounts with ‘A’ and ‘B’ be ‘x’ rupees ‘y’ rupees respectivelyA gives 4 rupees to B.A will have (x – 4)B will have (y + 4)

Given, y + 4 = 2(x – 4) = 2x – 8 2x – y = 12 -------------------(1)

Secondly, if B gives 15 rupees to A.B will have y – 15A will have x + 15

Given x + 15 = 10(y – 15) = 10y – 150 -x + 10y =165 ------(2)

(2) 2 -2x + 20y = 330 -------- (3) 2x – y = 12 -------- (1)

(3) + (1) 19y = 342

Put y = 18 in (1), we get 2x – 18 = 12 2x = 30

x = 15A’s original possession = Rs.15B’s original possession = Rs.18

7. Find two numbers which are such that one-fifth of the greater exceeds one-sixth of the smaller by ‘4’; and such that one-half of the greater plus one-quarter of the smaller equals ‘38’.Solution:

Let the two numbers be ‘x’ and ‘y’ sayLet x > y, say

Given 6x – 5y = 120 ----------- (1)

Given secondly 2x + y = 152 --------------- (2)

Solve for ‘x’ and ‘y’ from (1) & (2)(2) 5 10x + 5y = 760 ---------- (3)(1) + (3) 16x = 880

From (2), 110 + y = 152y = 152 – 110 = 42

The numbers are 55 and 42.

8. A man left Rs.1750 to be divided among his two daughters and four sons. Each daughter was to receive three times as much as a son. How much did each son and daughter receive?Solution:

Let a son’s share be x rupeesThe share of a daughter = 3x rupees.Total amount father left = Rs.1750, to be divided among 2 daughters and 4

sons.2 3x + 4x = 1750

10x = 1750x = 175 rupees

Each son’s share = Rs.175 and each daughter’s share = Rs.525.

9. The sum of a certain number and its square root is 90. Find the number.Solution:

Let the number be ‘x’, sayGiven ---------- (1)

Squaring x = (90 – x)2 = 8100 – 180x + x2

x2 – 181x + 8100 = 0 x2 – 100x – 81x + 8100 = 0 x(x – 100) – 81(x – 100) = 0 (x – 100) (x – 81) = 0 x = 100 or 81x = 100 is invalid , according to equation (1).Therefore x = 81.

10. In a family, eleven times the number of children is greater by 12 than twice the square of the number of children. How many children are there in the family?Solution:

Let the number of children in the family be ‘x’, sayGiven 11x = 2x2 + 12 2x2 – 11x + 12 = 0 2x2 – 8x – 3x + 12 = 0 2x(x – 4) –3(x – 4) = 0 x = 4 x = is invalid. So delete

The number of children in the family = 4

11. Find three consecutive positive integers such that the square of their sum exceeds the sum of their squares by 214.Solution:

Let the three consecutive +ve integers be x, x + 1, x + 2Given (x + x + 1 + x +2)2 = x2 +(x + 1)2 + (x + 2)2 + 214(3x + 3)2 = x2 + x2 + 2x + 1 + x2 + 4x + 4 + 214

9(x2 + 2x + 1) = 3x2 + 6x + 219 9x2 + 18x + 9 = 3x2 + 6x + 219 6x2 + 12x – 210 = 0 x2 + 2x –35 = 0 (Dividing through out by 6.) (x + 7) (x – 5) = 0

x = 5 or –7 [x = -7 is to be deleted. Since we have to take only +ve integers as per given data]The three consecutive positive integers are 5, 6, 7.

12. I have a certain number of apples to be divided equally among 18 children. If the number of apples and the number of children were increased by ‘2’, each child would get 5 apples less. How many apples have to distribute?Solution:

Let the number of apples be ‘x’, sayNumber of children = 18

Since each should get equal number of apples,No of apples for each child =

Now, when the number of apples and the number of children were increased by 2, each child gets 5 apples less.

20x – 18(x + 2) = 360 5 20x – 18x – 36 = 1800

2x = 1800 + 36 = 1836

The number of apples is ‘918’, so that equal distribution among ‘18’ children is quite likely.

13. A fraction is such that if ‘2’ is added to the numerator and ‘5’ to the denominator, the fraction becomes . If the numerator of the original fraction is trebled and the

denominator increased by 15, the resulting fraction is . Find the original fraction.Solution:

Let the original fraction be .

Given 2x + 4 = y + 5 2x – y = 1 -------------- (1)Also given 9x = y + 15 9x – y = 15 ------------- (2)(2) – (1) 7x = 14

x = 7From (1) we get y = 2x – 1 = 14 – 1 = 13Hence, the original fraction is

14. Two friends A and B start on a holiday together, A with Rs.380 and B with Rs.260. During the holiday, B spends 4 rupees more than A and when holidays end, A has 5 times as much as B. How much has each spent?Solution:

Let ‘x’ rupees be the amount spent by A.(x + 4) rupees would have been spent by B.

Initially A had Rs.380 and B had Rs.260.At the end of the holidays ‘A’ had an amount of 380 – x

‘B’ had an amount of 260 – (x + 4)i.e., 256 – x

Given, 380 – x = 5(256 – x)= 1280 – 5x

4x = 1280 – 380 = 900Rupees

‘A’ spent Rs.225 and ‘B’ spent Rs.229.

QUADRATIC EQUATIONS

(1) ax2 + bx + c = 0 is called a general quadratic equation. The solution for ‘x’ is

(2) If , are the roots of the equations, then + = ; (3) b2 – 4ac is called the ‘discriminant’ of the quadratic, denoted by (4) If > 0 and not a perfect square, the roots are real and unequal.(5) If > 0 and also a perfect square, the roots are unequal and rational(6) If = 0, the roots are equal and real(7) If < 0, the roots are imaginary

Problems:

1. Solve: 9x2+15x-14 = 0

Here ‘a’ = 9; b =15; c = -14x = = or =

2. Solve (x+5) (x-5) = 39x2 – 25 = 39x2 = 64x =

3. Solve:

Put

8y2 – 8 = 63y8y2- 63y – 8 = 08y2 – 64y +y –8 = 08y(y-8) + 1( y-8) = 0(y-8) (8y+1) = 0y=8 or y =

But, y = x

X= 8

Or, y = x3/2 = -

X = =

4. Solve

(x2-x+1)(a2+a+1) = (x2+x+1) (a2-a+1)x2(a2+a+1-a2+a-1) +x(-a2-a-1-a2+a-1) + (a2+a+1-a2+a-1) = 0 2ax2-2x(a2+1) +2a = 0ax2-x(a2+1) + a = 0ax2-a2x-x+a = 0ax(x-a) –1(x-a) = 0(x-a)(ax-1) = 0 x = a or

5. Show that (x-1)(x-3)(x-4)(x-6) + 10 is positive for real values of x

Consider, (x-1)(x-6)(x-3)(x-4) + 10 =(x2-7x+6) (x2-7x+12) + 10 Put x2-7x+6 = tt(t+6) + 10= t2+6t+10= (t2+6t+9) + 1= (t+3)2+1 = (x2-7x+9)2+1 which is clearly positive for real values of ‘x’.

6. A two digit number is less than 3 times the product of its digits by 8 and the digit in the ten’s place exceeds the digit in the unit’s place by ‘2’.Find the number.

Let the numbers have x in ten’s place and y in the unit place 10x+y = 3xy – 8……….(1)x-y = 2 …………...(2)

From (2) x= y+2In (1), 10(y+2)+y = 3y(y+2)-810y + 20 + y = 3y2+6y-83y2-5y-28 = 03y2-12y+7y-28 = 03y(y-4) + 7(y-4) = 0(y-4) (3y+7) = 0y=4 or - ( y = -7/3 is deleted) x = y+2 => x = 6The number is ‘64’.

7. A carpet whose length is times its width is laid on the floor of a rectangular room, with a margin of 1 foot all around. The area of the floor is 4 times that of the margin. Find the width of the room.

Let x and y be the length and breadth of the carpet.Given, x= 6x = 7y…………(1)Length of the floor = x+2Breadth of the floor = y+2Area of the margin = Area of the floor – Area of the carpet

=(x+2)(y+2) – xy

D C

S R 1 y P Q x A B

= xy +2x + 2y + 4 – xy= 2x+2y+4

Given, Area of the floor = 4 Area of the margin.(x+2)(y+2) = 4 (2x+2y+4)xy + 2x + 2y + 4 = 8x + 8y + 16xy – 6x – 6y – 12 = 0………..(2)

put, x = in (2)

7y2-78y – 72 = 07y2- 84y + 6y –72 = 07y(y-12) +6(7y-12) = 0(y-120 (7y+6) = 0y =12 ; y = - is deleted.

x = Width of the room = 12 feet Length of the room = 14 feet

Note: ‘10x+y’ is the value of the number

8. The sum of the first ‘x’ natural numbers is . How many numbers must be taken to get 351 as the sum?

x2+x = 702x2+x-702 = 0x2+27x-26x-702 = 0 x (x+27) – 26(x+27) = 0(x-26) (x-27) = 0x = 26 (x= -27 is to be deleted)The first 26 natural numbers must be taken to get a sum of ‘351’.

9. Solve: x + y = x2-y2 = 23

x+y= 23……..(1)x2 – y2 = 23……(2)

(x+y) (x-y) = 2323(x-y) = 23x-y = 1………(3)x+y = 23……….(1)

(1) + (3) , 2x = 24x = 12y= 11