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ERR - 9th - Maths [Linear Equations in One Variable]

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9th (Maths)
Linear Equations in One Variable
Solving Linear Equations without Brackets Solving Linear Equations without Brackets: Solving an equation means finding the value of the variable which satisfies it. Rules for Solving Linear Equations In One Variable: There are certain facts about equality. Rule 1: Same quantity (number) can be added to both sides of an equaton without changing the equality. Rule 2: Same quantity can be subtracted from both sides of an equation without changing the equality. Rule 3: Both sides of an equation may be multiplied by the same non - zero number without changing the equality. Rule 4: Both sides of an equation may be divided by the same non - zero number without changing the equality. These rules are used to solve the problems: Example: Solve the Following equation and verify the result : $large frac{x}{5}+11=frac{1}{15}$ Solution We have, $large frac{x}{5}+11=frac{1}{15}$ $large Rightarrow ;;frac{x}{5}+11-11=frac{1}{15}-11$ [ Subtracting 11 from both sides] $large Rightarrow ;;frac{x}{5}=frac{1}{15}-11$ $large Rightarrow ;;frac{x}{5}=frac{1-165}{15}$ $large Rightarrow ;;frac{x}{5}=-frac{164}{15}$ $large Rightarrow ;;5times frac{x}{5}=5times -frac{164}{15}$ $large Rightarrow ;;x=-frac{164}{3}$ $large Rightarrow ;;x=-frac{164}{3};is ;the ;solution ;of ;the; given ;equation.$ Verification Putting the vaue of x in LHS, we get $large =frac{x}{5}+11=frac{-164}{3}times frac{1}{5}+11=frac{-164}{15}+11=frac{-164+165}{15}=frac{1}{15}$ $large and, ; R.H.S; =frac{1}{15}$ $large Therefore ; L.H.S = R.H.S. ; For;x=frac{-164}{3}$ $large Hence;x=frac{-164}{3}$ , $large x=frac{-164}{3};is ;the ;solution ;of; the; given; equation.$
Solving Equation with Brackets Solving Equation with Brackets: There are situations in which both the sides of an equation contain both variable (unknown quantity) and constant (numerals). In such cases, we first simplify two sides to their simplest forms and then transpose (shift) terms containing variable on R.H,S to L.H.S and constant terms on L.H.S and R.H.S By transposing a term form one side to other side, we mean changing its sign and carrying it to the other side. In transposition the plus sign of the term changs into minus sign on oher side and vice-versa. The transposition method involves the following steps: Step I Obtain the linear euation. Step II Identify the variable (unknown quantity) and constants (numerals). Step III Simplify the L.H.S and R.H.S to their simplest forms by removing brackets. Step IV Transpose all terms containing variable on L.H.S and constant terms on R.H.S Note that the sign of the terms will change in shifting then form L.H.S to R.H.S and vice-versa. Step V SImplify L.H.S and R.H.S in the simplest form so that each side contains just one term. Step VI Solve the equation obtaineed in step V by dividing both sides by the coefficient of the variable on L.H.S Example: Solve: $large frac{x+2}{6}-left [ frac{11-x}{3}-frac{1}{4}right ]=frac{3x-4}{12}$ Solution We have, $large frac{x+2}{6}-left [ frac{11-x}{3}-frac{1}{4}right ]=frac{3x-4}{12}$ The denominators on two sides of the given equation are 6, 3, 4 and 12. Their LCM is 12. Multiplying both sides of the given equaton by 12, we get $large 12left (frac{x+2}{6} right )-12left ( frac{11-x}{3}-frac{1}{4} right )=12left ( frac{3x-4}{12} right )$ $large Rightarrow ;;2(x+2)-4(11-x)+12times frac{1}{4}=(3x-4)$ $large Rightarrow ;;2x+4-44+4x+3=3x-4$ $large Rightarrow ;;6x-37=3x-4$ $large Rightarrow ;;6x-3x=37-4$ [Transposing 3x to LHS and -37 to RHS] $large Rightarrow ;;3x=33$ $Rightarrow ;;x=frac{33}{3}$ [Dividing both sides by 3] $Rightarrow ;;x=11$ Check Subtituting x = 11 on both sides of the given equation, we get $L.H.S.=frac{x+2}{6}-left ( frac{11-x}{3}-frac{1}{4} right )$ $=frac{11+2}{6}-left ( frac{11-11}{3}-frac{1}{4} right )=frac{13}{6}-left ( 0-frac{1}{4} right )=frac{13}{6}+frac{1}{4}=frac{26+3}{12}=frac{29}{12}$ $and ;;R.H.S.=frac{3x-4}{12}=frac{3times 11-4}{12}=frac{33-4}{12}=frac{29}{12}$ Thus, for x = 11, we have L.H.S = R.H.S
Statement Sums involving Linear Equations Statement Sums involving Linear Equation: In these sums a certain situation is given we have to analyse the statement and then suppose a certain quantity. and will then try to identify a relation with the quantity we have supposed. The procedure to translate a word problem in the form of an equation is known as the formulation of the problem. Thus, the process of solving a word problem consists of two parts, namely, formulation and solution. The following steps should be followed to solve a word problem: Step I Read the problem carefully and note what is given and what is required. Step II Denote the unknown quantity by some letters, say x, y, z, etc. Step III Translate the statements of the problem into mathematical statements. Step IV Using the condition (s) given in the problem, form the equation Step V Solve the equation for the unknown. Step VI Check whether the solution satisfies the equation. Example: The numerator of a fraction is 4 less than the denominatr. If 1 is added to both its numerator and denominator, it becomes 1/2. FInd the fraction. Solution Let the denominator of the fraction be x. Then, Numerator of the fraction = x - 4 $large therefore$ Fraction $large =frac{x-4}{x}$ If 1 is added to both its numerator and denominator, the fraction becomes $large frac{1}{2}$ . $large therefore$ $large frac{x-4+1}{x+1}=frac{1}{2}$ $large Rightarrow frac{x-3}{x+1}=frac{1}{2}$ $large Rightarrow 2(x-3)=x+1$ [ Using cross - multiplication] $large Rightarrow 2x-6=x+1$ $large Rightarrow 2x-x=6+1$ $large Rightarrow x=7$ Putting x = 7 in (i), we get Fraction $large =frac{7-4}{7}=frac{3}{7}.$ Hence, the given fraction is $large frac{3}{7}.$ Example : How much pure alcohol be added to 400 ml of a 15% solution to make its strength 32%? Solution Let x ml pure alcohol be added to 400 ml of a 15% solution to make its strength 32%. Here, 15% solution means that there is 15 ml pure alcohol in a solution of 100 ml. Now, Quantity of alcohol in 100 ml solution = 15 ml $large therefore$ Quantity of alcohol in one ml solution $large =frac{15}{100};ml$ Quantity of alcohol in 400 ml solution $large =frac{15}{100}times 400;ml=60; ml$ Total quantity of the solution = (400 + x) ml Total quantity of alcohol in (400 + x) ml solution = (60 + x) ml $large therefore$ Quantity of alcohol in one ml $large =frac{60+x}{400+x};ml$ Quantity of alcohol in 100 ml $large large =frac{60+x}{400+x}times 100;ml$ $large Rightarrow$ Strength of the solution $large =left ( frac{60+x}{400+x} right )times$ 100% But, the strength of the solution is given as 32%. $large therefore ;;frac{60+x}{400+x}times 100=32$ $large Rightarrow ;;100(60+x)=32(400+x)$ [ Multiplying both sides by (400 + x)] $large Rightarrow ;;6000+100x=12800+32x$ $large Rightarrow ;;100x-32x=12800-6000$ $large Rightarrow ;;68x=6800$ $large Rightarrow ;;frac{68x}{68}=frac{6800}{68}$ $large Rightarrow ;;x=100$ Thus, 100 ml alcohol must be added to make 32% strength of the solution.
Solving Statement For Upstream And Downstream In these type of situations, while calculating upstream speed we subtract the speed of river from the speed of boat because the river is pushing the boat downwards even when the boat is moving upwards. Whereas when the boat is moving downstream the speed is calculated by adding the speed of river to the speed of boat. Let the speed of a boat ( or a body) in still water be x km/h and that of stream be y km/h, then Speed of boat downstream = (x +y ) km/h Speed of boat upstream = ( x - y) km/h Let the speed of boat in downstream and upstream be u km/h and v km/h, then Speed of boat in still water $=frac{1}{2}(u-v)km/h$ Illustration: A boat goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream is 1 km/hr. Calculate the speed of boat in still water. A. 15 Km/h B. 7.5 Km/h C. 19 Km/h D. 7 Km/h Answer: C Solution: Let the speed of boat in still water be x km/hr Speed of stream = 1 km/hr Speed upstream = x - 1 km/hr Speed Downstream = x + 1 km/hr As we know $large speed= frac{distance}{time}$ $Rightarrow distance= speed; X ;time$ Distance upstream = (x-1) X 10 = 10x - 10 Km Distance Downstream = (x+1) X 9 = 9x + 9 Km According to the question the distance is the same 10x-10 = 9x + 9 10x - 9x = 9+ 10 x = 19 km/hr So the speed of boat in still water = 19 km/h Hence the correct option is C. Illustration: A motor boat whose speed in still water is 15 km/h, goes 40 km down stream and comes back in a total time of 6 hours. Find the speed of stream. A. 5 Km/h B. 6 Km/h C. 15 Km/h D. 7 Km/h Answer: A Solution: Let the speed of stream = x km/h Speed upstream = 15 - x km/h Speed downstream = 15 + x km/h Distance = 40 km As we know that $speed= frac{distance}{time}$ $time= frac{distance}{speed}$ $time_{upstream}= frac{40}{15-x}$ $time_{downstream}= frac{40}{15+x}$ According to the question $frac{40}{15+x}+ frac{40}{15-x}=6$ $40left [ frac{1}{15+x}+ frac{1}{15-x} right ]=6$ $left [ frac{15-x+15+x}{(15+x)(15-x)} right ]=frac{6}{40}$ $left [ frac{30}{(15^2-x^2)} right ]=frac{6}{40}$ $left [ frac{1}{(225-x^2)} right ]=frac{6}{40;X;30}$ $left [ frac{1}{(225-x^2)} right ]=frac{1}{200}$ $200=225-x^2$ $200-225+x^2=0$ $x^2-25=0Rightarrow x^2=25Rightarrow x=pm 5$ The speed cannot be negative The speed of stream is x= 5 km/h Hence the correct option is A.
Solving Statements Involving Age Solving Statements Involving Age: While solving Problems Involving Age, we consider the age to be found as x. If in the statement it is mentioned after certain years then that value is added in the age of all the people. Similarly if it is mentioned before or ago certain years then that value is subtracted from the age of all the people. Example: Reena is four years older than Rohan. Five years ago, Rohan's age was four times Reena's age. Find the ages of Reena and Rohan. Solution Let Rohan's age be x years. Then, Reena's age is (x+4)years. Five years ago, Rohan's age was (x-5) years. Reena's age was (x+4-5) years = (x-1) years. It is given that five years ago Rohan's age was four times Reena's age. $large therefore ;;x-1=5(x-5)$ $Rightarrow ;;x-1=5x-25$ $Rightarrow ;;x-5x=-25+1$ $Rightarrow ;;-4x=-24$ $Rightarrow ;;frac{-4x}{-4}=frac{-24}{-4}$ $Rightarrow ;;x=6$ Hence, Rohan's age = 6 years Reena's age = (x+4) years = (6+4) years = 10 years. Example 2: After 12 years I shall be 3 times as old as I was 4 years ago. Find my present age. Solution Let my present age be x years. After 12 years my age will be (x+12) years. 4 years ago my age was (x - 4) years. It is given that after 12 years I shall be 3 times as old as I was 4 years ago. $therefore ;;x+12=3(x-4)$ $Rightarrow ;;x+12=3x-12$ $Rightarrow ;;x-3x=-12-12$ $Rightarrow ;;-2x=-24$ $Rightarrow ;;frac{-2x}{-2}=frac{-24}{-2}$ $Rightarrow ;;x=12$ Thus, my present age is 12 years.
Solving Statements For A Two Digit Number Solving Statements For a Two Digit Number: A two digit number consists of two digits one at unit's place and the other at ten's place. Then two express the number we will use the expanded form for conversion. Example: A number consists of two digit whose sum is 8. If 18 is added to the number its digits are reversed. Find the number. Solution Let ones digit be x. Since the sum of the digits is 8. Therefore, tens digit = 8- x. $large therefore$ Number $large =10times (8-x)+x=80-10x+x=80-9x$ ....(i) Now, Number obtained by reversing the digit The digit at units place will be = 8-x. The digit at ten's place will be = x. $large The; reversed ;Number =10times (x)+ 8-x=10x+8-x=9x+8$ It is given that if 18 is added to the number its digits are reversed. $large therefore$ Number + 18 = Reversed Number $large Rightarrow ;;80-9x+18=9x+8$ $large Rightarrow ;;98-9x=9x+8$ $large Rightarrow ;;98-8=9x+9x$ $large Rightarrow ;;90-18x$ $large Rightarrow ;;frac{18x}{18}=frac{90}{18}$ $large Rightarrow ;;x=5$ Putting the value of x in (i), we get Number $large =80-9times 5=80-45=35$
Statement sums involving Coins Statement sums involving Coins: While solving statements involving coins, we will multiply the value of the coin with number of coins to get the amount. Example: Saurabh has Rs 34 fifty paise and twenty-five paise coins. If the number of 25-paise coins be twice the number of 50-paise coins, how many coins of each kind does he have? Solution Let the number of 50-paise coins be x. Then, Number of 25-paise coins = 2x $large therefore$ Value of x t Fifty paise coins $large =50times x;;paise=Rsfrac{50times x}{100}=Rsfrac{x}{2}$ Value of 2x twenty-five paise coins $large =25times 2x$ paise = 50 x paise $large = Rs.;large frac{50x}{100}= Rs;frac{x}{2}$ $large Total ;value; of ;all; coins = Rsleft ( frac{x}{2}+frac{x}{2}right )=Rs ;;x$ But, the total value of the money is Rs 34 $large therefore ;;x=34$ Thus, number of 50-paise coins = 34 Number of twenty-five paise coins $large =2x=2times 34=68$
Solving Statement For Upstream And Downstream Solving Statements For Upstream and Downstream: In these type of situations, while calculating upstream speed we subtract the speed of river from the speed of boat because the river is pushing the boat downwards even when the boat is moving upwards. Whereas when the boat is moving downstream the speed is calculated by adding the speed of river to the speed of boat. Example: A boat goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream is 1 km/hr. Calculate the speed of boat in still water. Solution: Let the speed of boat in still water be x km/hr Speed of stream = 1 km/hr Speed upstream = x-1 km/hr Speed Downstream = x + 1 km/hr $large speed= frac{distance}{time}$ $large distance= speed; X ;time$ Distance upstream = (x-1) X 10 = 10x - 10 Km Distance Downstream = (x+1) X 9 = 9x + 9 Km According to the question the distance is the same 10x-10 = 9x + 9 10x - 9x = 9+ 10 x = 19 km/hr So the speed of boat in still watre = 19 km/hr.
Solving Statements Involving Age Solving Statements Involving Age: While solving Problems Involving Age, we consider the age to be found as x. If in the statement it is mentioned after certain years then that value is added in the age of all the people. Similarly if it is mentioned before or ago certain years then that value is subtracted from the age of all the people. Illustration: Reena is four years older than Rohan. Five years ago, Reena's age was five times Rohan's age. Find the ages of Reena and Rohan. Solution Let Rohan's age be x years. Then, Reena's age is (x+4)years. Six years ago, Rohan's age was (x-5) years. Reena's age was (x+4-5) years = (x-1) years. It is given that five years ago Reena's age was five times Rohan's age. $therefore ;;x-1=5(x-5)$ $Rightarrow ;;x-1=5x-25$ $Rightarrow ;;x-5x=-25+1$ $Rightarrow ;;-4x=-24$ $Rightarrow ;;frac{-4x}{-4}=frac{-24}{-4}$ $Rightarrow ;;x=6$ Hence, Rohan's age = 6 years Reena's age = (x+4) years = (6+4) years = 10 years. Illustration: After 12 years I shall be 3 times as old as I was 4 years ago. Find my present age. Solution Let my present age be x years. After 12 years my age will be (x+12) years. 4 years ago my age was (x - 4) years. It is given that after 12 years I shall be 3 times as old as I was 4 years ago. $therefore ;;x+12=3(x-4)$ $Rightarrow ;;x+12=3x-12$ $Rightarrow ;;x-3x=-12-12$ $Rightarrow ;;-2x=-24$ $Rightarrow ;;frac{-2x}{-2}=frac{-24}{-2}$ $Rightarrow ;;x=12$ Thus, my present age is 12 years. Illustration: One year age Ramesh was 8 times as old as his son. Now his age is equal to the square of his son's age. Find their present age. Solution: Let his son's age one year ago = x Ramesh's age is one year ago= 8x Present age of son = x+1 Present age of Ramesh = 8x + 1 According to the question: $large large Ramesh; present; age= (Son; present; age)^2$ $large 8x +1= (x+1)^2$ $large 8x +1= x^2+2x+1$ $large 8x +1- x^2-2x-1=0$ $large - x^2+6x=0$ $large x^2-6x=0$ $large x(x-6)=0$ either x = 0 or x = 6 As the age can not be zero so son's present age =x+1= 6+1=7 years Ramesh's age= 8x+1 = 8(6)+1=48+1=49 years
Solving Statements For A Two Digit Number Solving Statements For a Two Digit Number: A two digit number consists of two digits one at unit's place and the other at ten's place. Then two express the number we will use the expanded form for conversion. Example: A number consists of two digit whose sum is 8. If 18 is added to the number its digits are reversed. Find the number. Solution Let ones digit be x. Since the sum of the digits is 8. Therefore, tens digit = 8- x. $large therefore$ Number $large =10times (8-x)+x=80-10x+x=80-9x$ ....(i) Now, Number obtained by reversing the digit The digit at units place will be = 8-x. The digit at ten's place will be = x. $large The; reversed ;Number =10times (x)+ 8-x=10x+8-x=9x+8$ It is given that if 18 is added to the number its digits are reversed. $large therefore$ Number + 18 = Reversed Number $large Rightarrow ;;80-9x+18=9x+8$ $large Rightarrow ;;98-9x=9x+8$ $large Rightarrow ;;98-8=9x+9x$ $large Rightarrow ;;90-18x$ $large Rightarrow ;;frac{18x}{18}=frac{90}{18}$ $large Rightarrow ;;x=5$ Putting the value of x in (i), we get Number $large =80-9times 5=80-45=35$ Example: A two digit number is such that product of two digits is 14. When 45 is added to the number, the digits are reversed. Find the number. Solution: Let ones digit be x. Since the product of the digits is 14. $large Therefore,; tens; digit;=frac{14}{x}$ $large therefore$ Number $large =10times (frac{14}{x})+x=frac{140}{x}+x=frac{140+x^2}{x}$ ....(i) Now, Number obtained by reversing the digit $large The; digit; at ;units; place ;will; be = frac{14}{x}.$ The digit at ten's place will be = x. $large The; reversed ;Number =10x+ frac{14}{x}=frac{10x^2+14}{x}$ It is given that if 45 is added to the number its digits are reversed. $large therefore$ Number + 45 = Reversed Number $large frac{140+x^2}{x} +45 = frac{10x^2+14}{x}$ $large frac{140+x^2+45x}{x} = frac{10x^2+14}{x}$ $large 140+x^2+45x = 10x^2+14$ $large 140+x^2+45x - 10x^2-14=0$ $large -9x^2+45x + 126=0$ $large -9(x^2-5x -14)=0$ $large x^2-5x -14=0$ $large x^2-7x+2x -14=0$ $large x(x-7)+2(x-7)=0$ $large (x-7)(x+2)=0$ Either x = 7 or x = -2 As a digit can not be negative, So the digit at units place is 7 So the digit at tens place = 14/7 = 2 Hence the number = 27

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