Linear Equations in One Variable Video Lecture - (Maths) Class 8

0:0:0: hello students welcome to the lecture on
0:0:3: linear equations in one variable today
0:0:7: we will learn what are linear equations
0:0:9: in one variable how do we write them and
0:0:13: how do we solve them but before we dive
0:0:16: into that let me ask you a few
0:0:19: questions have you ever wondered how
0:0:22: much money you need to buy a certain
0:0:24: number of
0:0:25: chocolates or how many hours you need to
0:0:29: study to score a certain percentage in a
0:0:33: test these are all examples of
0:0:37: situations where you need to find the
0:0:39: value of an unknown
0:0:42: quantity and this is what we will learn
0:0:45: today but before learning how to solve a
0:0:49: linear equation in one variable we will
0:0:52: first learn about what are
0:0:55: variables
0:0:57: constants algebraic terms algebraic
0:1:2: expressions and algebraic
0:1:5: equations with the help of a simple
0:1:7: example let us first understand what are
0:1:10: variables and
0:1:13: constants suppose you want to buy X
0:1:16: number of
0:1:18: chocolates and each chocolate costs 10
0:1:22: Rupees then the total amount of money
0:1:25: you need to buy X chocolates will be 10
0:1:28: * X
0:1:30: or simply 10 x
0:1:33: rupees here x is a variable which means
0:1:38: it can take any value depending upon how
0:1:41: many chocolates you wish to
0:1:43: buy and 10 is a constant which means
0:1:48: that it does not change its
0:1:51: value so remember this variable changes
0:1:55: its value but constant never changes its
0:1:58: value
0:2:0: now what is a
0:2:2: term a multiplication of constants and
0:2:6: variables is called an algebraic
0:2:10: term for example in the previous example
0:2:15: 10 x was an algebraic
0:2:18: term similarly 5 y 3 Z 1 x Etc are all
0:2:29: algebraic
0:2:31: terms now what are algebraic
0:2:35: expressions when we write one or more
0:2:39: algebraic terms together separated by
0:2:42: the operations of plus or minus it
0:2:46: becomes an algebraic
0:2:49: expression for
0:2:52: example 10 x + 5
0:2:57: y 4 y - 9 + 3
0:3:3: Z
0:3:5: 2x + 7 y -
0:3:9: 5 Etc are all algebraic
0:3:15: expressions now what if we want to
0:3:18: compare two algebraic expressions using
0:3:20: an equal sign in
0:3:22: between for example what if I say that
0:3:26: the amount of money you need to buy EX
0:3:29: chocolates is equal to the 20 rupees in
0:3:32: your
0:3:34: pocket then we can very easily write
0:3:37: down that 10 x is equal to
0:3:41: 20 here 10 x is the algebraic expression
0:3:45: on the left side of the equal sign and
0:3:49: 20 is the algebraic expression on the
0:3:52: right side of the equal
0:3:54: sign this as a whole is an equation
0:4:0: now let's come to the name of the
0:4:3: chapter what is a linear equation in one
0:4:8: variable a linear equation in one
0:4:11: variable is a special type of equation
0:4:14: that has following
0:4:17: features as the name
0:4:19: suggests it has only one single variable
0:4:24: usually denoted by X Y or Zed
0:4:30: now what does linear
0:4:33: mean it means that the variable has
0:4:37: power of
0:4:39: one which means it is not raised to any
0:4:43: other exponent like 2 3
0:4:47: etc for example x - 7 is = 3 is a linear
0:4:54: equation but x² - 7 is = to 3 is not a
0:5:2: linear
0:5:3: equation because here x is raised to the
0:5:7: power of
0:5:9: two now let's come to the fun part how
0:5:14: to solve a linear equation in one
0:5:18: variable the main goal of solving a
0:5:20: linear equation in one variable is to
0:5:24: find the value of variable that makes
0:5:27: the left hand side that is the LHS of
0:5:31: the equation equal to the right hand
0:5:34: side that is rhs of the
0:5:37: equation for example if we have the
0:5:41: equation 10 x + 5 is equal to
0:5:47: 25 we want to find the value of x such
0:5:52: that LHS that is 10 x + 5 becomes equal
0:5:59: to the rhs that is
0:6:2: 25 now let's learn how to solve
0:6:7: this first of all we know that in an
0:6:11: equation the LHS is always equal to the
0:6:16: rhes so whatever operation it be plus
0:6:21: minus multiplication or division we do
0:6:25: on the LHS we have to do the exact same
0:6:29: on on the other
0:6:31: side let's look at the
0:6:33: example 10 x + 5 is =
0:6:38: 25 now here we need to find the value of
0:6:42: x how do we find it Well we will have to
0:6:47: remove the numbers that are with the X
0:6:50: on the LHS in this
0:6:53: equation 10 and 5 are there on the LHS
0:6:57: with x
0:7:0: let's first try to remove
0:7:2: five and how do we remove
0:7:5: it we know that + 5 - 5 is equal to
0:7:12: 0 so in Step 1 just subtract five from
0:7:17: the LHS and in the step two do the same
0:7:22: on the rhs so that the equation remains
0:7:26: balanced at all the times now the the
0:7:29: equation becomes 10 x + 5 - 5 is = to 25
0:7:37: -
0:7:38: 5 that is 10 x is = to
0:7:43: 20 now it's very easy to find X we just
0:7:48: need to remove 10 now that is on the
0:7:52: LHS and how do we remove that we know
0:7:56: that 10 upon 10 is equal to 1 1 so only
0:8:1: X will be left on the
0:8:4: LHS so in the step three just divide the
0:8:9: equation by 10 on both the sides that is
0:8:13: LHS as well as
0:8:16: rhs dividing with 10 on both these sides
0:8:20: we get 10 x upon 10 is equal to 20 upon
0:8:27: 10 that is
0:8:29: X is =
0:8:31: 2 now you might find this method big so
0:8:37: we have a shortcut method just shift all
0:8:41: the constant terms to the rhs and change
0:8:45: the sign this is called the method of
0:8:49: transposing let's understand how do we
0:8:52: apply this in the previous
0:8:54: question we had the equation 10 x + 5 is
0:8:59: equal to
0:9:1: 25 here the constant on the LHS is
0:9:6: 5 let's shift it to the
0:9:9: rhs the equation becomes 10 x is equal
0:9:15: to 25 -
0:9:18: 5 notice that how we changed the sign of
0:9:22: 5 from plus to
0:9:26: minus this is because behind this sces
0:9:30: we are actually subtracting five from
0:9:33: both these
0:9:35: sides but since 5 - 5 is 0 we are
0:9:40: skipping writing Min -5 on the
0:9:44: LHS so the equation now reduces to 10 x
0:9:49: is =
0:9:52: 20 now shift 10 from the LHS to the
0:9:57: rhs that becomes X is = to 20 upon
0:10:4: 10 as you can see when 10 was on the LHS
0:10:9: it was multiplied with
0:10:12: X but when we shifted it to rhs we
0:10:16: changed into
0:10:18: division so remember the following
0:10:21: things if a constant is added or
0:10:24: subtracted to a variable we shift it to
0:10:28: the r
0:10:29: and change its
0:10:32: sign if a constant is Multiplied to a
0:10:35: variable we shift it to the rhs and
0:10:39: divide
0:10:41: it if a constant is divided with a
0:10:45: variable we shift it to the rhs and
0:10:48: multiply
0:10:50: it so let's try some questions find the
0:10:54: solution of the following
0:10:57: problems first is 2x - 1 is = to 14 -
0:11:6: x here 2x - 1 is the LHS and 14 - x is
0:11:14: the
0:11:16: rhs now our aim is to shift all the
0:11:20: variable terms to LHS and all the
0:11:23: constant terms to the
0:11:26: rhs so shifting - 1 to rhs and - x to
0:11:32: LHS we get 2x + x is equal to 14 +
0:11:41: 1 now kindly notice that the signs of -
0:11:46: x and -1 have been changed because they
0:11:51: have been shifted from one side to the
0:11:54: other
0:11:55: side on solving both the sides the
0:11:58: equation reduces to 3x is =
0:12:4: 15 now we want the value of x so we will
0:12:9: again use the method of transposing and
0:12:11: move three to the rhs when we move 3 to
0:12:15: the rhs the equation
0:12:18: becomes X is equal to 15 upon
0:12:23: 3 and we all know that 15 / 3 = to 5
0:12:29: so finally the value of x turns out to
0:12:33: be
0:12:35: five now let's move to the next question
0:12:39: question says find the value of M and
0:12:43: the given equation is 3 m is = to 5 m -
0:12:49: 8 upon
0:12:51: 5 try solving this question on your own
0:12:56: first if your answer is 4 upon 5 well
0:13:1: done students you are absolutely
0:13:3: correct let's have a look at the
0:13:5: solution now here 3 m is the LHS and 5 m
0:13:12: - 8 upon 5 is the
0:13:16: rhs now just recall how we solved the
0:13:19: previous question our approach will be
0:13:22: to shift all the variable terms to the
0:13:25: LHS and all the constant terms to the
0:13:28: rhs
0:13:30: so shifting 5 m to the LHS we get 3 m -
0:13:36: 5 m is equal to - 8 upon
0:13:40: 5 the sign of 5 m changes and becomes
0:13:46: negative solving the LHS the equation
0:13:49: becomes - 2 m is equal to - 8 upon
0:13:55: 5 now since we have to find the value
0:13:58: Val of M we will divide both the sides
0:14:2: by Min
0:14:3: -2 on doing so we get - 2 m upon -2 is
0:14:10: equal to - 8 upon 5 into -1 upon
0:14:17: 2 on the LHS -2 and Min -2 gets
0:14:21: cancelled out and on the rhs - 8 upon 5
0:14:27: into - 1 upon 2 reduces to 4 upon 5
0:14:33: leaving us with our answer m is = to 4
0:14:38: upon
0:14:40: 5 now let's move to the next question so
0:14:45: question reads 6 x + 1 / 3 + 1 isal to x
0:14:54: - 3 /
0:14:57: 6 now this is a different type of
0:15:0: question here the variable parts are in
0:15:4: the form of
0:15:5: fractions we need to simplify this
0:15:8: equation to solve
0:15:10: it so in the first step we will take the
0:15:14: LCM of both the denominators on the LHS
0:15:18: and the
0:15:19: rhs so the numbers and the denominators
0:15:22: in the LHS and rhs are three and six
0:15:27: respectively what is the LCM of 3 and
0:15:29: six think
0:15:31: quickly yes you are correct it is
0:15:35: six now multiply both the sides with
0:15:39: six the equation
0:15:42: becomes 6 into 6 x + 1 / 3 + 6 into 1 is
0:15:52: = to 6 into x - 3 / 6
0:16:0: here the LHS is equal to 6 into 6 x + 1
0:16:6: whole / 3 + 6 into
0:16:11: 1 now 6 / 3 is 2 so the LHS becomes 2 *
0:16:19: 6 x + 1 +
0:16:22: 6 that
0:16:24: is 12 x + 2 +
0:16:27: 6 solving this we get our LHS is equal
0:16:32: to 12 x +
0:16:35: 8 now let's look at the
0:16:38: rhs 6 / 6 is 1 so the rhs simply becomes
0:16:45: x -
0:16:47: 3 now because this is an equation so LHS
0:16:52: must be equal to the
0:16:54: rhs therefore we can write down that 12
0:16:58: 12 x + 8 is = x -
0:17:3: 3 now we have simplified it just apply
0:17:8: the methods we used previously to solve
0:17:11: the questions shift all the variable
0:17:14: terms to the LHS and all the constant
0:17:18: terms to the
0:17:19: rhs the equation
0:17:22: becomes 12 x - x is equal to - - 3 -
0:17:31: 8 solving both the sides we get 11x is
0:17:37: equal to
0:17:40: -1 now we just need the value of x so we
0:17:45: divide both the sides by
0:17:49: 11 this becomes 11x upon 11 is equal to
0:17:55: -1 upon
0:17:57: 11 11 / 11 is 1 so X becomes equal to
0:18:6: -1 so you saw how easily we could solve
0:18:10: this
0:18:11: question now with this we have come to
0:18:14: the end of this video and we learned
0:18:16: some interesting ways to solve a linear
0:18:19: equation in one
0:18:21: variable we hope that you have
0:18:23: understood the concept of linear
0:18:25: equations in one variable now here's a
0:18:29: secret on how you can study effectively
0:18:32: with the edurev
0:18:34: app you can learn with chapter notes
0:18:37: watch video lectures and solve ncrt
0:18:40: based McQ tests of all this chapter on
0:18:44: eduv and that's not all you also get
0:18:48: amazing courses for class 11th and much
0:18:51: more for your class 11
0:18:53: preparation thank you