Linear equation in two variables
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Linear equation in two variables

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A project by Ananya Gupta et.al. of 9th grade for KV OFD Raipur Dehradun

A project by Ananya Gupta et.al. of 9th grade for KV OFD Raipur Dehradun

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    Linear equation in two variables Linear equation in two variables Presentation Transcript

    • A project made by : Ananya Gupta Priya Srivastava Manisha Negi Muskan Sharma Class IX CKV OFD Raipur, Dehradun
    • A common form of a linear equation in the two variablesx and y is y= mx + cwhere m and c designate constants. The origin of thename "linear" comes from the fact that the set ofsolutions of such an equation forms a straight linein the plane. In this particular equation, the constant mdetermines the slope or gradient of that line, and theconstant term “c" determines the point at which the linecrosses the y-axis, otherwise known as the y-intercept.Since terms of linear equations cannot contain productsof distinct or equal variables, nor any power(other than 1)or other function of a variable, equations involving termssuch as xy, x2, y1/3, and sin(x) are nonlinear.
    • A system of linear equations is two or more linearequations that are being solved simultaneously.In general, a solution of a system in two variablesis an ordered pair that makes BOTH equations true.In other words, it is where the two graphs intersect, whatthey have in common. So if an ordered pair is a solutionto one equation, but not the other, then it is NOT asolution to the system.A consistent system is a system that has at least one solution.An inconsistent system is a system that has no solution.The equations of a system are dependent if ALL the solutionsof one equation are also solutions of the other equation.In other words, they end up being the same line.The equations of a system are independent if theydo not share ALL solutions. They can have one point incommon, just not all of them.
    • In this presentation we will be lookingat systems that have two equations andtwo unknowns.We will look at solving them in threedifferent ways: 1-by graphing, 2-by thesubstitution method, and 3-by theelimination method.So, lets go ahead and look at thesesystems.
    • The Graph Method
    • There are three possible outcomes that we mayencounter when working with these systems:•one solution•no solution•infinite solutionsOne SolutionIf the system in two variables has one solution, it is an ordered pair that is asolution to BOTH equations.No SolutionIf the two lines are parallel to each other, they will never intersect. This meansthey do not have any points in common. In this situation, you would have no solution.Infinite SolutionsIf the two lines end up lying on top of each other, then there is an infinitenumber of solutions.
    • 4 5One Solution No Solution Infinite Solutions
    • Solving by GraphingStep 1: Graph the first equation. Unless the directions tell us differently, we canuse any "legitimate" way to graph the line.Let the Equation be x+y=5Solve the equation for yFor this let us make “y” the subject of the equation y = 5 – xNow substituting different values for x we can find respective values of y. x= -2 -1 0 1 2 y= 7 6 5 4 3Now with these different pairs of values of x and y we can plot the line on a graph.Similarly we can solve the second equation 2x – y = -2 , or y = 2x + 2the results are in the following table- x= -2 -1 0 1 2 y= -2 0 2 4 6
    • Step 2: Plot the two equations on a graph, with similar co-ordinatesNow we have something like this …..Step 3: Find the solution. We need to ask ourselves, is there any place that the twolines intersect, and if so, where? The answer is yes, they intersect at (1, 4).Step 4: Check the proposed ordered pair solution in BOTH equations.You will find that if you plug the ordered pair (1, 4) into BOTH equations of theoriginal system, that this is a solution to BOTH of them.The solution to this system is (1, 4). OK Got it/ Please repeat
    • The Substitution Method
    • Solve by the Substitution MethodStep 1: Simplify if needed.This would involve things like removing ( ) and removing fractions.To remove ( ): just use the distributive property.To remove fractions: since fractions are another way to write division, and the inverseof divide is to multiply, you remove fractions by multiplying both sides by the LCD ofall of your fractions.Step 2: Solve one equation for either variable.If you need to solve for a variable, then try to pick one that has a 1 as a coefficient.That way when you go to solve for it, you wont have to divide by a number and runthe risk of having to work with a fractionStep 3: Substitute what you get for step 2 into the other equation.This is why it is called the substitution method. Make sure that you substitute theexpression into the OTHER equation, the one you didnt use in step 2.This will give you one equation with one unknown.
    • Step 4: Solve for the remaining variable . Solve the equation set up in step 3 for the variable that is left. If your variable drops out and you have a FALSE statement, that means your answer is no solution. If your variable drops out and you have a TRUE statement, that means your answer is infinite solutions, which would be the equation of the line.Step 5: Solve for second variable. If you come up with a value for the variable in step 4, that means the two equations have one solution. Plug the value found in step 4 into any of the equations in the problem and solve for the other variable.Step 6: Check the proposed ordered pair solution in BOTH original equations. You can plug in the proposed solution into BOTH equations. If it makes BOTH equations true, then you have your solution to the system. If it makes at least one of them false, you need to go back and redo the problem.
    • Whatever you do to an equation, do the S A M E thingto B O T H sides of that equation!
    • Example: Solve the following equations 3(x+y) = 30/2 and 1/3(x-y) = -1/3Solution:Step 1: Simplify if needed. 3(x+y) = 30/2 1/3(x-y) = -1/3 Or 3x+3y = 30/2 (distributive property) or 1/3x-1/3y = -1/3 (distributive property) Or 3x+3y = 15 (solving 30/2 = 15) or x-y = -1 (multiply both sides by 3 ) Or x+y = 5 (divide both sides by 3)Step 2: Solve one equation for either variable. x+y = 5 Or x = (5-y)Step 3: Substitute what you get for step 2 into the other equation. x-y = -1 Or (5-y)-y = -1 (Substitute x = (5-y) in second equation) Or 5-2y = -1 Or -2y = -1-5 = -6 (Subtract 5 form both sides) Or 2y = 6 (Multiply both sides by -1) Or y=3 (Divide both sides by 2)
    • Step 4: Solve for the remaining variable . x-y = -1 substituting y = 3 in this equation, we have; x-3 = -1 (Add 3 to both sides) x=2 Now when we have got the solution i.e. x = 2 and y = 3, let’s check it. For this simply substitute the values of x and y in any of the equations. x+y = 5 Or 2+3 = 5 Or 5 = 5 Which is correct. Hence the answer to the above problem is x=2 y=3 OK Got it/ Please repeat
    • The Elimination Method Hee..he..he.. 120 kg 75
    • Solve by the Elimination by Addition MethodStep 1: Simplify and put both equations in the form Ax + By = C if needed.Step 2: Multiply one or both equations by a number that will create oppositecoefficients for either x or y if needed.In that process, we need to make sure that one of the variables drops out, leaving uswith one equation and one unknown. The only way we can guarantee that is if we areadding opposites. The sum of opposites is 0.If neither variable drops out, then we are stuck with an equation with two unknownswhich is unsolvable.Step 3: Add equations.Step 4: Solve for remaining variable.Solve the equation found in step 3 for the variable that is left.If both variables drop out and you have a FALSE statement, that meansyour answer is no solution.If both variables drop out and you have a TRUE statement, that meansyour answer is infinite solutions, which would be the equation of the line.
    • Step 5: Solve for second variable.If you come up with a value for the variable in step 4, that means the twoEquations have one solution. Plug the value found in step 4 into any of theequations in the problem and solve for the other variable.Step 6: Check the proposed ordered pair solution in BOTH original equations.You can plug the proposed solution into BOTH equations. If it makes BOTH equationstrue, then you have your solution to the system.If it makes at least one of them false, you need to go back and redo the problem.Let’s make it clear with an example:Solve the following equations x/3+5y = 26 4x+3y/2 = 38/2
    • Solution:Step 1: Simplify and put both equations in the form Ax + By = C if needed. x/3+5y = 26 …(1) 4x+3y/2 = 39/2 …(2)multiplying eq. (1) by 3 and eq.(2) by 2 we have x+15y = 78 …(1) 8x+3y = 39 …(2)Step 2: Multiply one or both equations by a number that will create oppositecoefficients for either x or y if needed. Here multiplying eq. (2) with -5 will solve the purpose. x+15y = 78 …(1) -40x-15y = -195 …(2)Step 3: Add equations. x+15y = 78 -40x-15y = -195 -39x = -117Step 4: Solve for remaining variable. -39x *(-1) = -117*(-1) or 39x = 117 or 39x/39 = 117/39 or x = 3
    • Step 5: Solve for second variable.Plug-in x = 3 in any of the equations and solve for y x+15y = 78 …(1) or 3+15y = 78 or 3+15y-3 = 78-3 or 15y = 75 or 15y/15 = 75/15 or y =5The answer to the given pair of equations is x = 3 ; y = 5Step 6: Check the proposed ordered pair solution in BOTH original equations. x/3+5y = 26 4x+3y/2 = 39/2 plug-in x = 3 and y = 5 in this equation 3/3+5*5 = 26 4*3+3*5/2 = 39/2 or 1+25 = 26 4*3*2+3*5 /2*2 = 39/2*2 or 26 = 26 24+15 = 39 or 39 = 39The ordered pair (3, 5) is a solution to BOTH the equations. (3,5) is a solution to the system. OK Got it/ Please repeat/ Repeat the presentation
    • A PROJECT BY ANANYA GUPTA PRIYA SRIVASTAVAAcknowledgements MANISHA NEGI &We are thankful to: MUSKAN SHARMAwww.wikipedia.comwww.watmu.edu OFOur Principal Shri. T.P. Gaur K.V. O.F.D.Maths Teacher Mr. A.A. Ansari RAIPUROur Parents DAHRADUN UTTARAKHAND