Algebra+tutorial Linear+eqn

Algebra Tutorial-solving system of linear equations A syste m of line ar e quations is re ally a pair of e quations, re pre se nting two straigh t line s .[We a re now inte re ste d in two variable s, say x and Y only.] The two straight line s may inte rse ct or cross at a point , say x = x1 and y = y1. Now the va lue s x1,y1 [coordinate s ina graph of x ve rsus y] is the solution for this syste m of e qua tions.It me ans that the point of inte rse ction of two line s have a common point fo r both the line s. Why we ta ke up this case of two e quations in two variable s. We ll, the y are inte rconne cte d. Le t x be the price of orange s and y ,the price of apple s. S uppose 3 orange s and 6 apple s cost $6. S uppose 4 orange s and 4 apple s cost $6. We want to find the cost of one ora nge and one apple . Le t us se t up the e quations. Le t x be the price of one orange and y the price of one apple . The n

3 x +6 y =6 4x +4y =6

------(1) ------(2)

Now we ca n solve for x and y, taking the two e quations toge the r or 'simultane ously". We ll, the se two e quations are actually two straight line s. Re write the first e quation: 6y = - 3x + 6 or y = - 0.5 x + 1 This is a stra ight line in the form: y = ax + b whe re a is the slope and b is the inte rce pt. S imilarly, the se cond e quation is : 4y = 6 - 4x or y = -x + 1.5 We can draw the se two straight line s and note the point of inte rse ction.This is the gra phica l me thod. But he re we use two othe r me thods: Elimination method I have numbe re d the e quations as (1) and (2). Look at the two e quations.The first e qua tion has 3x te rm and the se cond has 4x te rm. We wish to have the same first x te rm in both the e quations.S o I multiply first e quation by 4 and the se cond e quation by 3: 12 x + 24 y = 24 12 x + 12 Y = 18 ---------------------S ubtract : 0 x + 12 y = 6 or 12 y = 6 y= 0.5 The cost of apple is $ .5 only. Having known y value , we find the x value using the e quation 1: 3 x + 6 x .5 = 6 3 x +3 =6 3 x = 6 -3 = 3 x =1 The cost of orange is $1 We can che ck with e quation 2 the re sult we got : is corre ct. S ome practice proble ms we can try!

1 Solve :

2 x- y=6 x + 3y = 3

4 x 1 + 4 x .5 = 4 + 2 = 6 ....th e re sult

Let us keep the first equation as such , but multiply all the terms in second equation by 2. We write: equation (1) x 1 equation (2) x 2 Subtract: or y = 0. Then using equation (1)

2x - y = 6 2x + 6y = 6 -----------------0 - 7y = 0

2 x- y=6 2 x- 0 =6 2x =6 or The solution is : x = 3, y = 0

x =3

Note what if the two lines representing the equations are parallel...well, no solution exists. Take the following example: x - 2y = 4 2x - 4y = 8 Try solving this : you will get 0 = 0. Not a surprise because the two equations are the same.There is only one equation as such. 2 Solve : x + y = 3 x- y=1 3 Solve: 3 x - y = 7 2 x + 2y = 10

[ Ans: x=2, y=1]

[ Ans: x= 3 y=2]

Substitution method This is another method you can try also. Consider this set of linear equations:

x + 2y = 7 2x + 3y =12

First look at the problem carefully.The frist equation appears simpler to handle. Let us write x in terms of Y : x = 7 - 2y Now, put this expression in equation 2 or substitute for x in this equation: 2x + 3 y = 12 2 ( 7 - 2y ) + 3y = 12 Expand this equation: 14 - 4 y + 3 y = 12 - y =12 - 14 = -2 or Y= 2 As a last step, use equation 1 to get x:

x + 2y = 7 x + 2 .2 = 7 X=3 Now check by putting x and y in both the equations.! 3 + 2.2 = 7 2.3 + 3.2 = 12 ------------------------------------------------------------------------------------Which method to use? --elimination method or substitution method...You can have your own preference..Try both and find which is comfortable for you.But if a method is specified in a test ,use that method only.

Graphical method The third method is to draw graphs--draw straight lines for the two equations using graph (or quadrille ) paper.Find the point where the two lines inersect.Its coordinates (x,y) is the solution. Why would you use the graph method? "A picture is worth thousand words" --people say.Sometimes you can visualize the problem and grasp better with a graph.Business people prefer graphs so they can watch the solution emerge. In general,they have an aversion or allergic to equations. A graph is great for presentations.The same problem as above, you can put in graph and show it to your class or friends. First convert the equation in the form of y = ax + b before you go to graph paper. A Business Problem Many business problems end up in decision-making.Here Algebra and set of linear equations are helpful for many problems.Let us try a simple one. John wants to start a moving business with either small trucks or big trucks. For each truck he has to pay hiring charges , a fixed amount per day and spend for gas depending on the miles or distance traveled. For large trucks, the hiring charge is $100 per day, and mileage is $0.10 per mile.If he travels x miles per day then the cost : Cost-big truck y = 100 + 0.10 x Note that this is a straight line equation. Let us write another equation for small truck: the hiring charge is more: $150 per day. The mileage cost is less: $0.08 per mile Cost -small truck z = 150 + 0.08 X Which is preferable or low cost solution? This will depend on the distance traveled x. Let us find the distance at which both the costs are equal.You can draw the the two lines and see the point of intersection. or solve by equations: y =z 100 + 0.10 x = 150 + 0.08 x Subtract 0.08x from both sides: 100 + 0.10 x - 0.08 x = 150 Now subtract 100 from both sides: 0.10 x - 0.08 x = 150 - 100 0.02 x = 50 x = 50/0.02 = 2500 miles Well , if the distance traveled exceeds 2500miles, small truck will be profitable; For shorter distances, bigger truck is less costlier. Such problems are regularly solved in almost all business. 'Do it yourself' problems: 1. Amy has to decide to buy or to rent a computer for a special project.The rental charge is $10 per day and you have to pay $24 as service charge.If she buys ,she has to pay the interest charge per day which works out to $ 4 per day and maintenance cost of $ 60, a one-time payment. The project may last for 5 to 15 days;Amy does not know the number of days at this point. Let x be the number of days computer required: For rental : rental cost = 24 + 10 X For buying: buy cost = 4 X + 50

Find X when both are equal: Solving:

24 + 10 x = 4 x + 60 10 x - 4x = 60 - 24 6x = 36 x=6 Well, if Amy's project is lasting more that 6 days, it is better to buy than rent. If the number of project days is 7, rental cost = 24 + 10x7 = 94 buy cost = 4x7 +60 = 88 You save $6 every day if you buy and use the computer.If you rent ,it is going to cost more.If your project ends in 6 days or less, you should rent only. Try this with graphical method too. 2. A classic problem: Jane is setting up a small bakery to make muffins and sell in the local restaurant.The set up cost for Jane [for ovens,dough mixer ,pot and pans) is $400. The cost of making each muffin is $ 0.5 0nly. She hopes to sell each muffin at a price of $1. How many muffins she should sell before she starts getting the profit from her small business.? { This is the standard 'break-even point' problem ] [Ans: 800 ] 3. Amanda can reach Denver from San Francisco by road or by air.The cost of air travel is given by the equation: Air tariff: A = 50 + 0.05 x where x is the distance. By car ,you have to spend for gas: C = 0.01 X Find the distance for which car is economical to use. [ ans x= 1250] Note: This tutorial and similar ones on Algebra are based on my experience in tutoring school children in Palo Alto,California.