Solving linear equation
- 2. FABULOUS FIVE 0017 Rija Arshad 0030 Sania Ijaz 0031 Anam Zahid 0048 Alisha Asghar 0141 Sidra Basharat
- 3. LINEAR EQUATIONS AND FUNCTIONS Solution of Linear Equation & Inequalities in one variable. Functions , notation operation with function. Linear function graph, slopes, equations. Solution of system of linear equations in three variables. Applications of functions in business and economics.
- 4. SOLUTION OF LINEAR EQUATIONS & INEQUALITIES IN ONE VARIABLE INTRODUCTION Linear equations were invented in 1843 by Irish mathematician Sir William Rowan Hamilton. He was born in 1805 and died in 1865. Sir Hamilton made important contributions to mathematics.
- 5. DIFFERENTIATE BETWEEN EQUALITY & IN-EQUALITY EQUATION:- An equation is a mathematical statement wherein two expressions are set equal to each other. FOR EXAMPLE:- 𝟐 𝟑 𝒙 − 𝟏 𝟐 𝒙 = 𝒙 + 𝟏 𝟔 IN-EQUALITY:- A Formal statement of inequality between two quantities usually separated by a sign of inequality (as < , > or ≠ OR signifying respectively is less than, is greater than, or is not equal to). FOR EXAMPLE:- 𝟑𝒙 + 𝟏 < 𝟓𝒙 − 𝟒 𝟗 − 𝟕𝒙 > 𝟏𝟗 − 𝟐𝒙
- 6. PROPERTIES OF EQUALITY ADDITION PROPERTY The equation formed by adding the same quantity to both side of an equation is equivalent to the original equation. Example:- 𝒙 − 𝟒 = 𝟔 is equivalent to 𝒙 = 𝟏𝟎 SUBSTITUTION PROPERTY The equation formed by substituting one expression for an equal expression is equivalent to original equation. Example:- 𝟑 𝒙 − 𝟑 − 𝟏 𝟐 𝟒𝒙 − 𝟏𝟖 = 𝟒 Is equivalent to 𝟑𝒙 − 𝟗 − 𝟐𝒙 + 𝟗 = 𝟒 & 𝒕𝒐 𝒙 = 𝟒 The solution set is {4} MULTIPLICATION PROPERTY The equation forms by multiplying both side of an equation by the same non zero quantity is equivalent to the original equation. Example:- 𝟏 𝟑 𝒙 = 𝟔 𝑖𝑠 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑡𝑜 𝟑 𝟏 𝟑 𝒙 = 𝟑 𝟔 𝒐𝒓 𝒕𝒐 𝒙 = 𝟏𝟖
- 7. TO SOLVE LINEAR EQUATION IN ONE VARIABLE 1. SIMPLIFY BOTH SIDE OF EQUATION. 2. USE THE ADDITION AND SUBSTRACTION PROPERTY. 3. SIMPLIFY BOTH SIDE OF THE EQUATIONS. 4. DIVIDE BOTH SIDE OF THE EQUATION BY THE COEFFICIENT OF THE VARIABLE.
- 8. EXAMPLE:- SOLVE:- 𝑥 + 1 = 3 𝑥 − 5 𝑥 + 1 = 3 𝑥 − 5 (Original equation) 𝑥 + 1 = 3𝑥 − 15 (simplify right hand side) 𝑥 = 3𝑥 − 15 − 1 𝑥 = 3𝑥 − 16 (by subtracting 1 from 16) −2𝑥 = −16 (by subtracting 1 from 3x) 𝑥 = − 16 −2 𝒙 = 𝟏𝟖 (by dividing -2) The solution is 8 Check :- 8+1=3(8-5) 9=3(3) 9=9
- 9. SOLUTION OF LINEAR EQUATION FUTURE VALUE OF AN INVESTMENT :- The future value of a simple interest investment is given by S= p+ prt ,where p is the principal invested, r is the annual interest rate (as a decimal), and t is the time in years, at what simple interest r must p=1500 dollars be invested so that the future value is $2940 after 8 year. Solution:- Entering the values S=2904, P=1500, and t into S=P +prt gives 2940 = 1500 + 1500(r)(8) 2940=1500+12,000r 2940-1500=12,000r 1440=12,000r 1440/12,000 = r 0.12 = r So, the interest rate is 0.12 or 12 % .
- 10. SOLUTION OF LINEAR EQUATION :- VOTING Example:- Using data from 1952-2004, the percent p of the eligible U.S. population voting is presidential selection has been estimated to be p=63.20 – 0.26x Where x is the number of years past 1950. according to this model/ in what election year is the % voting equal to 55.4% ? Solution:- 55.4=63.20 – 0.26x -7.8= - 0.26x 30=x
- 11. SOLVING OF LINEAR EQUATION PROFIT :- SUPPOSE THAT THE RELATIONSHIP BETWEEN A FIRM’S PROFIT P AND THE NUMBER X OF ITEM SOLD CAN BE DESCRIBED BY THE EQUATION. 5x – 4p = 1200 Find the profit when 240 units are sold. when, p = 𝟓 𝟒 𝒙 − 𝟑𝟎𝟎 = 𝟓 𝟒 𝟐𝟒𝟎 − 𝟑𝟎𝟎 p = 0 It means profit is zero when the firm produced 240 units.
- 12. SOLVING OF LINEAR IN-EQUALIIES IN ONE VARIABLE :- DEFINITION :- A linear inequality in one variable is a sentence of the form ax + b < 0 , a 6= 0. EXAMPLE :- Solve x + 2 < 4 x + 2 < 4 x < 4 – 2 x < 2 The graph of this solution is as follow:- -1 0 1 2 3 5
- 13. GRAPH OF LINEAR EQUATION AND IN-EQUALITIES :-
- 14. FUNCTION It is a relationship between a set of inputs and a set of outputs with the property that each input related to exactly that output Example:- A depend on “X” where A is the area and formula is A =𝑥2 , here A is a function of x If Y depends on x then Y is a function of x y= f(x)
- 15. TYPES OF FUNCTIONS 1- Algebraic function polynomial function linear function quadratic function identity function constant function rational function 2- Trigonometric function 3- Inverse trigonometric function 4- Exponential function 5- Logarithm function
- 16. LINEAR FUNCTION
- 17. EXAMPLE:- The total cost of producing a product is given by C(x)=300x+0.1𝑥2+1200 Where x represents the number of unit produced. Find the total cost of producing 10 units:- x=10 C(x)=300x+0.1𝑥2 +1200 C(10)=300(10)+0.1(10)2+1200 C(10)=3000+10+1200 C(10)=4210 when we’ll produce 10 units the total cost will be 4210
- 18. APPLICATION OF FUNCTION IN DAILY LIFE MONEY AS A FUNCTION OF TIME. YOU NEVER HAVE MORE THEN ONE AMOUNT OF MONEY AT ANY TIME BECAUSE YOU CAN ALWAYS ADD EVERYTHING TO GIVE ONE TOTAL AMOUNT BY UNDERSTANDING HOW YOUR MONEY CHANGES OVER TIME, YOU CAN PLAN TO SPEND YOUR MONEY SENSIBLY.
- 19. LINEAR FUNCTION A linear function involves a record variable like y , and a variable like x whose highest power is 1. EXAMPLE:- Y=2X+4 Y=5X+25 Y=3X+12
- 20. LINEAR FUNCTION DOMAIN All the x-coordinates in the function’s ordered pairs Example:- {3.2.5} RANGE All the y-coordinates in the function’s ordered pair Example:- {6,8,3}
- 21. GRAPH OF LINEAR FUNCTION
- 22. EXAMPLE:- Depreciation:- A business property is purchased for $ 122,880 and depreciated over a its value y is related to the number of months of service x by the equation 𝟒𝟎𝟗𝟔𝒙 + 𝟒𝒚 = 𝟒𝟗𝟏520 Find the x-intercept and the y-intercept and use them to sketch the graph Solution:- for x-intercept , y=0 gives 4096𝑥 = 491,520 𝐱 = 𝟏𝟐𝟎 Thus 120 is the x-intercept for y-intercept , x=0 gives 4𝑦 = 491,520 𝒚 = 𝟏𝟐𝟐, 𝟖𝟖𝟎 Thus 122,880 is the y-intercept
- 23. LINEAR FUNCTION EXAMPLE IN REAL LIFE
- 24. ‘’APPLICATION OF FUNCTIONS IN BUSINESS AND ECONOMICS’’ Here are some applications of Functions: Profit & Break-Even Point Supply, Demand & Market Equilibrium 1. Profit Function: The profit is the net proceeds, or what remains the revenue when costs are subtracted. Profit= revenue-cost
- 25. EXAMPLE OF PROFIT FUNCTION: Suppose that profit function for a product is linear and marginal profit is $5. if the profit is $200 when 125 units are sold, write the equation of the profit function. Solution: The marginal profit gives us the slope of the line representing the profit function. Using this slope(m=5) and the point(125,000) in the point-slope formula P-P1=m(x-x1) gives P-200=5(x-125) or P=5x-425
- 26. 2. BREAK-EVEN POINT: In break-even point is the number of item x at which break-even occurs. In break-even point revenue is equal to cost. Formula: Revenue=Cost In Break-Even point PROFIT = ZERO LOSS = ZERO EXAMPLE:- 4P=81x-29970 4(0)=81x-29970 29970=81x 29970/81=x x=370.
- 27. 3. SUPPLY & DEMAND: Supply: The law of Supple states that the quantity supplied for sale will increase as the prices of the product increase. Demand: The law of Demand states that the quantity demanded increases as the prices decreases and vise versa.
- 28. EXAMPLE OF DEMAND AND SUPPLY IN DAILY LIFE
- 29. 4. MARKET EQUILIBRIUM: IN market-equilibrium supply is equal to demand. Market equilibrium occurs when the quantity of a commodity demanded is equal to the quantity supplied. Example: Find the equilibrium point for the following supply and demand function. Demand: p= -3q+36 Supply: p=4q+1 At market equilibrium, the demand price equals the supply price. Thus, demand=supply -3q+36=4q+1 35=7q 35/7=q q=5 Putting the value of q in equation 2 , you’ll find the value of P=21 q=5 p=21 So the market-equilibrium point is (5,21)
- 30. ‘’SOLUTION OF SYSTEM OF LINEAR EQUATION’’ It is a collection of 2 or more linear equation involving same set of variables that you deal all together at once. For Example: x+2y=4 3x+5y=7
- 31. METHODS OF SOLUTION OF SYSTEM OF LINEAR EQUATION There are 2 methods of solving of system of linear equation: Elimination Method. Substitution Method.
- 32. SUBSTITUTION METHOD NO SOLUTION -4x+8y=9 x-2y=3 By multiplying equation 2 with 4 4(x-2y)=4(3) 4x-8y=12 -4x+8y=9 4x-8y=12 0x+0y=21 0=21 NO SOLUTION SOLUTION 5x+4y=1 3x-6y=2 By multiplying equation 1 with 3 By multiplying equation 2 with 2 3(5x+4y)=3(1), 2(3x-6y)=2(2) 15x+12y=3 (eq 3) , 6x-12y=4 (eq 4) By adding equation 3 and 4 15x+12y=3 6x-12y=7 21x =7 x=7/21 x=1/3 By putting the value in equation 2 3(1/3)-6y=2 1-6y=2 Y= -1/6
- 33. Elimination Method Example: x + y=335 10x+7y=2741 Solution: multiplying equation 1 with -10. -10(x + y)=335(-10) -10x-10y=-3350 -10x-10y=-3350 10x+7y=2741 -3y=-609 y=-609/3 y=203 Applying the value of y in equation 1. X+203=335 x=335-203 x=132
- 34. EXAMPLE OF SYSTEM OF LINEAR EQUATION IN REAL LIFE
- 35. EXAMPLE OF SYSTEM OF LINEAR EQUATION:-