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### abused animals

1. 1. SOLVE SYSTEMS OF EQUATIONSA system of equations refers to "n" unknowns for "n" equations.Example: 3x - y = 8 x+y=8Solutions to a system of equations is an ordered pair (x, y) in which when you substitute thevalues for x and y into both equations it yields a true statement for both equations.Example: Is (1, 2) a solution of the system, y - x =1 and 2x + y = 4?substitute (1, 2) into y - x = 1: 2-1=1 1=1 true statementsubstitute (1, 2) into 2x + y = 4: 2(1) + 2 = 4 4=4 true statementTherefore, (1,2) is a solution to the system.There are many different methods to solve systems of equations. Three methods will bedemonstrated.SOLVE SYSTEMS OF EQUATIONS BY GRAPHING METHOD.Solve x + y = 8 x-y=2Both equations are linear. Therefore, both their graphs will be lines. That is why a graph can beused to find the solution.STEPS:1) Graph both linear equations on the same coordinate plane.2) The point, (x, y), at which the two lines intersect is the solution to the system of equation.NOTE: There are three possible types of solutions to a system.1) There is only one point of intersection. This occurs when the two lines intersect at one point.2) There is no solution. This occurs when the two lines are parallel and dont intersect at anypoint.3) There is an infinite number of solutions. This occurs when the two lines are graphed on topof each other and so they intersect at every point on the line.
2. 2. SOLVE SYSTEMS OF EQUATIONS BY SUBSTITUTION METHODSTEPS:1) Choose one equation to solve for one variable.2) Substitute the chosen equation into the unused equation and solve.3) Substitute the value found in step 2 into any one of the original equations and solve for theremaining variable.4) The solution can be written as an ordered pair, (x, y).NOTE: There are three possible types of solutions to a system.1) There is a unique solution. This occurs when a value for both variables are found.2) There is no solution. This occurs when both variables cancel out and a false statementremains.3) There is an infinite number of solutions. This occurs when both variables cancel out and atrue statement remains.EXAMPLES:1) Solve x = 2y 3x - y = -5step 1: x = 2y is the chosen equation and is already solved for "x"step 2: substitute into unused: 3(2y) - y = -5 solve: 6y - y = -5 5y = -5 y = -1step 3: substitute y = -1 into x = 2y then x = 2(-1) x = -2The solution is (-2, -1); a unique solution.2) Solve x - y = -4 7x + 5y = -28step 1: choose x - y = -4 and solve for x; x = y - 4.step 2: substitute into unused: 7(y - 4) + 5y = -28 solve: 7y - 28 + 5y = -28 12y - 28 = -28 12y = 0 y=0step 3: substitute y = 0 into x = y - 4 then x = 0 - 4 x = -4The solution is (-4, 0); a unique solution.
3. 3. 1 1 33) Solve x− y = 3 4 12 2x + y = 9The top equation will be multiplied by the LCD of 12 to yield 4x - 3y = 3.step 1: 2x + y = 9 is the chosen equation and solve for y, then y = 9 - 2x.step 2: substitute into unused: 4x - 3(9 - 2x) = 3 solve: 4x - 27 + 6x = 3 10x - 27 = 3 10x = 30 x=3step 3: substitute x = 3 into y = 9 - 2x then y = 9 - 2(3) y= 9-6 y=3The solution is (3, 3); a unique solution.4) Solve 3x - 15y = 15 x = 5y + 5step 1: x = 5y + 5 is the chosen equation and already solved for x.step 2: substitute into unused: 3(5y + 5) - 15y = 15 solve: 15y + 15 -15y = 15 15 = 15The variable canceled out and 15 = 15 is a true statement. Therefore, the solution is infinitenumber of solutions.
4. 4. SOLVE SYSTEMS OF EQUATIONS BY ELIMINATION METHODSTEPS:1) Choose one variable to eliminate when like terms are collected.2) Multiply one or both equations by values so that the coefficients of the chosen variable areadditive inverses.3) Collect like terms and solve.4) Substitute the value found in step 4 into any one of the original equations and solve for theother variable.5) The solution can be written as an ordered pair, (x, y).NOTE: There are three possible types of solutions to a system.1) There is a unique solution. This occurs when a value for both variables are found.2) There is no solution. This occurs when both variables cancel out and a false statementremains.3) There is an infinite number of solutions. This occurs when both variables cancel out and atrue statement remains.EXAMPLES:1) Solve 3x - y = 4 2x + y = 1step 1: The variable y has been chosen to eliminate.step 2: There is no need to multiply since -y and +y are additive inverses already.step 3: collect like terms: 5x = 5 solve: x=1step 4: substitute x =1 into 2x + y = 1, then 2(1) + y = 1 2+y=1 y = -1The solution is (1, -1); a unique solution.2) Solve 6x - 10y = 5 2x - 3y = 1step 1: The variable x is chosen to eliminate.step 2: Multiply the bottom equation by -3 so the additive inverse is 6x and -6x. 6x - 10y = 5 6x - 10y = 5 -3(2x - 3y = 1) -6x + 9y = -3step 3: collect like terms: -y = 2 solve: y = -2step 4: substitute y = -2 into 2x - 3y = 1, then 2x - 3(-2) = 1 2x + 6 = 1 2x = 7 x = 7/2The solution is (7/2, -2); a unique solution.
5. 5. 3) Solve 5x - 4y = 1 2x + 3y = 5step 1: The variable y is chosen to eliminate.step 2: Multiply the top equation by 3 and the bottom equation by 4 so the additive inverses are-12y and 12y. 3(5x - 4y = 1) 15x - 12y = 3 4(2x + 3y = 5) 8x + 12y = 20step 3: collect like terms: 23x = 23 solve: x=1step 4: substitute x =1 into 2x + 3y = 5, then 2(1) + 3y = 5 2 + 3y = 5 3y = 3 y=1The solution is (1, 1); a unique solution.4) Solve 2x - 8y = 9 x - 4y = -6step 1: The variable x is chosen to eliminate.step 2: Multiply the bottom equation by -2 so the additive inverses are 2x and -2x. 2x - 8y = 9 2x - 8y = 9 -2(x - 4y = -6) -2x + 8y = 12step 3: collect like terms: 0 = 21Both variables cancelled out and 0 = 21 is a false statement. Therefore, the solution is nosolution.
6. 6. Systems of equations prove helpful when solving word problems.WORD PROBLEM EXAMPLES:1) Kerry asked a bank teller to cash a \$390 check using \$20 bills and \$50 bills. If the teller gaveher a total of 15 bills, how many of each type of bill did she receive?Let x = amount of \$20 bills y = amount of \$50 billsSet up a system of two equations. total number of bills is 15 then x + y = 15 total amount of money is \$390 then 20x + 50y = 390solve using elimination method: -20(x + y) = 15 -20x - 20y = -300 20x + 50y = 390 20x + 50y = 390 30y = 90 y=3substitute y = 3 into x + y = 15, then x + 3 = 15, then x = 12The teller gave Kerry 12-\$20 bills and 3-\$50 bills.2) A financial planner wants to invest \$8000, some in stocks earning 15% annually and the restin bonds earning 6% annually. How much should be invested at each rate to get a return of \$930annually from the two investments?Let x = amount invested at 15% y = amount invested at 6%Set up a system of equation.total amount to be invested is \$8000 x + y = 8000total amount of interest is \$930 0.15x + 0.06y = 930solve using substitution method:x = 8000 - y then 0.15(8000-y) + 0.06y = 930 1200 - 0.15y + 0.06y = 930 -0.09y = -270 y = 3000substitute y = 3000 into x = 8000 - y, then x = 5000.\$5000 should be invested at the 15% rate and \$3000 should be invested at the 6% rate.
7. 7. 3) Matthew flew his ultra light plane to a nearby town against a head wind of15 km/h in 2 hours 20 minutes. The return trip with a tail wind of 15 km/h took1 hour 24 minutes. Find the planes air speed and the distance to the nearby town.Definitions:The air speed refers to the speed of an aircraft in still air (the speedometer reading).The wind speed refers to the speed of the wind.The head wind refers to the wind blowing against the aircraft.The tail wind refers to the wind blowing with the aircraft.Recall the formula: Distance = Rate x TimeNOTE: The rate of a plane: the air speed plus tail wind the air speed minus the head windLet d = the distance to the town and return trip. a = the air speedset up the equation:to the town: d = 2 1/3 (a - 15)return trip: d = 1 2/5 (a + 15)solve by substitution: 7/3(a - 15) = 7/5(a + 15) 7/3a - 35 = 7/5a + 21 14/15a = 56 a = 60The air speed is 60 km/h.To find the distance: d = 7/3(60-15) d = 105The distance is 105 km.