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East Campus, CB 117 361-698-1579 Math Learning Center SOLVING SYSTEMS OF EQUATIONS West Campus, HS1 203 361-698-1860 This handout covers four methods to solve a system of equations: 1) Graphing 2) Substitution 3) Elimination or addition method 4) TI-83/TI-84 Matrix Solution For Example: 6π₯π₯ + 3π¦π¦ = 12 5π₯π₯ + π¦π¦ = 7 Method 1 β Graphing To solve by graphing, the two lines must be graphed on the same rectangular coordinate system. To graph each line, you need to find at least two points on each line. Set up and x & y chart for each. The point at which the two graphs intersect is the solution. (select a value for one variable and solve for the other, Remember: these are just my choices) For 6π₯π₯ + 3π¦π¦ = 12 x y 0 4 2 0 For x 0 1.4 5π₯π₯ + π¦π¦ = 7 y 7 0 Method 2 β Substitution To solve by substitution, select one equation (it does not matter which equation) and solve for one of the variables (it does not matter which variable). Then substitute that expression in to the other equation. For 6π₯π₯ + 3π¦π¦ = 12, Letβs solve for y. 6π₯π₯ + 3π¦π¦ = 12 3π¦π¦ = β6π₯π₯ + 12 β6π₯π₯ β 6π₯π₯ 3π¦π¦ 3 = β6π₯π₯+12 π¦π¦ = 3 β6π₯π₯ 3 + 12 3 π¦π¦ = β2π₯π₯ + 4 Since I solved for y, Iβm going to use this to substitute in to the y variable of the second equation. For 5π₯π₯ + π¦π¦ = 7 5π₯π₯ + (β2π₯π₯ + 4) = 7 5π₯π₯ + -2π₯π₯ + 4 = 7 Now solve for x. 3π₯π₯ + 4 = 7 3π₯π₯ = 3 β4 β 4 3π₯π₯ 3 3 =3 π₯π₯ = 1 Now that we know what x is, use this to solve for y by using one of the original equations (it does not matter which equation). Letβs use 6π₯π₯ + 3π¦π¦ = 12. 6(1) + 3π¦π¦ = 12 3π¦π¦ 6 6 + 3π¦π¦ = 12 3π¦π¦ = 6 =3 π¦π¦ = 2 3 β6 β6 Therefore, our answer is (1,2) which happens to be the point of intersection. East Campus, CB 117 361-698-1579 Math Learning Center West Campus, HS1 203 361-698-1860 Method 3: Elimination or Addition Method The object of the elimination method is to add the equations such that like terms add up to zero. If no like terms add up to zero, then multiplication is used to alter the system of equations such that like terms do add up to zero. If the equations 6π₯π₯ + 3π¦π¦ = 12 were added now, no like terms would add up to zero. + 5π₯π₯ + π¦π¦ = 7 11π₯π₯ + 4π¦π¦ = 19 Therefore, multiplication is needed to alter the equations. To do this legally, remember that whatever you do to one side of an equation you do the same thing on the other side of the equation. To eliminate a variable do the following: 1) Decide which variable you want to eliminate (it does not matter which variable) 6π₯π₯ + 3π¦π¦ = 12 5π₯π₯ + π¦π¦ = 7 Letβs pick the x terms. 2) Looking at the variable from each equation, find the Least Common Multiple of the coefficients πππ₯π₯ + 3π¦π¦ = 12 πππ₯π₯ + π¦π¦ = 7 The Least Common Multiple of 6 and 5 is 30. 3) Now multiply each equation by the number which will make the coefficients on the x variables 30 ππ(πππ₯π₯ + 3π¦π¦ = 12) 30π₯π₯ + 15π¦π¦ = 60 βππ(πππ₯π₯ + π¦π¦ = 7) + β 30π₯π₯ β 6π¦π¦ = β42 Notice that if these equations were added, the x terms would still not add to zero, therefore one of the factors, 5 or 6, which was used to multiply with the equations must be negative. Letβs choose the 6 to be negative. ππ(πππ₯π₯ + 3π¦π¦ = 12) βππ(πππ₯π₯ + π¦π¦ = 7) 30π₯π₯ + 15π¦π¦ = 60 + β 30π₯π₯ β 6π¦π¦ = β42 9π¦π¦ = 18 9π¦π¦ 9 = 18 9 π¦π¦ = 2 4) Now substitute this value into one of the original equations (it does not matter which one) to solve for x. Letβs use 6π₯π₯ + 3π¦π¦ = 12. 6π₯π₯ 6 6π₯π₯ + 3(2) = 12 6π₯π₯ + 6 = 12 6π₯π₯ = 6 =6 π₯π₯ = 1 6 β6 β 6 Therefore, our answer is (1,2) which happens to be the point of intersection. East Campus, CB 117 361-698-1579 Math Learning Center West Campus, HS1 203 361-698-1860 Method 4: TI-83 / TI- 84 Matrix Solution GRAPHING CALCULATOR GUASS-JORDAN METHOD (reduced-row echelon form) 1. Write the augmented matrix 2. 2nd x-1 for MATRIX* > > EDIT, 1: [A] 3. Input the number of rows and then the number of columns, input the values from the augmented matrix 4. 2nd MODE to QUIT 5. MATRIX > MATH, Choose: B: rref( 6. MATRIX, NAMES, Choose: 1: [A] Example: Solve the system. x + y β 4z = 10 2x β 3y + z = 7 3x β y β z = 12 The augmented matrix is: 1 1 β4 10 οΏ½2 β3 1 7οΏ½ 3 β1 β1 12 Define the size of the matrix. Rows by columns. *Press 2nd x-1 for MATRIX and scroll over to EDIT *Hit Enter Now the screen looks like this. We still need to input the name of the matrix. Go to MATRIX and scroll over to MATH To leave this screen press 2nd MODE which is QUIT Go to MATRIX and select 1: [A] Scroll down to option B: rref( Now hit ENTER This last matrix corresponds to the system: x=3 y = -1 z = -2 TADA! * On the TI-83 the MATRX button is used rather than 2nd x-1 for MATRIX Answer: {(3, -1, -2)} Turn decimals into fractions by going to MATH, then select option 1: > Frac, and then ENTER.