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**Solving Systems of three variables**

Use the answers in the original equations to solve the remaining variables. Solve the new system of equations by elimination or substitution. Eliminate one of the variables using 2 different equations 2 times

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Even though there is not a set order or rules to these problems I will try an give you some guidelines. 1.) You want to get three equations with 3 variables down to 2 equations with 2 variables. a) on the easier problems you might be able to eliminate one variable from 2 of the 3 equations and the third equation might already be missing the same variable. b) The hardest of problems will require you to eliminate one variable from any 2 of the 3 equations then using elimination again eliminate that same variable using the equation you did not use in the first time with one of the equations you did use the first time. 2.) Now you have 2 equations with 2 variables so use elimination or substitution to solve for one variable 3.) Once you have solved for one variable substitute that value into one of the 2 equations with 2 variables and solve for the 2nd variable. 4.) Lastly use the 2 values you have found in one of the original 3 equations and solve for the last variable.

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x + 2y – 3z = 15 2x – 2z = 6 x + z = 3 Now use x = 3 to substitute in to solve for the z. x 2 2x – 2z = 6 2x + 2z = 6 3 + z = 3 (3,6,0) 4x = 12 z = 0 Now use x = 3 and z = 0 to solve for y. x = 3 3 + 2y -3(0)= 15 3 +2y + 0 = 15 2y = 12 y = 6

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**I know it can be long but its nothing to bang your head over.**

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**Use the elimination method**

EXAMPLE 1 Use the elimination method Solve the system. 4x + 2y + 3z = 1 Equation 1 2x – 3y + 5z = –14 Equation 2 6x – y + 4z = –1 Equation 3 SOLUTION STEP 1 Rewrite the system as a linear system in two variables. 4x + 2y + 3z = 1 12x – 2y + 8z = –2 Add 2 times Equation 3 to Equation 1. 16x z = –1 New Equation 1

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**Use the elimination method**

EXAMPLE 1 Use the elimination method 2x – 3y + 5z = –14 Add – 3 times Equation 3 to Equation 2. –18x + 3y –12z = 3 –16x – 7z = –11 New Equation 2 STEP 2 Solve the new linear system for both of its variables. 16x + 11z = –1 Add new Equation 1 and new Equation 2. –16x – 7z = –11 4z = –12 z = –3 Solve for z. x = 2 Substitute into new Equation 1 or 2 to find x.

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**Use the elimination method**

EXAMPLE 1 Use the elimination method STEP 3 Substitute x = 2 and z = – 3 into an original equation and solve for y. 6x – y + 4z = –1 Write original Equation 3. 6(2) – y + 4(–3) = –1 Substitute 2 for x and –3 for z. y = 1 Solve for y. Solution ( 2,-3,1)

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**3x + y +z = 14 -x + 2y – 3z = -9 5x - y + 5z = 30**

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#12 pg odd

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SOLVING SYSTEMS ALGEBRAICALLY SECTION 3-2. SOLVING BY SUBSTITUTION 1) 3x + 4y = 12 STEP 1 : SOLVE ONE EQUATION FOR ONE OF THE VARIABLES 2) 2x + y = 10.

SOLVING SYSTEMS ALGEBRAICALLY SECTION 3-2. SOLVING BY SUBSTITUTION 1) 3x + 4y = 12 STEP 1 : SOLVE ONE EQUATION FOR ONE OF THE VARIABLES 2) 2x + y = 10.

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