Bellringer 1. Solve by Elimination 2. Solve by Substitution

3-6 Systems with Three Variables Objective: To solve systems in three variables by elimination and substitution

Objectives Solving Three Variable Systems by Elimination Solving Three Variable Systems by Substitution

Solutions No Solution Infinite SolutionsOne Solution

Elimination Steps Step 1: Choose one of the equations to pair with each of the other two equations to eliminate one of the variables. Step 2: From Step 1 you should have two new equations with only two variables. Pair the new equations together and eliminate one more variable. You should have solved for your first variable. Step 3: Substitute your solved variables backwards to find the values of the two remaining variables.

Solve the system by elimination. Step 2: Write the two new equations as a system. Solve for x and y. –5x + 3y + 2z = 11 8x – 5y + 2z= –55 4x – 7y – 2z= – Step 1:Pair the equations to eliminate z, because the terms are already additive inverses. 8x – 5y + 2z = –55–5x + 3y + 2z= 11 4x – 7y – 2z = –29Add.4x – 7y – 2z= –29 12x – 12y= –84 –x – 4y= – x – 12y= –84 –12x – 48y= –216 – 60y= –300 y= Multiply equation by 12 to make it an additive inverse. Add. 5 Solving by Elimination

(continued) The solution of the system is (–2, 5, –7). Step 3: Substitute the values for x and y into one of the original equations (,, or ) and solve for z. 123 –5x + 3y + 2z= 11 –5(–2) + 3(5) + 2z = z= 11 2z= –14 z= –7 1 12x – 12y= –84 12x – 12(5) = –84 Substitute the value of y. 12x= –24 x= –2 4 Continued

(continued) Check: Show that (–2, 5, –7) makes each equation true. –5x + 3y + 2z= 118x – 5y + 2z= –55 –5(–2) + 3(5) + 2(–7) 118(–2) – 5(5) + 2(–7) – – 1411–16 – 25 – 14–55 11= 11 –55= –55 4x – 7y – 2z= –29 4(–2) – 7(5) – 2(–7) –29 –8 – –29 –29= –29 Continued

Step 1: Pair the equations to eliminate x. Solve the system by elimination. x + 2y + 5z= 1 –3x + 3y + 7z= 4 –8x + 5y + 12z= x + 2y + 5z= 1 3x + 6y + 15z = 3Multiply by 3. –3x + 3y + 7z= 4–3x + 3y + 7z = 4 9y + 22z = 7 x + 2y + 5z= 1 8x + 16y + 40z = 8Multiply by 8. –8x + 5y + 12z= 11–8x + 5y + 12z = 11 21y + 52z = Step 2: Write the two new equations as a system. Solve for y and z. 9y + 22z= 763y + 154z = 49Multiply by 7. 21y + 52z= 19–63y – 156z = –57Multiply by –3. –2z = –8 z = Solving Equivalent Systems

(continued) 9y + 22z= 7 9y + 22(4) = 7Substitute the value of z. y= –9 The solution of the system is (–1, –9, 4). Step 3: Substitute the values for y and z into one of the original equations (,, or ) and solve for x. 123 x + 2y + 5z= 1 x + 2(–9) + 5(4) = 1 x= –1 1 1 Continued

Solve the system by substitution. 12x + 7y + 5z= 16 –2x + y – 14z= –9 –3x – 2y + 9z= – Step 1: Choose one equation to solve for one of its variables. Step 2: Substitute the expression for x into each of the other two equations. –3x – 2y + 9z = –12Solve the third equation for x. –3x – 2y = –9z – 12 –3x = 2y – 9z – 12 x = – y + 3z ( ) 12x + 7y + 5z= – y + 3z y + 5z= 16 –8y + 36z y + 5z= 16 –y + 41z + 48= 16 –y + 41z= – –2x + y – 14z= – 9 –2 – y + 3z y – 14z = – 9 y – 6z – 8 + y – 14z= – 9 y – 20z – 8= – 9 y – 20z= – 1 ( ) Solving by Substitution

(continued) Step 3: Write the two new equations as a system. Solve for y and z. –y + 41z= –32 –y + 41(–1) = –32Substitute the value of z. –y – 41= –32 –y= 9 y= –9 4 4 –y + 41z= –32– y + z= –Multiply by. y – 20z= –1 y – 20z= –1 z=– z = – Continued

(continued) –2x + y – 14z= –9 –2x + (–9) – 14(–1) = –9 –2x – = –9 –2x + 5= –9 –2x= –14 x= 7 2 Step 4: Substitute the values for y and z into one of the original equations (,, or ) and solve for x. 123 The solution of the system is (7, –9, –1) Continued

You have $10,000 in a savings account. You want to take most of the money out and invest it in stocks and bonds. You decide to invest nine times as much as you leave in the account. You also decide to invest five times as much in stocks as in bonds. How much will you invest in stocks, how much in bonds, and how much will you leave in savings? Relate: money in stocks + money in bonds + money in savings = $10,000 money in stocks + money in bonds = 9 money in savings money in stocks = 5 money in bonds Define:Let k = amount invested in stocks. Let b = amount invested in bonds. Let s = amount left in savings. Real World Example

(continued) Step 1: Substitute 5b for k in equations and. Simplify. k + b + s = k + b = 9s 5b + b + s = b + b = 9s 6b + s = b = 9s Step 2: Write the two new equations as a system. Solve for b and s. 6b + s= b + s= b – 9s= 0–6b + 9s= 0Multiply by –1. 10s= s= Write: k + b + s = k + b = 9 s k = 5 b Continued

(continued) You should invest $7,500 in stocks, $1,500 in bonds, and leave $1,000 in savings. 6b + s = 1,0000 6b = 1,0000 Substitute s. 6b = 9,000 b = 1,500 4 k = 5b k = 5(1,500) k = 7,500 3 Step 3: Substitute the value of b into equation and solve for k. 3 Continued

Homework Pg 157 # 1,2,10,11