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Equations w/ Substitution Date________________ Period____  \nSolve each system by substitution.  \n1) −x − y − 3z = −9 2) x = −4z − 19 z = −3x − 1 y = 5x + z − 4 x = 5 y − z + 23 −5 y − z = 25  \n3) y = x + z + 5 4) −2 y + 5z = −3 z = −3 y − 3 y = −5x − 4z − 5  \n2x − y = −4 x = 4z + 4  \n5) y = x + 4z − 5 6) x = 3 y − 3z + 8  \n4x + 3 y − 2z = 5 z = 4x + 5 y − 14 z = −2x + 2 3 y + 2z = 14  \n7) −5x − 3 y + z = −4 8) −4x + 2z = 14  \n−2x − 2 y + 2z = 4 y = x + z + 12 z = x + 5 −2x − 4z = 22  \n9) 3x − 3 y = −6 10) x = −5 y + 4z + 1 z = −3x − 3 y + 9 x − 2 y + 3z = 1  \n−4x + 5 y + z = 8 2x + 3 y − z = 2  \n11) a − 2b + c = −6 12) −2x + 3 y + 5z = −21 a + 5c = −12 −4z = 20  \n−a + 6b + 4c = 3 6x − 3 y = 0  \n13) 2x − 4z = 20 14) x + 3 y = −17  \n−3x + y − 4z = 20 3x = −6  \n−4x + 2 y + 3z = −15 4x − 3 y + 6z = 25  \n15) 5r + 4s − 6t = −24 16) −5r + 5s + 3t = −23  \n−2s + 2t = 0 −5r + 3s − 3t = −11 s − t = 2 −6r + 6t = −12  \nKuta Software-Infinite Algebra 2 Name   \nSolving Systems of Three Equations w/ Substitution Date  Period   \nSolve each system by substitution.  \n1) −x − y − 3z = −9 z = −3x − 1 x = 5 y − z + 23  \n(−2, −4, 5)  \n3) y = x + z + 5 z = −3 y − 3  \n2x − y = −4  \n(−2, 0, −3)  \n5) y = x + 4z − 5  \n4x + 3 y − 2z = 5 z = −2x + 2  \n(0, 3, 2)  \n7) −5x − 3 y + z = −4  \n−2x − 2 y + 2z = 4 z = x + 5  \n(0, 3, 5)  \n2) x = −4z − 19 y = 5x + z − 4  \n−5 y − z = 25  \n( 1, −4, −5)  \n4) −2 y + 5z = −3 y = −5x − 4z − 5 x = 4z + 4  \n(0, −1, −1)  \n6) x = 3 y − 3z + 8 z = 4x + 5 y − 14  \n3 y + 2z = 14  \n(2, 2, 4)  \n8) −4x + 2z = 14 y = x + z + 12  \n−2x − 4z = 22  \n(−5, 4, −3)  \n9) 3x − 3 y = −6 z = −3x − 3 y + 9  \n−4x + 5 y + z = 8  \n( 1, 3, −3)  \n11) a − 2b + c = −6 a + 5c = −12  \n−a + 6b + 4c = 3  \n(3, 3, −3)  \n13) 2x − 4z = 20  \n−3x + y − 4z = 20  \n−4x + 2 y + 3z = −15  \n(0, 0, −5)  \n15) 5r + 4s − 6t = −24  \n−2s + 2t = 0 s − t = 2  \nNo unique solution  \n10) x = −5 y + 4z + 1 x − 2 y + 3z = 1  \n2x + 3 y − z = 2  \nNo unique solution  \n12) −2x + 3 y + 5z = −21  \n−4z = 20  \n6x − 3 y = 0  \n( 1, 2, −5)  \n14) x + 3 y = −17  \n3x = −6  \n4x − 3 y + 6z = 25  \n(−2, −5, 3)  \n16) −5r + 5s + 3t = −23  \n−5r + 3s − 3t = −11  \n−6r + 6t = −12  \n( 1, −3, −1)  \nCreate your own worksheets like this one with Infinite Algebra 2. Free trial available [at KutaSoftware.com](at KutaSoftware.com)","cbCaisQaHBzgLB0I","https://ap.wps.com/l/cbCaisQaHBzgLB0I","pdf",32885,"English","# Solving Systems of Three Equations w/ Substitution\n## Substitution problems by number\n## Solutions in coordinate form and “No unique solution” cases","[{\"question\":\"What method is used to solve each system on the worksheet?\",\"answer\":\"Each system is solved using the substitution method, as instructed on the worksheet.\"},{\"question\":\"What do the worksheet’s coordinate triples represent?\",\"answer\":\"They represent the solution to the system as ordered values for the three variables (such as x, y, z or r, s, t).\"},{\"question\":\"What does “No unique solution” mean in these problems?\",\"answer\":\"It indicates the system does not have exactly one solution, typically due to equations being dependent and producing either no solution or infinitely many solutions.\"}]","Solving Systems of Three Equations w/ Substitution - Solve each system by substitution | PDF",1791426163]