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Solving Linear Systems Using Substitution HW 3 Solve the linear systems using the substitution method. Make sure to write your final answer as (x,y).
1) π¦ = β2π₯ + 5 π₯ β 8π¦ = β23
2) 6π₯ + 4π¦ = 16
π¦ = 2π₯ β 3
3) π¦ = β6π₯ β 20 β3π₯ β 3π¦ = β18
4) 7π₯ + 5π¦ = β13
π₯ = 4π¦ β 16
5) 3π₯ + π¦ = 9 2π₯ + 3π¦ = β16
6) β5π₯ β 5π¦ = β20
β2π₯ + π¦ = 7 Solving Linear Systems Using Substitution HW 3 Solve the linear systems using the substitution method. Make sure to write your final answer as (x,y).
1) π¦ = β2π₯ + 5 π₯ β 8π¦ = β23
2) 6π₯ + 4π¦ = 16
π¦ = 2π₯ β 3
3) π¦ = β6π₯ β 20 β3π₯ β 3π¦ = β18
4) 7π₯ + 5π¦ = β13
π₯ = 4π¦ β 16
5) 3π₯ + π¦ = 9 2π₯ + 3π¦ = β16
6) β5π₯ β 5π¦ = β20
β2π₯ + π¦ = 7 Solving Linear Systems Using Substitution HW 3 Solve the linear systems using the substitution method. Make sure to write your final answer as (x,y).
1) π¦ = β2π₯ + 5 π₯ β 8π¦ = β23
2) 6π₯ + 4π¦ = 16
π¦ = 2π₯ β 3
3) π¦ = β6π₯ β 20 β3π₯ β 3π¦ = β18
4) 7π₯ + 5π¦ = β13
π₯ = 4π¦ β 16
5) 3π₯ + π¦ = 9 2π₯ + 3π¦ = β16
6) β5π₯ β 5π¦ = β20
β2π₯ + π¦ = 7
Solving Linear Systems Using Substitution HW 3 Solve the linear systems using the substitution method. Make sure to write your final answer as (x,y).
1) π¦ = β2π₯ + 5 π₯ β 8π¦ = β23
2) 6π₯ + 4π¦ = 16
π¦ = 2π₯ β 3
3) π¦ = β6π₯ β 20 β3π₯ β 3π¦ = β18
4) 7π₯ + 5π¦ = β13
π₯ = 4π¦ β 16
5) 3π₯ + π¦ = 9 2π₯ + 3π¦ = β16
6) β5π₯ β 5π¦ = β20
β2π₯ + π¦ = 7 Solving Linear Systems Using Substitution HW 3 Solve the linear systems using the substitution method. Make sure to write your final answer as (x,y).
1) π¦ = β2π₯ + 5 π₯ β 8π¦ = β23
2) 6π₯ + 4π¦ = 16
π¦ = 2π₯ β 3
3) π¦ = β6π₯ β 20 β3π₯ β 3π¦ = β18
4) 7π₯ + 5π¦ = β13
π₯ = 4π¦ β 16
5) 3π₯ + π¦ = 9 2π₯ + 3π¦ = β16
6) β5π₯ β 5π¦ = β20
β2π₯ + π¦ = 7 Solving Linear Systems Using Substitution HW 3 Solve the linear systems using the substitution method. Make sure to write your final answer as (x,y).
1) π¦ = β2π₯ + 5 π₯ β 8π¦ = β23
2) 6π₯ + 4π¦ = 16
π¦ = 2π₯ β 3
3) π¦ = β6π₯ β 20 β3π₯ β 3π¦ = β18
4) 7π₯ + 5π¦ = β13
π₯ = 4π¦ β 16
5) 3π₯ + π¦ = 9 2π₯ + 3π¦ = β16
6) β5π₯ β 5π¦ = β20
β2π₯ + π¦ = 7