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• Create an equation: x*0.22 + y*0.02 = 20*0.04 x = gallons of cream y = gallons of milk Now, since x + y = 20, we can replace y with 20 - x: x*0.22 + (20 - x)*0.02 = 20*0.04 Now simplify and solve it: 11x + 20 - x = 40 10x = 20 x = 20/10…

  in algebra mixture problem? Comment by computerguy103 June 2021
• Create an equation: x*0.22 + y*0.02 = 20*0.04 x = gallons of cream y = gallons of milk Now, since x + y = 20, we can replace y with 20 - x: x*0.22 + (20 - x)*0.02 = 20*0.04 Now simplify and solve it: 11x + 20 - x = 40 10x = 20 x = 20/10…

  in algebra mixture problem? Comment by computerguy103 May 2021
• Create an equation: x*0.22 + y*0.02 = 20*0.04 x = gallons of cream y = gallons of milk Now, since x + y = 20, we can replace y with 20 - x: x*0.22 + (20 - x)*0.02 = 20*0.04 Now simplify and solve it: 11x + 20 - x = 40 10x = 20 x = 20/10…

  in algebra mixture problem? Comment by computerguy103 May 2021
• Create an equation: x*0.22 + y*0.02 = 20*0.04 x = gallons of cream y = gallons of milk Now, since x + y = 20, we can replace y with 20 - x: x*0.22 + (20 - x)*0.02 = 20*0.04 Now simplify and solve it: 11x + 20 - x = 40 10x = 20 x = 20/10…

  in algebra mixture problem? Comment by computerguy103 May 2021
• Create an equation: x*0.22 + y*0.02 = 20*0.04 x = gallons of cream y = gallons of milk Now, since x + y = 20, we can replace y with 20 - x: x*0.22 + (20 - x)*0.02 = 20*0.04 Now simplify and solve it: 11x + 20 - x = 40 10x = 20 x = 20/10…

  in algebra mixture problem? Comment by computerguy103 May 2021