SOLUTION: solve the system of equations using gaussian elimination or gauss-jordan elimination. 1. x-6y=23 2x-6y=28 2. Ellen wishes to mix candy worth \$3.44 per pound with candy w

Algebra ->  Algebra  -> Matrices-and-determiminant -> SOLUTION: solve the system of equations using gaussian elimination or gauss-jordan elimination. 1. x-6y=23 2x-6y=28 2. Ellen wishes to mix candy worth \$3.44 per pound with candy w      Log On

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 Question 419050: solve the system of equations using gaussian elimination or gauss-jordan elimination. 1. x-6y=23 2x-6y=28 2. Ellen wishes to mix candy worth \$3.44 per pound with candy worth \$9.96 per pound to form 24 pounds of a mixture worth \$8.33 per pound. How many pounds of the more expensive candy should she use? Answer by mananth(12270)   (Show Source): You can put this solution on YOUR website!1. x-6y=23 2x-6y=28 .................. x-6y=23 .............1 2x-6y=28.............2 multiply (1)by 1 Multiply (2) by -1 x-6y=23 -2x+6y= -28 Add the two equations -x=-5 /-1 x=5 plug value of x in (1) x-6y=23 5-6y=23 -6y=23-5 -6y=18 y=-3 ..... 2. Ellen wishes to mix candy worth \$3.44 per pound with candy worth \$9.96 per pound to form 24 pounds of a mixture worth \$8.33 per pound. How many pounds of the more expensive candy should she use? 3.44 ---------------- x 9.96 ---------------- 24-x 8.33 ---------------- 24 .................... 3.44x+9.96(24-x)=8.33*24 3.44x+239.04-9.96x=199.92 3.44x-9.96x=199.92-239.04 -6.52x =-39.12 /-6.52 x=6 pounds of \$3.44 candy 18 pounds of \$9.96 candy