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

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