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 ->
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
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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?
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
