Question
Please, help me to solve this problems by Cramer's rule
Question number :
1. 5/x + 3/y = 19
2/x + 1/y = 7
Question number :
2. log x + 2 log y = 4
2 log x - 3 log y = 1
The Answer for number 1 :
D = -1
Dx = -2
Dy = -3
The answer for number 2 :
D = -5
Dx = -14
Dy = -3
Please, explain me the steps to get both of that answer..
Question number :
1. 5/x + 3/y = 19
2/x + 1/y = 7
Question number :
2. log x + 2 log y = 4
2 log x - 3 log y = 1
The Answer for number 1 :
D = -1
Dx = -2
Dy = -3
The answer for number 2 :
D = -5
Dx = -14
Dy = -3
Please, explain me the steps to get both of that answer..
Answers
Steve
Your text explains Cramer's Rule, surely. Read it where it discusses taking determinants of coefficients. Or google cramers rule for lots of examples.
Fo the first,
D =
|5 3|
|2 1| = 5-6 = -1
Dx =
|19 3|
|7 1| = 19-21 = -2
x = Dx/D = 2
and similarly for the other parts
Fo the first,
D =
|5 3|
|2 1| = 5-6 = -1
Dx =
|19 3|
|7 1| = 19-21 = -2
x = Dx/D = 2
and similarly for the other parts
Damon
let u = 1/x
let v = 1/y
5 3 |u| 19
2 1 |v| 7
D = 5*1-3*2 = 5-6 = -1
Du = 19*1-3*7 = 19-21 = -2
Dv = 5*7-19*2 = 35-38 = -3
then u = Du/D = -2/-1 = 2
so
x = 1/u = 1/2
and v = Dv/D = -3/-1 = 3
so
y = 1/3
You have typos I think
I assume the second one is actually
log x + 2 log y = 4
log x - 3 log y = 1
let u = log x and v = log y
1 +2 |u| 4
1 -3 |v| 1
D = 1(-3)-2(1) = -3-2 = -5
etc
let v = 1/y
5 3 |u| 19
2 1 |v| 7
D = 5*1-3*2 = 5-6 = -1
Du = 19*1-3*7 = 19-21 = -2
Dv = 5*7-19*2 = 35-38 = -3
then u = Du/D = -2/-1 = 2
so
x = 1/u = 1/2
and v = Dv/D = -3/-1 = 3
so
y = 1/3
You have typos I think
I assume the second one is actually
log x + 2 log y = 4
log x - 3 log y = 1
let u = log x and v = log y
1 +2 |u| 4
1 -3 |v| 1
D = 1(-3)-2(1) = -3-2 = -5
etc