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
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..
2 answers
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
1 answer