Asked by Iman
                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
                    Answered by
            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
                    Answered by
            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
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