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..
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
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
Certainly! To solve the given problems using Cramer's rule, we need to find the values of x and y that satisfy the system of equations.
Here are the step-by-step explanations for both question numbers:
Question number 1:
1. Rewrite the equations in the form Ax + By = C:
5/x + 3/y = 19 -> 5xy + 3xy = 19xy -> 5y + 3x = 19xy (Equation 1)
2/x + 1/y = 7 -> 2y + x = 7xy (Equation 2)
2. Set up the determinant D (denoted as D):
D = (coefficient of x in Equation 1 * coefficient of y in Equation 2) - (coefficient of y in Equation 1 * coefficient of x in Equation 2)
= (3 * 2) - (5 * 1)
= 6 - 5
= 1
3. Set up the determinant Dx:
Dx = (constant term in Equation 1 * coefficient of y in Equation 2) - (coefficient of y in Equation 1 * constant term in Equation 2)
= (19 * 2) - (5 * 7)
= 38 - 35
= 3
4. Set up the determinant Dy:
Dy = (coefficient of x in Equation 1 * constant term in Equation 2) - (constant term in Equation 1 * coefficient of x in Equation 2)
= (3 * 7) - (5 * 2)
= 21 - 10
= 11
5. Calculate the values of x and y:
x = Dx / D = 3 / 1 = 3
y = Dy / D = 11 / 1 = 11
Therefore, the solution to the system of equations in question number 1 is x = 3 and y = 11.
Question number 2:
1. Rewrite the equations in the form M log(x) + N log(y) = P:
log(x) + 2log(y) = 4 -> log(x) + log(y^2) = 4 -> log(xy^2) = 4 (Equation 1)
2log(x) - 3log(y) = 1 -> log(x^2) - log(y^3) = 1 -> log(x^2/y^3) = 1 (Equation 2)
2. Convert the logarithmic equations into exponential form:
Equation 1 becomes xy^2 = 10^4 -> xy^2 = 10000
Equation 2 becomes x^2/y^3 = 10^1 -> x^2/y^3 = 10
3. Use Cramer's rule to calculate the determinants:
- Set up the determinant D:
D = (coefficient of x in Equation 1 * coefficient of y in Equation 2) - (coefficient of y in Equation 1 * coefficient of x in Equation 2)
= (1 * -3) - (2 * 1)
= -3 - 2
= -5
- Set up the determinant Dx:
Dx = (constant term in Equation 1 * coefficient of y in Equation 2) - (coefficient of y in Equation 1 * constant term in Equation 2)
= (10000 * -3) - (2 * 1)
= -30000 - 2
= -30002
- Set up the determinant Dy:
Dy = (coefficient of x in Equation 1 * constant term in Equation 2) - (constant term in Equation 1 * coefficient of x in Equation 2)
= (1 * 1) - (10000 * -3)
= 1 - (-30000)
= 30001
4. Calculate the values of x and y:
x = Dx / D = -30002 / -5 = 6000.4
y = Dy / D = 30001 / -5 = -6000.2
Therefore, the solution to the system of equations in question number 2 is x ≈ 6000.4 and y ≈ -6000.2.
Please note that the answers for question number 2 are approximate due to the use of logarithmic operations.