Use Cramer's rule to solve each system of equations. If D = 0,...

Question

Use Cramer's rule to solve each system of equations. If D = 0, then use another method to determine the solution set. See Examples 5–7.

(1/2)x + (1/3)y = 2

(3/2)x - (1/2)y = -12

Accepted Answer

Hello, everyone. We are asked to solve the system of equations by using Kramer's rule. If Kramer's rule does not work because D equals zero, we are asked to find the solution using a different method. The system of equations we are given is two equations. The first one being one third, X plus one half, Y equals two. The second is five halves, X minus three halves, Y equals negative 27. Our answer choices are A since D equals zero, Kramer's rule does not apply and it's the empty set B, the set negative 86 C, the set negative 68 D, the set negative 44. First recall that in Kramer's rule, the first thing we need to do is find the determinant of the matrix that is comprised of the values that are in front of the variables. So the coefficients, so we have the first row is one third, one half. The second row is five halves, negative three halves. Recall that the determinant of A two by two matrix is found by multiplying the diagonals starting in the top left corner and then subtracting the dia the product of the diagonals from the top right corner. So we will have one third multiplied by negative three halves minus one half, multiplied by five halves. And this is gonna be our D. So when I multiply, I recall I can cross reduce. So I am left with negative one half minus 5/ and I'm gonna rewrite this. So they have a common denominator. So I have negative 2/4 minus 5/ and this is negative 7/4 and that's our D. So since D is not zero, we can continue using Kramer's rule. Next, I wanna find the determinant for X, which means I'm going to rewrite a two by two matrix instead of the coefficients for X. In that first column, I'm gonna put our solutions. So two negative 27 and then the second column is the same. It's the coefficients from Y. So one half, negative three halves and again, I'm gonna multiply the diagonals. So two multiplied by negative three halves minus negative multiplied by one half. OK. Cross reducing, I get negative three and then plus 27 halves. And I'm going to change negative three into negative six halves just to make this a little easier to work with um negative six halves plus 27 halves. We'll give us 21 halves. So 21/2. So that's D of X. Next, we need to find the of why? So again, we're gonna make a two by two matrix and in this case, we'll keep the coefficients from X. So one third in five halves is our first column and the solutions will be our second column. So two and negative 27 again, cross multiplying here, I get one third multiplied by negative minus two, multiplied by five halves. Cross reducing is definitely helpful here. So I end up with negative nine minus five. So this is negative 14. So next to actually solve this, I need to do D of X divided by D. So D FX was 21 halves and that's gonna be divided by D which was negative 7/ recall that division could be rewritten as 21/2 divided by negative 7/4. And when we divide fractions, we multiply by the reciprocal. So 21 0 halves multiplied by negative 4/7. And again, cross reducing is definitely something I want to do when I can because this gives me that D FX is negative six. So we know X is negative six. Next, I'm gonna do D F Y. So again, it would be D F Y over D and we have negative 14, negative 14 divided by negative 7/4 which I can rewrite as negative 14/1. There would be times the reciprocal of negative 7/4 which is negative 4/7 cross reducing. I get this as a positive eighth. So my Y value is eight. So this means that our solution set is the point negative 68. And when I check that with my answer choices that is answer choice c have a nice day.

Use Cramer's rule to solve each system of equations. If D = 0, then use another method to determine the solution set. See Examples 5–7.
(1/2)x + (1/3)y = 2
(3/2)x - (1/2)y = -12

Key Concepts

Cramer's Rule

Determinant of a 2x2 Matrix

Alternative Methods for Solving Systems When D = 0

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