Question

100 POINTS AND TO WHOEVER SOLVES FIRST NEED NOW

Using Cramer’s Rule, what is the value of x in the system of linear equations below?
3X+4Y=12
x-6=-18

A). 0
B). 3
C). There is no solution to the system because |Ax|=0
D). There are multiple solutions to the system because |Ax|=0

Answer 1

Using Cramer's Rule, the value of x in the given system of linear equations is approximately -5.09.

Step-by-step explanation:

To solve the system of linear equations using Cramer's Rule, we need to find the determinants of the coefficient matrix and the matrices obtained by replacing the corresponding column of the coefficient matrix with the constant matrix. Let's find the determinants:

Determinant of the coefficient matrix = |3 -4| = (3)(-6) - (4)(1) = -18 - 4 = -22

Determinant of the matrix obtained by replacing the x column = |-18 -4| = (-18)(-6) - (-4)(1) = 108 - (-4) = 112

Now, using Cramer's Rule:

x = Determinant of the matrix obtained by replacing the x column / Determinant of the coefficient matrix

x = 112 / -22 = -5.09

Therefore, the value of x in the given system of linear equations is approximately -5.09.

Answer 2

`\large\boxed{A.\ 0}`

Step-by-step explanation:

`\left\{\begin{array}{ccc}ax+by=c\\dx+ey=f\end{array}\right\\\\D=\left|\begin{array}{ccc}a&b\\d&e\end{array}\right|=ae-bd\\\\D_x=\left|\begin{array}{ccc}c&b\\f&e\end{array}\right|=ce-fb\\\\D_y=\left|\begin{array}{ccc}a&c\\d&f\end{array}\right|=fa-cd\\\\\text{If}\ D\\eq 0\ \text{then the system of equations has one solution}\ x=(D_x)/(D)\ and\ y=(D_y)/(D)\\\\\text{If}\ D=0\ \text{and}\ D_x=0\ \text{and}\ D_y=0\ \text{then the system of equations has}\\\text{infinitely many solutions.}\\\\\text{If}\ D=0\ \text{and}\ D_x\\eq0\ \text{or}\ D_y\\eq0\ \text{then the system of equations has}\\\text{no solutions.}`

`\text{We have}\\\\\left\{\begin{array}{ccc}3x+4y=12\\x-6y=-18\end{array}\right\\\\D=\left|\begin{array}{ccc}3&4\\1&-6\end{array}\right|=(3)(-6)+(1)(4)=-18+4=-14\\\\D_x=\left|\begin{array}{ccc}12&4\\-18&-6\end{array}\right|=(12)(-6)-(-18)(4)=-72+72=0\\\\x=(D_x)/(D)\to x=(0)/(-14)=0`