Given the system of equations: 2x – y = –2 x = 14 + 2y What is the value of the system determinant? What is the value of the y−determinant? What is the value of the x−determinant? What is the solution to the system of equations?

Respuesta :

2x - y = -2

x = 14 + 2y

Plug in the x equation

2(14+2y) - y = -2

Distribute

28 + 4y - y = -2

Combine variables

28 + 3y = -2

Subtract 28 on both sides

3y = -30

Divide the y

y = -10

ANSWER 1

The given system  of equations is:

[tex]2x-y=-2[/tex]

[tex]x=14+2y[/tex]

Let us rewrite the second system to obtain,

[tex]2x-y=-2[/tex]

[tex]x-2y=14[/tex]

We need to apply the Cramer's rule. So we write out the coefficient matrices to obtain:

[tex]\left[\begin{array}{cc}2&-1\\1&-2\end{array}\right][/tex]


The answer column is;

[tex]\left[\begin{array}{c}-2&14\end{array}\right][/tex]


The value of the y-determinant is denoted by [tex]D_y[/tex]. To find this we replace the coefficient determinant with answer-column values in y-column to get,

[tex]D_y=\left|\begin{array}{cc}2&-2\\1&14\end{array}\right|[/tex]

Recall that,

If [tex]A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right][/tex]

then, [tex]det_{A}=\left|\begin{array}{cc}a&b\\c&d\end{array}\right|=ad-bc[/tex]

This implies that,

[tex]D_y=2\times14-1\times(-2)[/tex]

[tex]D_y=28+2[/tex]

[tex]D_y=30[/tex]


ANSWER 2

The value of the x-determinant is denoted by [tex]D_x[/tex]. To find this we replace the coefficient determinant with answer-column values in x-column to get,

[tex]D_x=\left|\begin{array}{cc}-2&-1\\14&-2\end{array}\right|[/tex]


This implies that,

[tex]D_x=-2\times-2-14\times-1[/tex]

[tex]D_x=4+14[/tex]

[tex]D_x=18[/tex]


ANSWER 3

To find the solution to the system of equations.


We need to find the determinant of the coefficient matrix.

[tex]D=\left|\begin{array}{cc}2&-1\\1&-2\end{array}\right|[/tex]


This implies that,

[tex]D=2\times(-2)-1\times-1[/tex]


[tex]D=-4+1[/tex]


[tex]D=-3[/tex]


Cramer's rule says that,

[tex]x=\frac{D_x}{D}[/tex]


[tex]\Rightarrow x=\frac{18}{-3}[/tex]


[tex]\Rightarrow x=-6[/tex]


and


[tex]y=\frac{D_y}{D}[/tex]


[tex]\Rightarrow y=\frac{30}{-3}[/tex]


[tex]\Rightarrow y=-10[/tex]


Therefore the solution is [tex]x=-6[/tex] and  [tex]y=-10[/tex].