Will give braineliest

Mathematics

2 answers:

OleMash [197]4 years ago

7 0

Answer:

$3Fx + 3Gy = 3H\\Qx + Ry = S$

Step-by-step explanation:

Given:

The system of equations given are:

$Fx + Gy = H \\Qx + Ry = S$

If an equation is multiplied by a number throughout, then the equation remains the same by multiplication property of equality. The multiplication property of equality says that if both sides of the equation is multiplied by the same number, then the new equation is equivalent to the previous equation.

For example: If $x+y =2$ is multiplied by 2 on both sides, then the new equation will be $2x+2y=4$ which is equivalent to $x+y =2$ .

So, from among all the choices given, the first choice represents the above mentioned property. The equation $Fx + Gy = H$ is multiplied by 3 on both sides. So, the equation $3Fx + 3Gy = 3H$ is equivalent to $Fx + Gy = H$ .

The second equation $Qx + Ry = S$ has not been changed. So, the equivalent system of equations to the given system is:

$3Fx + 3Gy = 3H\\Qx + Ry = S$

o-na [289]4 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

Find the value of the determinant using the method of expansion by minors; expand on the third row

valkas [14]

For the matrix

$\begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}$

the determinant using the method of expansion by minors, expanding on the third row is:

$\det \begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}=a_{31}\det \begin{bmatrix}{a_{12}} & {a_{13}} & {} \\ {a_{22}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}-a_{32}\det \begin{bmatrix}{a_{11}} & {a_{13}} & {} \\ {a_{21}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}+a_{33}\det \begin{bmatrix}{a_{12}} & {a_{12}} & {} \\ {a_{22}} & {a_{22}} & {} \\ {} & {} & {}\end{bmatrix}$

Answer:

First, we compute the determinants of the minors:

$\begin{gathered} \det \begin{bmatrix}{0} & {4} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=0+4=4, \\ \det \begin{bmatrix}{1} & {4} & {} \\ {1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=3-4=-1, \\ \det \begin{bmatrix}{1} & {0} & {} \\ {1} & {-1} & {} \\ {} & {} & {}\end{bmatrix}=-1-0=-1. \end{gathered}$

Therefore: