Question Solve the matrix equation $\left(\begin{array}{ll}6 & 1 \\ 4 & 3\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}-3 \\ 5\end{array}\right)$ for $x$ and $y$ .

Studdy Solution

STEP 1

Assumptions1. We are given a system of linear equations in matrix form $AX = B$ , where $A$ is the matrix of coefficients, $X$ is the column vector of variables, and $B$ is the column vector of constants. . The system of equations is consistent and has a unique solution.

STEP 2

First, let's write down the system of equations represented by the given matrix equation.
$\begin{bmatrix}6 &1 \\4 &\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}- \\5\end{bmatrix}$ This represents the system of equations $6x + y = -$ $4x +y =5$

STEP 3

We can solve this system of equations using substitution or elimination. Here, we will use the elimination method. Multiply the first equation by3 and the second equation by1 to make the coefficients of $y$ in both equations the same.
$18x +3y = -9$ $x +3y =5$

STEP 4

Now, subtract the second equation from the first to eliminate $y$ .
$(18x +3y) - (4x +3y) = -9 -$

STEP 5

implify the left and right sides of the equation.
$14x = -14$

STEP 6

olve for $x$ by dividing both sides of the equation by14.
$x = -14 /14 = -1$

STEP 7

Now that we have the value of $x$ , we can substitute it into the first equation of our original system to solve for $y$ .
$6(-1) + y = -3$

STEP 8

implify the left side of the equation and solve for $y$ .
$-6 + y = -3$ $y = -3 +6 =3$ The solution to the system of equations is $x = -1$ and $y =3$ .

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Question Solve the matrix equation $\left(\begin{array}{ll}6 & 1 \\ 4 & 3\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}-3 \\ 5\end{array}\right)$ for $x$ and $y$ .

Studdy Solution

STEP 1

Assumptions1. We are given a system of linear equations in matrix form $AX = B$ , where $A$ is the matrix of coefficients, $X$ is the column vector of variables, and $B$ is the column vector of constants. . The system of equations is consistent and has a unique solution.

STEP 2

First, let's write down the system of equations represented by the given matrix equation.
$\begin{bmatrix}6 &1 \\4 &\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}- \\5\end{bmatrix}$ This represents the system of equations $6x + y = -$ $4x +y =5$

STEP 3

We can solve this system of equations using substitution or elimination. Here, we will use the elimination method. Multiply the first equation by3 and the second equation by1 to make the coefficients of $y$ in both equations the same.
$18x +3y = -9$ $x +3y =5$

STEP 4

Now, subtract the second equation from the first to eliminate $y$ .
$(18x +3y) - (4x +3y) = -9 -$

STEP 5

implify the left and right sides of the equation.
$14x = -14$

STEP 6

olve for $x$ by dividing both sides of the equation by14.
$x = -14 /14 = -1$

STEP 7

Now that we have the value of $x$ , we can substitute it into the first equation of our original system to solve for $y$ .
$6(-1) + y = -3$

STEP 8

implify the left side of the equation and solve for $y$ .
$-6 + y = -3$ $y = -3 +6 =3$ The solution to the system of equations is $x = -1$ and $y =3$ .

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Question Solve the matrix equation (6143)(xy)=(−35)\left(\begin{array}{ll}6 & 1 \\ 4 & 3\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}-3 \\ 5\end{array}\right)(64​13​)(xy​)=(−35​) for xxx and yyy.

Studdy Solution

STEP 1

Assumptions1. We are given a system of linear equations in matrix form AX=BAX = BAX=B, where AAA is the matrix of coefficients, XXX is the column vector of variables, and BBB is the column vector of constants. . The system of equations is consistent and has a unique solution.

STEP 2

STEP 3

We can solve this system of equations using substitution or elimination. Here, we will use the elimination method. Multiply the first equation by3 and the second equation by1 to make the coefficients of yyy in both equations the same.18x+3y=−918x +3y = -918x+3y=−9x+3y=5x +3y =5x+3y=5

STEP 4

Now, subtract the second equation from the first to eliminate yyy.(18x+3y)−(4x+3y)=−9−(18x +3y) - (4x +3y) = -9 -(18x+3y)−(4x+3y)=−9−

STEP 5

implify the left and right sides of the equation.14x=−1414x = -1414x=−14

STEP 6

olve for xxx by dividing both sides of the equation by14.x=−14/14=−1x = -14 /14 = -1x=−14/14=−1

STEP 7

Now that we have the value of xxx, we can substitute it into the first equation of our original system to solve for yyy.6(−1)+y=−36(-1) + y = -36(−1)+y=−3

STEP 8

implify the left side of the equation and solve for yyy.−6+y=−3-6 + y = -3−6+y=−3y=−3+6=3y = -3 +6 =3y=−3+6=3The solution to the system of equations is x=−1x = -1x=−1 and y=3y =3y=3.

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Question Solve the matrix equation $\left(\begin{array}{ll}6 & 1 \\ 4 & 3\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}-3 \\ 5\end{array}\right)$ for $x$ and $y$ .

Assumptions1. We are given a system of linear equations in matrix form $AX = B$ , where $A$ is the matrix of coefficients, $X$ is the column vector of variables, and $B$ is the column vector of constants. . The system of equations is consistent and has a unique solution.

We can solve this system of equations using substitution or elimination. Here, we will use the elimination method. Multiply the first equation by3 and the second equation by1 to make the coefficients of $y$ in both equations the same.
$18x +3y = -9$ $x +3y =5$

Now, subtract the second equation from the first to eliminate $y$ .
$(18x +3y) - (4x +3y) = -9 -$

implify the left and right sides of the equation.
$14x = -14$

olve for $x$ by dividing both sides of the equation by14.
$x = -14 /14 = -1$

Now that we have the value of $x$ , we can substitute it into the first equation of our original system to solve for $y$ .
$6(-1) + y = -3$

implify the left side of the equation and solve for $y$ .
$-6 + y = -3$ $y = -3 +6 =3$ The solution to the system of equations is $x = -1$ and $y =3$ .