Solved on Dec 13, 2023

Find the y-value of the solution to the $2 \times 2$ matrix equation $\left[\begin{array}{ll} 3 & 2 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \end{array}\right]$ .

STEP 1

Assumptions
1. We are given a matrix equation of the form $A\vec{x} = \vec{b}$ , where $A$ is a $2 \times 2$ matrix, $\vec{x}$ is a column vector with variables $x$ and $y$ , and $\vec{b}$ is a column vector with constants.
2. We need to solve for the variable $y$ in the vector $\vec{x}$ .
The matrix equation is: $\left[\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{c} 3 \\ 4 \end{array}\right]$

STEP 2

Write down the system of linear equations represented by the matrix equation.
$\begin{cases} 3x + 2y = 3 \\ 2x + y = 4 \end{cases}$

STEP 3

We will use the method of substitution or elimination to solve the system of equations. Let's first try to isolate $y$ in the second equation.
$y = 4 - 2x$

STEP 4

Substitute the expression for $y$ from the second equation into the first equation.
$3x + 2(4 - 2x) = 3$

STEP 5

Distribute the $2$ into the parentheses.
$3x + 8 - 4x = 3$

STEP 6

Combine like terms.
$-x + 8 = 3$

STEP 7

Add $x$ to both sides to isolate the constant term on one side.
$8 = x + 3$

STEP 8

Subtract $3$ from both sides to solve for $x$ .
$x = 8 - 3$

STEP 9

Calculate the value of $x$ .
$x = 5$

STEP 10

Now that we have the value of $x$ , substitute it back into the expression for $y$ that we found in STEP_3.
$y = 4 - 2(5)$

STEP 11

Calculate the value of $y$ .
$y = 4 - 10$

STEP 12

Solve for $y$ .
$y = -6$
The $y$ -value of the solution to the matrix equation is $-6$ .

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Find the y-value of the solution to the $2 \times 2$ matrix equation $\left[\begin{array}{ll} 3 & 2 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \end{array}\right]$ .

STEP 1

STEP 2

Write down the system of linear equations represented by the matrix equation.
$\begin{cases} 3x + 2y = 3 \\ 2x + y = 4 \end{cases}$

STEP 3

We will use the method of substitution or elimination to solve the system of equations. Let's first try to isolate $y$ in the second equation.
$y = 4 - 2x$

STEP 4

Substitute the expression for $y$ from the second equation into the first equation.
$3x + 2(4 - 2x) = 3$

STEP 5

Distribute the $2$ into the parentheses.
$3x + 8 - 4x = 3$

STEP 6

Combine like terms.
$-x + 8 = 3$

STEP 7

Add $x$ to both sides to isolate the constant term on one side.
$8 = x + 3$

STEP 8

Subtract $3$ from both sides to solve for $x$ .
$x = 8 - 3$

STEP 9

Calculate the value of $x$ .
$x = 5$

STEP 10

Now that we have the value of $x$ , substitute it back into the expression for $y$ that we found in STEP_3.
$y = 4 - 2(5)$

STEP 11

Calculate the value of $y$ .
$y = 4 - 10$

STEP 12

Solve for $y$ .
$y = -6$
The $y$ -value of the solution to the matrix equation is $-6$ .

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Find the y-value of the solution to the 2×22 \times 22×2 matrix equation [3221][xy]=[34]\left[\begin{array}{ll} 3 & 2 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \end{array}\right][32​21​][xy​]=[34​].

STEP 1

STEP 2

Write down the system of linear equations represented by the matrix equation.{3x+2y=32x+y=4 \begin{cases} 3x + 2y = 3 \\ 2x + y = 4 \end{cases} {3x+2y=32x+y=4​

STEP 3

We will use the method of substitution or elimination to solve the system of equations. Let's first try to isolate yyy in the second equation.y=4−2x y = 4 - 2x y=4−2x

STEP 4

Substitute the expression for yyy from the second equation into the first equation.3x+2(4−2x)=3 3x + 2(4 - 2x) = 3 3x+2(4−2x)=3

STEP 5

Distribute the 222 into the parentheses.3x+8−4x=3 3x + 8 - 4x = 3 3x+8−4x=3

STEP 6

Combine like terms.−x+8=3 -x + 8 = 3 −x+8=3

STEP 7

Add xxx to both sides to isolate the constant term on one side.8=x+3 8 = x + 3 8=x+3

STEP 8

Subtract 333 from both sides to solve for xxx.x=8−3 x = 8 - 3 x=8−3

STEP 9

Calculate the value of xxx.x=5 x = 5 x=5

STEP 10

Now that we have the value of xxx, substitute it back into the expression for yyy that we found in STEP_3.y=4−2(5) y = 4 - 2(5) y=4−2(5)

STEP 11

Calculate the value of yyy.y=4−10 y = 4 - 10 y=4−10

STEP 12

Solve for yyy.y=−6 y = -6 y=−6The yyy-value of the solution to the matrix equation is −6-6−6.

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Find the y-value of the solution to the $2 \times 2$ matrix equation $\left[\begin{array}{ll} 3 & 2 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \end{array}\right]$ .

Write down the system of linear equations represented by the matrix equation.
$\begin{cases} 3x + 2y = 3 \\ 2x + y = 4 \end{cases}$

We will use the method of substitution or elimination to solve the system of equations. Let's first try to isolate $y$ in the second equation.
$y = 4 - 2x$

Substitute the expression for $y$ from the second equation into the first equation.
$3x + 2(4 - 2x) = 3$

Distribute the $2$ into the parentheses.
$3x + 8 - 4x = 3$

Combine like terms.
$-x + 8 = 3$

Add $x$ to both sides to isolate the constant term on one side.
$8 = x + 3$

Subtract $3$ from both sides to solve for $x$ .
$x = 8 - 3$

Calculate the value of $x$ .
$x = 5$

Now that we have the value of $x$ , substitute it back into the expression for $y$ that we found in STEP_3.
$y = 4 - 2(5)$

Calculate the value of $y$ .
$y = 4 - 10$

Solve for $y$ .
$y = -6$
The $y$ -value of the solution to the matrix equation is $-6$ .