Solve the Matrix Equation [[5,11,3]]-[[2,0,2]]=[[3,x,1]]

$[\begin{array}{c} 5 & 11 & 3 \end{array}] - [\begin{array}{c} 2 & 0 & 2 \end{array}] = [\begin{array}{c} 3 & x & 1 \end{array}]$

Simplify the right side of the equation.

Subtract the corresponding elements.

$X = [\begin{array}{c} 5 - 2 & 11 - 0 & 3 - 2 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} 5 - 2 & 11 - 0 & 3 - 2 \end{array}]$ .

Simplify $5 - 2$ .

$X = [\begin{array}{c} 3 & 11 - 0 & 3 - 2 \end{array}]$

Simplify $11 - 0$ .

$X = [\begin{array}{c} 3 & 11 & 3 - 2 \end{array}]$

Simplify $3 - 2$ .

$X = [\begin{array}{c} 3 & 11 & 1 \end{array}]$

Solve.

$3 = 3, x = 11, 1 = 1$

Find the function rule.

Check if the function rule is linear.

To find if the table follows a function rule, check to see if the values follow the linear form $y = a x + b$ .

$y = a x + b$

Build a set of equations from the table such that $y = a x + b$ .

$3 = a (5 + 2) + b 1 = a (3 + 2) + b$

Calculate the values of $a$ and $b$ .

Solve for $b$ in $3 = a \cdot 7 + b$ .

Rewrite the equation as $a \cdot 7 + b = 3$ .

$a \cdot 7 + b = 3$

$1 = a \cdot 5 + b$

Move $7$ to the left of $a$ .

$7 a + b = 3$

$1 = a \cdot 5 + b$

Subtract $7 a$ from both sides of the equation.

$b = 3 - 7 a$

$1 = a \cdot 5 + b$

$b = 3 - 7 a$

$1 = a \cdot 5 + b$

Replace all occurrences of $b$ with $3 - 7 a$ in each equation.

Replace all occurrences of $b$ in $1 = a \cdot 5 + b$ with $3 - 7 a$ .

$1 = a \cdot 5 + 3 - 7 a$

$b = 3 - 7 a$

Simplify $1 = a \cdot 5 + 3 - 7 a$ .

Simplify the left side.

Remove parentheses.

$1 = a \cdot 5 + 3 - 7 a$

$b = 3 - 7 a$

$1 = a \cdot 5 + 3 - 7 a$

$b = 3 - 7 a$

Simplify the right side.

Simplify $a \cdot 5 + 3 - 7 a$ .

Move $5$ to the left of $a$ .

$1 = 5 a + 3 - 7 a$

$b = 3 - 7 a$

Subtract $7 a$ from $5 a$ .

$1 = - 2 a + 3$

$b = 3 - 7 a$

$1 = - 2 a + 3$

$b = 3 - 7 a$

$1 = - 2 a + 3$

$b = 3 - 7 a$

$1 = - 2 a + 3$

$b = 3 - 7 a$

$1 = - 2 a + 3$

$b = 3 - 7 a$

Solve for $a$ in $1 = - 2 a + 3$ .

Rewrite the equation as $- 2 a + 3 = 1$ .

$- 2 a + 3 = 1$

$b = 3 - 7 a$

Move all terms not containing $a$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$- 2 a = 1 - 3$

$b = 3 - 7 a$

Subtract $3$ from $1$ .

$- 2 a = - 2$

$b = 3 - 7 a$

$- 2 a = - 2$

$b = 3 - 7 a$

Divide each term in $- 2 a = - 2$ by $- 2$ and simplify.

Divide each term in $- 2 a = - 2$ by $- 2$ .

$\frac{- 2 a}{- 2} = \frac{- 2}{- 2}$

$b = 3 - 7 a$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 a}{- 2} = \frac{- 2}{- 2}$

$b = 3 - 7 a$

Divide $a$ by $1$ .

$a = \frac{- 2}{- 2}$

$b = 3 - 7 a$

$a = \frac{- 2}{- 2}$

$b = 3 - 7 a$

$a = \frac{- 2}{- 2}$

$b = 3 - 7 a$

Simplify the right side.

Divide $- 2$ by $- 2$ .

$a = 1$

$b = 3 - 7 a$

$a = 1$

$b = 3 - 7 a$

$a = 1$

$b = 3 - 7 a$

$a = 1$

$b = 3 - 7 a$

Replace all occurrences of $a$ with $1$ in each equation.

Replace all occurrences of $a$ in $b = 3 - 7 a$ with $1$ .

$b = 3 - 7 \cdot 1$

$a = 1$

Simplify the right side.

Simplify $3 - 7 \cdot 1$ .

Multiply $- 7$ by $1$ .

$b = 3 - 7$

$a = 1$

Subtract $7$ from $3$ .

$b = - 4$

$a = 1$

$b = - 4$

$a = 1$

$b = - 4$

$a = 1$

$b = - 4$

$a = 1$

List all of the solutions.

$b = - 4, a = 1$

Calculate the value of $y$ using each $x$ value in the relation and compare this value to the given $y$ value in the relation.

Calculate the value of $y$ when $a = 1$ , $b = - 4$ , and $x = 5 + 2$ .

Simplify each term.

Multiply $5 + 2$ by $1$ .

$y = 5 + 2 - 4$

Add $5$ and $2$ .

$y = 7 - 4$

Subtract $4$ from $7$ .

$y = 3$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 5 + 2$ . This check passes since $y = 3$ and $y = 3$ .

$3 = 3$

Calculate the value of $y$ when $a = 1$ , $b = - 4$ , and $x = 3 + 2$ .

Simplify each term.

Multiply $3 + 2$ by $1$ .

$y = 3 + 2 - 4$

Add $3$ and $2$ .

$y = 5 - 4$

Subtract $4$ from $5$ .

$y = 1$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 3 + 2$ . This check passes since $y = 1$ and $y = 1$ .

$1 = 1$

Since $y = y$ for the corresponding $x$ values, the function is linear.

The function is linear

Since all $y = y$ , the function is linear and follows the form $y = x - 4$ .

$y = x - 4$

Find $x$ .

Use the function rule equation to find $x$ .

$x = (11 + 0) - 4$

Simplify.

$x = 7$

Solve the Matrix Equation [[5,11,3]]-[[2,0,2]]=[[3,x,1]]

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