Solve the Matrix Equation [[3,7],[2,5]][[x],[y]]=[[14],[10]]

$[\begin{array}{c} 3 & 7 \\ 2 & 5 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 14 \\ 10 \end{array}]$

Step 1

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} 3 x + 7 y \\ 2 x + 5 y \end{array}] = [\begin{array}{c} 14 \\ 10 \end{array}]$

Step 2

Write as a linear system of equations.

$3 x + 7 y = 14 2 x + 5 y = 10$

Step 3

Solve the system of equations.

Solve for $x$ in $3 x + 7 y = 14$ .

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

$3 x = 14 - 7 y$

$2 x + 5 y = 10$

Divide each term in $3 x = 14 - 7 y$ by $3$ and simplify.

Divide each term in $3 x = 14 - 7 y$ by $3$ .

$\frac{3 x}{3} = \frac{14}{3} + \frac{- 7 y}{3}$

$2 x + 5 y = 10$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{14}{3} + \frac{- 7 y}{3}$

$2 x + 5 y = 10$

Divide $x$ by $1$ .

$x = \frac{14}{3} + \frac{- 7 y}{3}$

$2 x + 5 y = 10$

$x = \frac{14}{3} + \frac{- 7 y}{3}$

$2 x + 5 y = 10$

$x = \frac{14}{3} + \frac{- 7 y}{3}$

$2 x + 5 y = 10$

Simplify the right side.

Move the negative in front of the fraction.

$x = \frac{14}{3} - \frac{7 y}{3}$

$2 x + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$2 x + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$2 x + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$2 x + 5 y = 10$

Replace all occurrences of $x$ with $\frac{14}{3} - \frac{7 y}{3}$ in each equation.

Replace all occurrences of $x$ in $2 x + 5 y = 10$ with $\frac{14}{3} - \frac{7 y}{3}$ .

$2 (\frac{14}{3} - \frac{7 y}{3}) + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Simplify the left side.

Simplify $2 (\frac{14}{3} - \frac{7 y}{3}) + 5 y$ .

Simplify each term.

Apply the distributive property.

$2 (\frac{14}{3}) + 2 (- \frac{7 y}{3}) + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Multiply $2 (\frac{14}{3})$ .

Combine $2$ and $\frac{14}{3}$ .

$\frac{2 \cdot 14}{3} + 2 (- \frac{7 y}{3}) + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Multiply $2$ by $14$ .

$\frac{28}{3} + 2 (- \frac{7 y}{3}) + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$\frac{28}{3} + 2 (- \frac{7 y}{3}) + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Multiply $2 (- \frac{7 y}{3})$ .

Multiply $- 1$ by $2$ .

$\frac{28}{3} - 2 \frac{7 y}{3} + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Combine $- 2$ and $\frac{7 y}{3}$ .

$\frac{28}{3} + \frac{- 2 (7 y)}{3} + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Multiply $7$ by $- 2$ .

$\frac{28}{3} + \frac{- 14 y}{3} + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$\frac{28}{3} + \frac{- 14 y}{3} + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Move the negative in front of the fraction.

$\frac{28}{3} - \frac{14 y}{3} + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$\frac{28}{3} - \frac{14 y}{3} + 5 y = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

To write $5 y$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{28}{3} - \frac{14 y}{3} + 5 y \cdot \frac{3}{3} = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Combine $5 y$ and $\frac{3}{3}$ .

$\frac{28}{3} - \frac{14 y}{3} + \frac{5 y \cdot 3}{3} = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Combine the numerators over the common denominator.

$\frac{28}{3} + \frac{- 14 y + 5 y \cdot 3}{3} = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Combine the numerators over the common denominator.

$\frac{28 - 14 y + 5 y \cdot 3}{3} = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Multiply $3$ by $5$ .

$\frac{28 - 14 y + 15 y}{3} = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Add $- 14 y$ and $15 y$ .

$\frac{28 + y}{3} = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$\frac{28 + y}{3} = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$\frac{28 + y}{3} = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$\frac{28 + y}{3} = 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Solve for $y$ in $\frac{28 + y}{3} = 10$ .

Multiply both sides of the equation by $3$ .

$3 (\frac{28 + y}{3}) = 3 \cdot 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$3 (\frac{28 + y}{3}) = 3 \cdot 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Rewrite the expression.

$28 + y = 3 \cdot 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$28 + y = 3 \cdot 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

$28 + y = 3 \cdot 10$

$x = \frac{14}{3} - \frac{7 y}{3}$

Simplify the right side.

Multiply $3$ by $10$ .

$28 + y = 30$

$x = \frac{14}{3} - \frac{7 y}{3}$

$28 + y = 30$

$x = \frac{14}{3} - \frac{7 y}{3}$

$28 + y = 30$

$x = \frac{14}{3} - \frac{7 y}{3}$

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

Subtract $28$ from both sides of the equation.

$y = 30 - 28$

$x = \frac{14}{3} - \frac{7 y}{3}$

Subtract $28$ from $30$ .

$y = 2$

$x = \frac{14}{3} - \frac{7 y}{3}$

$y = 2$

$x = \frac{14}{3} - \frac{7 y}{3}$

$y = 2$

$x = \frac{14}{3} - \frac{7 y}{3}$

Replace all occurrences of $y$ with $2$ in each equation.

Replace all occurrences of $y$ in $x = \frac{14}{3} - \frac{7 y}{3}$ with $2$ .

$x = \frac{14}{3} - \frac{7 (2)}{3}$

$y = 2$

Simplify the right side.

Simplify $\frac{14}{3} - \frac{7 (2)}{3}$ .

Combine the numerators over the common denominator.

$x = \frac{14 - 7 \cdot 2}{3}$

$y = 2$

Simplify the expression.

Multiply $- 7$ by $2$ .

$x = \frac{14 - 14}{3}$

$y = 2$

Subtract $14$ from $14$ .

$x = \frac{0}{3}$

$y = 2$

Divide $0$ by $3$ .

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

List all of the solutions.

$x = 0, y = 2$

$x = 0, y = 2$

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