Linear Algebra Examples

$x + y = 3$ , $2 x - 8 y = 19$

Subtract $y$ from both sides of the equation.

$x = - y + 3$

$2 x - 8 y = 19$

Replace all occurrences of $x$ with the solution found by solving the last equation for $x$ . In this case, the value substituted is $- y + 3$ .

$x = - y + 3$

$2 (- y + 3) - 8 y = 19$

Solve for $y$ in the second equation.

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Simplify the left side.

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Simplify each term.

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Apply the distributive property.

$x = - y + 3$

$2 (- y) + 2 \cdot 3 - 8 y = 19$

Multiply $- 1$ by $2$ .

$x = - y + 3$

$- 2 y + 2 \cdot 3 - 8 y = 19$

Multiply $2$ by $3$ .

$x = - y + 3$

$- 2 y + 6 - 8 y = 19$

$x = - y + 3$

$- 2 y + 6 - 8 y = 19$

Subtract $8 y$ from $- 2 y$ .

$x = - y + 3$

$- 10 y + 6 = 19$

$x = - y + 3$

$- 10 y + 6 = 19$

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

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Subtract $6$ from both sides of the equation.

$x = - y + 3$

$- 10 y = - 6 + 19$

Add $- 6$ and $19$ .

$x = - y + 3$

$- 10 y = 13$

$x = - y + 3$

$- 10 y = 13$

Divide each term by $- 10$ and simplify.

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Divide each term in $- 10 y = 13$ by $- 10$ .

$x = - y + 3$

$- \frac{10 y}{- 10} = \frac{13}{- 10}$

Simplify the left side of the equation by cancelling the common factors.

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Reduce the expression by cancelling the common factors.

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Factor $10$ out of $10 y$ .

$x = - y + 3$

$- \frac{10 (y)}{- 10} = \frac{13}{- 10}$

Move the negative one from the denominator of $\frac{y}{- 1}$ .

$x = - y + 3$

$- (- 1 \cdot y) = \frac{13}{- 10}$

$x = - y + 3$

$- (- 1 \cdot y) = \frac{13}{- 10}$

Simplify the expression.

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Multiply $- 1$ by $y$ .

$x = - y + 3$

$- (- 1 y) = \frac{13}{- 10}$

Rewrite $- 1 y$ as $- y$ .

$x = - y + 3$

$y = \frac{13}{- 10}$

$x = - y + 3$

$y = \frac{13}{- 10}$

$x = - y + 3$

$y = \frac{13}{- 10}$

Move the negative in front of the fraction.

$x = - y + 3$

$y = - \frac{13}{10}$

$x = - y + 3$

$y = - \frac{13}{10}$

$x = - y + 3$

$y = - \frac{13}{10}$

Replace all occurrences of $y$ with the solution found by solving the last equation for $y$ . In this case, the value substituted is $- \frac{13}{10}$ .

$x = - (- \frac{13}{10}) + 3$

$y = - \frac{13}{10}$

Simplify $- (- \frac{13}{10}) + 3$ .

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Simplify $- (- \frac{13}{10})$ .

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Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{13}{10}) + 3$

$y = - \frac{13}{10}$

Multiply $\frac{13}{10}$ by $1$ .

$x = \frac{13}{10} + 3$

$y = - \frac{13}{10}$

$x = \frac{13}{10} + 3$

$y = - \frac{13}{10}$

To write $\frac{3}{1}$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$x = \frac{13}{10} + \frac{3}{1} \cdot \frac{10}{10}$

$y = - \frac{13}{10}$

Write each expression with a common denominator of $10$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{13}{10} + \frac{3 \cdot 10}{1 \cdot 10}$

$y = - \frac{13}{10}$

Multiply $10$ by $1$ .

$x = \frac{13}{10} + \frac{3 \cdot 10}{10}$

$y = - \frac{13}{10}$

$x = \frac{13}{10} + \frac{3 \cdot 10}{10}$

$y = - \frac{13}{10}$

Combine the numerators over the common denominator.

$x = \frac{13 + 3 \cdot 10}{10}$

$y = - \frac{13}{10}$

Simplify the numerator.

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Multiply $3$ by $10$ .

$x = \frac{13 + 30}{10}$

$y = - \frac{13}{10}$

Add $13$ and $30$ .

$x = \frac{43}{10}$

$y = - \frac{13}{10}$

$x = \frac{43}{10}$

$y = - \frac{13}{10}$

$x = \frac{43}{10}$

$y = - \frac{13}{10}$

The solution can be shown in both point and equation forms.

Point Form:

$(\frac{43}{10}, - \frac{13}{10})$

Equation Form: $x = \frac{43}{10} y = - \frac{13}{10}$

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