Finite Math Examples

$x - y = 9$ , $x + y = 6$

Add $y$ to both sides of the equation.

$x = y + 9$

$x + y = 6$

Replace all occurrences of $x$ with the solution found by solving the last equation for $x$ . In this case, the value substituted is $y + 9$ .

$x = y + 9$

$(y + 9) + y = 6$

Solve for $y$ in the $s e c o n d$ equation.

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Add $y$ and $y$ .

$x = y + 9$

$2 y + 9 = 6$

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

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Subtract $9$ from both sides of the equation.

$x = y + 9$

$2 y = - 9 + 6$

Add $- 9$ and $6$ .

$x = y + 9$

$2 y = - 3$

$x = y + 9$

$2 y = - 3$

Divide each term by $2$ and simplify.

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Divide each term in $2 y = - 3$ by $2$ .

$x = y + 9$

$\frac{2 y}{2} = - \frac{3}{2}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$x = y + 9$

$\frac{2 y}{2} = - \frac{3}{2}$

Divide $y$ by $1$ .

$x = y + 9$

$y = - \frac{3}{2}$

$x = y + 9$

$y = - \frac{3}{2}$

$x = y + 9$

$y = - \frac{3}{2}$

$x = y + 9$

$y = - \frac{3}{2}$

Replace all occurrences of $y$ with the solution found by solving the last equation for $y$ . In this case, the value substituted is $- \frac{3}{2}$ .

$x = (- \frac{3}{2}) + 9$

$y = - \frac{3}{2}$

Simplify the right side.

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To write $\frac{9}{1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = - \frac{3}{2} + \frac{9}{1} \cdot \frac{2}{2}$

$y = - \frac{3}{2}$

Write each expression with a common denominator of $2$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = - \frac{3}{2} + \frac{9 \cdot 2}{1 \cdot 2}$

$y = - \frac{3}{2}$

Multiply $2$ by $1$ .

$x = - \frac{3}{2} + \frac{9 \cdot 2}{2}$

$y = - \frac{3}{2}$

$x = - \frac{3}{2} + \frac{9 \cdot 2}{2}$

$y = - \frac{3}{2}$

Combine the numerators over the common denominator.

$x = \frac{- 1 \cdot 3 + 9 \cdot 2}{2}$

$y = - \frac{3}{2}$

Simplify the numerator.

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Multiply $- 1$ by $3$ .

$x = \frac{- 3 + 9 \cdot 2}{2}$

$y = - \frac{3}{2}$

Multiply $9$ by $2$ .

$x = \frac{- 3 + 18}{2}$

$y = - \frac{3}{2}$

Add $- 3$ and $18$ .

$x = \frac{15}{2}$

$y = - \frac{3}{2}$

$x = \frac{15}{2}$

$y = - \frac{3}{2}$

$x = \frac{15}{2}$

$y = - \frac{3}{2}$

The solution to the system of equations can be represented as a point.

$(\frac{15}{2}, - \frac{3}{2})$

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