Finite Math Examples

$2 x + y = - 2$ , $x + 2 y = 2$

Solve for $x$ in the first equation.

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Since $y$ does not contain the variable to solve for, move it to the right side of the equation by subtracting $y$ from both sides.

$2 x = - y - 2$

$x + 2 y = 2$

Divide each term by $2$ and simplify.

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

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

$x + 2 y = 2$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

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

$x + 2 y = 2$

Divide $x$ by $1$ to get $x$ .

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

$x + 2 y = 2$

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

$x + 2 y = 2$

Simplify each term.

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Divide $2$ by $2$ to get $1$ .

$x = - \frac{y}{2} - 1 \cdot 1$

$x + 2 y = 2$

Multiply $- 1$ by $1$ to get $- 1$ .

$x = - \frac{y}{2} - 1$

$x + 2 y = 2$

$x = - \frac{y}{2} - 1$

$x + 2 y = 2$

$x = - \frac{y}{2} - 1$

$x + 2 y = 2$

$x = - \frac{y}{2} - 1$

$x + 2 y = 2$

Replace all occurrences of $x$ with the solution found by solving the last equation for $x$ . In this case, the value substituted is $- \frac{y}{2} - 1$ .

$x = - \frac{y}{2} - 1$

$(- \frac{y}{2} - 1) + 2 y = 2$

Solve for $y$ in the second equation.

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Simplify the left side.

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

$x = - \frac{y}{2} - 1$

$- \frac{y}{2} + \frac{2 y}{1} \cdot \frac{2}{2} - 1 = 2$

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

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Combine.

$x = - \frac{y}{2} - 1$

$- \frac{y}{2} + \frac{2 y \cdot 2}{1 \cdot 2} - 1 = 2$

Multiply $2$ by $1$ to get $2$ .

$x = - \frac{y}{2} - 1$

$- \frac{y}{2} + \frac{2 y \cdot 2}{2} - 1 = 2$

$x = - \frac{y}{2} - 1$

$- \frac{y}{2} + \frac{2 y \cdot 2}{2} - 1 = 2$

Combine the numerators over the common denominator.

$x = - \frac{y}{2} - 1$

$\frac{- y + 2 y \cdot 2}{2} - 1 = 2$

Simplify each term.

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Simplify the numerator.

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Factor $y$ out of $- y + 2 y \cdot 2$ .

$x = - \frac{y}{2} - 1$

$\frac{y (- 1 + 2 \cdot 2)}{2} - 1 = 2$

Multiply $2$ by $2$ to get $4$ .

$x = - \frac{y}{2} - 1$

$\frac{y (- 1 + 4)}{2} - 1 = 2$

Add $- 1$ and $4$ to get $3$ .

$x = - \frac{y}{2} - 1$

$\frac{y \cdot 3}{2} - 1 = 2$

$x = - \frac{y}{2} - 1$

$\frac{y \cdot 3}{2} - 1 = 2$

Move $3$ to the left of the expression $y \cdot 3$ .

$x = - \frac{y}{2} - 1$

$\frac{3 \cdot y}{2} - 1 = 2$

Multiply $3$ by $y$ to get $3 y$ .

$x = - \frac{y}{2} - 1$

$\frac{3 y}{2} - 1 = 2$

$x = - \frac{y}{2} - 1$

$\frac{3 y}{2} - 1 = 2$

$x = - \frac{y}{2} - 1$

$\frac{3 y}{2} - 1 = 2$

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

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Since $- 1$ does not contain the variable to solve for, move it to the right side of the equation by adding $1$ to both sides.

$x = - \frac{y}{2} - 1$

$\frac{3 y}{2} = 1 + 2$

Add $1$ and $2$ to get $3$ .

$x = - \frac{y}{2} - 1$

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

$x = - \frac{y}{2} - 1$

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

Solve for $y$ .

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Multiply both sides of the equation by $2$ .

$x = - \frac{y}{2} - 1$

$3 y = 3 \cdot (2)$

Simplify the right side.

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Multiply $3$ by $2$ to get $3 (2)$ .

$x = - \frac{y}{2} - 1$

$3 y = 3 (2)$

Multiply $3$ by $2$ to get $6$ .

$x = - \frac{y}{2} - 1$

$3 y = 6$

$x = - \frac{y}{2} - 1$

$3 y = 6$

Divide each term by $3$ and simplify.

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Divide each term in $3 y = 6$ by $3$ .

$x = - \frac{y}{2} - 1$

$\frac{3 y}{3} = \frac{6}{3}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$x = - \frac{y}{2} - 1$

$\frac{3 y}{3} = \frac{6}{3}$

Divide $y$ by $1$ to get $y$ .

$x = - \frac{y}{2} - 1$

$y = \frac{6}{3}$

$x = - \frac{y}{2} - 1$

$y = \frac{6}{3}$

Divide $6$ by $3$ to get $2$ .

$x = - \frac{y}{2} - 1$

$y = 2$

$x = - \frac{y}{2} - 1$

$y = 2$

$x = - \frac{y}{2} - 1$

$y = 2$

$x = - \frac{y}{2} - 1$

$y = 2$

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

$x = - \frac{2}{2} - 1$

$y = 2$

Simplify the right side.

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Simplify each term.

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Divide $2$ by $2$ to get $1$ .

$x = - 1 \cdot 1 - 1$

$y = 2$

Multiply $- 1$ by $1$ to get $- 1$ .

$x = - 1 - 1$

$y = 2$

$x = - 1 - 1$

$y = 2$

Subtract $1$ from $- 1$ to get $- 2$ .

$x = - 2$

$y = 2$

$x = - 2$

$y = 2$

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

$(- 2, 2)$

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