Algebra Examples

$y = \frac{2}{x + 3}$ , $x - 2 y = 0$

Step 1

Replace all occurrences of $y$ with $\frac{2}{x + 3}$ in each equation.

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Step 1.1

Replace all occurrences of $y$ in $x - 2 y = 0$ with $\frac{2}{x + 3}$ .

$x - 2 \frac{2}{x + 3} = 0$

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

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify $x - 2 \frac{2}{x + 3}$ .

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Step 1.2.1.1

Simplify each term.

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Step 1.2.1.1.1

Multiply $- 2 \frac{2}{x + 3}$ .

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Step 1.2.1.1.1.1

Combine $- 2$ and $\frac{2}{x + 3}$ .

$x + \frac{- 2 \cdot 2}{x + 3} = 0$

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

Step 1.2.1.1.1.2

Multiply $- 2$ by $2$ .

$x + \frac{- 4}{x + 3} = 0$

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

$x + \frac{- 4}{x + 3} = 0$

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

Step 1.2.1.1.2

Move the negative in front of the fraction.

$x - \frac{4}{x + 3} = 0$

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

$x - \frac{4}{x + 3} = 0$

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

Step 1.2.1.2

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

$x \cdot \frac{x + 3}{x + 3} - \frac{4}{x + 3} = 0$

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

Step 1.2.1.3

Simplify terms.

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Step 1.2.1.3.1

Combine $x$ and $\frac{x + 3}{x + 3}$ .

$\frac{x (x + 3)}{x + 3} - \frac{4}{x + 3} = 0$

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

Step 1.2.1.3.2

Combine the numerators over the common denominator.

$\frac{x (x + 3) - 4}{x + 3} = 0$

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

$\frac{x (x + 3) - 4}{x + 3} = 0$

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

Step 1.2.1.4

Simplify the numerator.

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Step 1.2.1.4.1

Apply the distributive property.

$\frac{x \cdot x + x \cdot 3 - 4}{x + 3} = 0$

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

Step 1.2.1.4.2

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot 3 - 4}{x + 3} = 0$

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

Step 1.2.1.4.3

Move $3$ to the left of $x$ .

$\frac{x^{2} + 3 \cdot x - 4}{x + 3} = 0$

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

Step 1.2.1.4.4

Factor $x^{2} + 3 x - 4$ using the AC method.

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Step 1.2.1.4.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $3$ .

$- 1, 4$

Step 1.2.1.4.4.2

Write the factored form using these integers.

$\frac{(x - 1) (x + 4)}{x + 3} = 0$

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

$\frac{(x - 1) (x + 4)}{x + 3} = 0$

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

$\frac{(x - 1) (x + 4)}{x + 3} = 0$

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

$\frac{(x - 1) (x + 4)}{x + 3} = 0$

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

$\frac{(x - 1) (x + 4)}{x + 3} = 0$

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

$\frac{(x - 1) (x + 4)}{x + 3} = 0$

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

Step 2

Solve for $x$ in $\frac{(x - 1) (x + 4)}{x + 3} = 0$ .

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Step 2.1

Set the numerator equal to zero.

$(x - 1) (x + 4) = 0$

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

Step 2.2

Solve the equation for $x$ .

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Step 2.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x + 4 = 0$

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

Step 2.2.2

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 2.2.2.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

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

Step 2.2.2.2

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 1$

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

Step 2.2.3

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.2.3.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

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

Step 2.2.3.2

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$x = - 4$

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

Step 2.2.4

The final solution is all the values that make $(x - 1) (x + 4) = 0$ true.

$x = 1, - 4$

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

$x = 1, - 4$

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

$x = 1, - 4$

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

Step 3

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

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Step 3.1

Replace all occurrences of $x$ in $y = \frac{2}{x + 3}$ with $1$ .

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

$x = 1$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $\frac{2}{(1) + 3}$ .

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Step 3.2.1.1

Add $1$ and $3$ .

$y = \frac{2}{4}$

$x = 1$

Step 3.2.1.2

Cancel the common factor of $2$ and $4$ .

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Step 3.2.1.2.1

Factor $2$ out of $2$ .

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

$x = 1$

Step 3.2.1.2.2

Cancel the common factors.

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Step 3.2.1.2.2.1

Factor $2$ out of $4$ .

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

$x = 1$

Step 3.2.1.2.2.2

Cancel the common factor.

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

$x = 1$

Step 3.2.1.2.2.3

Rewrite the expression.

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

$x = 1$

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

$x = 1$

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

$x = 1$

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

$x = 1$

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

$x = 1$

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

$x = 1$

Step 4

Replace all occurrences of $x$ with $- 4$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = \frac{2}{x + 3}$ with $- 4$ .

$y = \frac{2}{(- 4) + 3}$

$x = - 4$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{2}{(- 4) + 3}$ .

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Step 4.2.1.1

Add $- 4$ and $3$ .

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

$x = - 4$

Step 4.2.1.2

Divide $2$ by $- 1$ .

$y = - 2$

$x = - 4$

$y = - 2$

$x = - 4$

$y = - 2$

$x = - 4$

$y = - 2$

$x = - 4$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(1, \frac{1}{2})$

$(- 4, - 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, \frac{1}{2}), (- 4, - 2)$

Equation Form:

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

$x = - 4, y = - 2$

Step 7

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$