Precalculus Examples

$4 x - 8 y = 16$ , $3 x - 6 y = 12$

Step 1

Solve for $x$ in $4 x - 8 y = 16$ .

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Step 1.1

Add $8 y$ to both sides of the equation.

$4 x = 16 + 8 y$

$3 x - 6 y = 12$

Step 1.2

Divide each term in $4 x = 16 + 8 y$ by $4$ and simplify.

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Step 1.2.1

Divide each term in $4 x = 16 + 8 y$ by $4$ .

$\frac{4 x}{4} = \frac{16}{4} + \frac{8 y}{4}$

$3 x - 6 y = 12$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $4$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{16}{4} + \frac{8 y}{4}$

$3 x - 6 y = 12$

Step 1.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{16}{4} + \frac{8 y}{4}$

$3 x - 6 y = 12$

$x = \frac{16}{4} + \frac{8 y}{4}$

$3 x - 6 y = 12$

$x = \frac{16}{4} + \frac{8 y}{4}$

$3 x - 6 y = 12$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Divide $16$ by $4$ .

$x = 4 + \frac{8 y}{4}$

$3 x - 6 y = 12$

Step 1.2.3.1.2

Cancel the common factor of $8$ and $4$ .

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Step 1.2.3.1.2.1

Factor $4$ out of $8 y$ .

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

$3 x - 6 y = 12$

Step 1.2.3.1.2.2

Cancel the common factors.

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Step 1.2.3.1.2.2.1

Factor $4$ out of $4$ .

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

$3 x - 6 y = 12$

Step 1.2.3.1.2.2.2

Cancel the common factor.

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

$3 x - 6 y = 12$

Step 1.2.3.1.2.2.3

Rewrite the expression.

$x = 4 + \frac{2 y}{1}$

$3 x - 6 y = 12$

Step 1.2.3.1.2.2.4

Divide $2 y$ by $1$ .

$x = 4 + 2 y$

$3 x - 6 y = 12$

$x = 4 + 2 y$

$3 x - 6 y = 12$

$x = 4 + 2 y$

$3 x - 6 y = 12$

$x = 4 + 2 y$

$3 x - 6 y = 12$

$x = 4 + 2 y$

$3 x - 6 y = 12$

$x = 4 + 2 y$

$3 x - 6 y = 12$

$x = 4 + 2 y$

$3 x - 6 y = 12$

Step 2

Replace all occurrences of $x$ with $4 + 2 y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $3 x - 6 y = 12$ with $4 + 2 y$ .

$3 (4 + 2 y) - 6 y = 12$

$x = 4 + 2 y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $3 (4 + 2 y) - 6 y$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Apply the distributive property.

$3 \cdot 4 + 3 (2 y) - 6 y = 12$

$x = 4 + 2 y$

Step 2.2.1.1.2

Multiply $3$ by $4$ .

$12 + 3 (2 y) - 6 y = 12$

$x = 4 + 2 y$

Step 2.2.1.1.3

Multiply $2$ by $3$ .

$12 + 6 y - 6 y = 12$

$x = 4 + 2 y$

$12 + 6 y - 6 y = 12$

$x = 4 + 2 y$

Step 2.2.1.2

Combine the opposite terms in $12 + 6 y - 6 y$ .

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Step 2.2.1.2.1

Subtract $6 y$ from $6 y$ .

$12 + 0 = 12$

$x = 4 + 2 y$

Step 2.2.1.2.2

Add $12$ and $0$ .

$12 = 12$

$x = 4 + 2 y$

$12 = 12$

$x = 4 + 2 y$

$12 = 12$

$x = 4 + 2 y$

$12 = 12$

$x = 4 + 2 y$

$12 = 12$

$x = 4 + 2 y$

Step 3

Remove any equations from the system that are always true.

$x = 4 + 2 y$

Step 4

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