Precalculus Examples

$y - 2 = 4 (x + 3)$ , $y - 2 = 4 x + 12$

Step 1

Simplify $4 (x + 3)$ .

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Step 1.1

Rewrite.

$y - 2 = 0 + 0 + 4 (x + 3)$

$y - 2 = 4 x + 12$

Step 1.2

Simplify by adding zeros.

$y - 2 = 4 (x + 3)$

$y - 2 = 4 x + 12$

Step 1.3

Apply the distributive property.

$y - 2 = 4 x + 4 \cdot 3$

$y - 2 = 4 x + 12$

Step 1.4

Multiply $4$ by $3$ .

$y - 2 = 4 x + 12$

Step 2

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

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Step 2.1

Add $2$ to both sides of the equation.

$y = 4 x + 12 + 2$

$y - 2 = 4 x + 12$

Step 2.2

Add $12$ and $2$ .

$y = 4 x + 14$

$y - 2 = 4 x + 12$

$y = 4 x + 14$

$y - 2 = 4 x + 12$

Step 3

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

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Step 3.1

Add $2$ to both sides of the equation.

$y = 4 x + 12 + 2$

$y = 4 x + 14$

Step 3.2

Add $12$ and $2$ .

$y = 4 x + 14$

Step 4

Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.

Step 5

Nothing further can be done with this topic. Please check the expression entered or try another topic.

$y - 2 = 4 (x + 3), y - 2 = 4 x + 12$

Step 6