Algebra Examples

$y + 2 = 3 (x - 1)$ $y = - 2 x + 10$

Step 1

Replace all occurrences of $y$ with $- 2 x + 10$ in each equation.

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Step 1.1

Replace all occurrences of $y$ in $y + 2 = 3 (x - 1)$ with $- 2 x + 10$ .

$(- 2 x + 10) + 2 = 3 (x - 1)$

$y = - 2 x + 10$

Step 1.2

Simplify $(- 2 x + 10) + 2 = 3 (x - 1)$ .

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Step 1.2.1

Simplify the left side.

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Step 1.2.1.1

Add $10$ and $2$ .

$- 2 x + 12 = 3 (x - 1)$

$y = - 2 x + 10$

$- 2 x + 12 = 3 (x - 1)$

$y = - 2 x + 10$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Simplify $3 (x - 1)$ .

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Step 1.2.2.1.1

Apply the distributive property.

$- 2 x + 12 = 3 x + 3 \cdot - 1$

$y = - 2 x + 10$

Step 1.2.2.1.2

Multiply $3$ by $- 1$ .

$- 2 x + 12 = 3 x - 3$

$y = - 2 x + 10$

$- 2 x + 12 = 3 x - 3$

$y = - 2 x + 10$

$- 2 x + 12 = 3 x - 3$

$y = - 2 x + 10$

$- 2 x + 12 = 3 x - 3$

$y = - 2 x + 10$

$- 2 x + 12 = 3 x - 3$

$y = - 2 x + 10$

Step 2

Solve for $x$ in $- 2 x + 12 = 3 x - 3$ .

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Step 2.1

Move all terms containing $x$ to the left side of the equation.

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Step 2.1.1

Subtract $3 x$ from both sides of the equation.

$- 2 x + 12 - 3 x = - 3$

$y = - 2 x + 10$

Step 2.1.2

Subtract $3 x$ from $- 2 x$ .

$- 5 x + 12 = - 3$

$y = - 2 x + 10$

$- 5 x + 12 = - 3$

$y = - 2 x + 10$

Step 2.2

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

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Step 2.2.1

Subtract $12$ from both sides of the equation.

$- 5 x = - 3 - 12$

$y = - 2 x + 10$

Step 2.2.2

Subtract $12$ from $- 3$ .

$- 5 x = - 15$

$y = - 2 x + 10$

$- 5 x = - 15$

$y = - 2 x + 10$

Step 2.3

Divide each term in $- 5 x = - 15$ by $- 5$ and simplify.

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Step 2.3.1

Divide each term in $- 5 x = - 15$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 15}{- 5}$

$y = - 2 x + 10$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $- 5$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 15}{- 5}$

$y = - 2 x + 10$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 15}{- 5}$

$y = - 2 x + 10$

$x = \frac{- 15}{- 5}$

$y = - 2 x + 10$

$x = \frac{- 15}{- 5}$

$y = - 2 x + 10$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $- 15$ by $- 5$ .

$x = 3$

$y = - 2 x + 10$

$x = 3$

$y = - 2 x + 10$

$x = 3$

$y = - 2 x + 10$

$x = 3$

$y = - 2 x + 10$

Step 3

Replace all occurrences of $x$ with $3$ in each equation.

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Step 3.1

Replace all occurrences of $x$ in $y = - 2 x + 10$ with $3$ .

$y = - 2 \cdot 3 + 10$

$x = 3$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $- 2 \cdot 3 + 10$ .

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Step 3.2.1.1

Multiply $- 2$ by $3$ .

$y = - 6 + 10$

$x = 3$

Step 3.2.1.2

Add $- 6$ and $10$ .

$y = 4$

$x = 3$

$y = 4$

$x = 3$

$y = 4$

$x = 3$

$y = 4$

$x = 3$

Step 4

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

$(3, 4)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(3, 4)$

Equation Form:

$x = 3, y = 4$

Step 6