Algebra Examples

$- 3 x - 2 y = 5$ $- 2 x + 1 = y$

Step 1

Rewrite the equation as $y = - 2 x + 1$ .

$y = - 2 x + 1$

$- 3 x - 2 y = 5$

Step 2

Replace all occurrences of $y$ with $- 2 x + 1$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $- 3 x - 2 y = 5$ with $- 2 x + 1$ .

$- 3 x - 2 (- 2 x + 1) = 5$

$y = - 2 x + 1$

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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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.

$- 3 x - 2 (- 2 x) - 2 \cdot 1 = 5$

$y = - 2 x + 1$

Step 2.2.1.1.2

Multiply $- 2$ by $- 2$ .

$- 3 x + 4 x - 2 \cdot 1 = 5$

$y = - 2 x + 1$

Step 2.2.1.1.3

Multiply $- 2$ by $1$ .

$- 3 x + 4 x - 2 = 5$

$y = - 2 x + 1$

$- 3 x + 4 x - 2 = 5$

$y = - 2 x + 1$

Step 2.2.1.2

Add $- 3 x$ and $4 x$ .

$x - 2 = 5$

$y = - 2 x + 1$

$x - 2 = 5$

$y = - 2 x + 1$

$x - 2 = 5$

$y = - 2 x + 1$

$x - 2 = 5$

$y = - 2 x + 1$

Step 3

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

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Step 3.1

Add $2$ to both sides of the equation.

$x = 5 + 2$

$y = - 2 x + 1$

Step 3.2

Add $5$ and $2$ .

$x = 7$

$y = - 2 x + 1$

$x = 7$

$y = - 2 x + 1$

Step 4

Replace all occurrences of $x$ with $7$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = - 2 x + 1$ with $7$ .

$y = - 2 \cdot 7 + 1$

$x = 7$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- 2 \cdot 7 + 1$ .

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Step 4.2.1.1

Multiply $- 2$ by $7$ .

$y = - 14 + 1$

$x = 7$

Step 4.2.1.2

Add $- 14$ and $1$ .

$y = - 13$

$x = 7$

$y = - 13$

$x = 7$

$y = - 13$

$x = 7$

$y = - 13$

$x = 7$

Step 5

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

$(7, - 13)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(7, - 13)$

Equation Form:

$x = 7, y = - 13$

Step 7