Algebra Examples

$\frac{1}{2} x = 10 - \frac{1}{2} y$ $8 = x - y$

Step 1

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

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Step 1.1

Rewrite the equation as $x - y = 8$ .

$x - y = 8$

$\frac{1}{2} x = 10 - \frac{1}{2} y$

Step 1.2

Add $y$ to both sides of the equation.

$x = 8 + y$

$\frac{1}{2} x = 10 - \frac{1}{2} y$

$x = 8 + y$

$\frac{1}{2} x = 10 - \frac{1}{2} y$

Step 2

Replace all occurrences of $x$ with $8 + y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $\frac{1}{2} x = 10 - \frac{1}{2} y$ with $8 + y$ .

$\frac{1}{2} \cdot (8 + y) = 10 - \frac{1}{2} y$

$x = 8 + y$

Step 2.2

Simplify $\frac{1}{2} \cdot (8 + y) = 10 - \frac{1}{2} y$ .

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Step 2.2.1

Simplify the left side.

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Step 2.2.1.1

Simplify $\frac{1}{2} \cdot (8 + y)$ .

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Step 2.2.1.1.1

Apply the distributive property.

$\frac{1}{2} \cdot 8 + \frac{1}{2} y = 10 - \frac{1}{2} y$

$x = 8 + y$

Step 2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.1.1.2.1

Factor $2$ out of $8$ .

$\frac{1}{2} \cdot (2 (4)) + \frac{1}{2} y = 10 - \frac{1}{2} y$

$x = 8 + y$

Step 2.2.1.1.2.2

Cancel the common factor.

$\frac{1}{2} \cdot (2 \cdot 4) + \frac{1}{2} y = 10 - \frac{1}{2} y$

$x = 8 + y$

Step 2.2.1.1.2.3

Rewrite the expression.

$4 + \frac{1}{2} y = 10 - \frac{1}{2} y$

$x = 8 + y$

$4 + \frac{1}{2} y = 10 - \frac{1}{2} y$

$x = 8 + y$

Step 2.2.1.1.3

Combine $\frac{1}{2}$ and $y$ .

$4 + \frac{y}{2} = 10 - \frac{1}{2} y$

$x = 8 + y$

$4 + \frac{y}{2} = 10 - \frac{1}{2} y$

$x = 8 + y$

$4 + \frac{y}{2} = 10 - \frac{1}{2} y$

$x = 8 + y$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

Combine $y$ and $\frac{1}{2}$ .

$4 + \frac{y}{2} = 10 - \frac{y}{2}$

$x = 8 + y$

$4 + \frac{y}{2} = 10 - \frac{y}{2}$

$x = 8 + y$

$4 + \frac{y}{2} = 10 - \frac{y}{2}$

$x = 8 + y$

$4 + \frac{y}{2} = 10 - \frac{y}{2}$

$x = 8 + y$

Step 3

Solve for $y$ in $4 + \frac{y}{2} = 10 - \frac{y}{2}$ .

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Step 3.1

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

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Step 3.1.1

Add $\frac{y}{2}$ to both sides of the equation.

$4 + \frac{y}{2} + \frac{y}{2} = 10$

$x = 8 + y$

Step 3.1.2

Combine the numerators over the common denominator.

$4 + \frac{y + y}{2} = 10$

$x = 8 + y$

Step 3.1.3

Add $y$ and $y$ .

$4 + \frac{2 y}{2} = 10$

$x = 8 + y$

Step 3.1.4

Cancel the common factor of $2$ .

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Step 3.1.4.1

Cancel the common factor.

$4 + \frac{2 y}{2} = 10$

$x = 8 + y$

Step 3.1.4.2

Divide $y$ by $1$ .

$4 + y = 10$

$x = 8 + y$

$4 + y = 10$

$x = 8 + y$

$4 + y = 10$

$x = 8 + y$

Step 3.2

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

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Step 3.2.1

Subtract $4$ from both sides of the equation.

$y = 10 - 4$

$x = 8 + y$

Step 3.2.2

Subtract $4$ from $10$ .

$y = 6$

$x = 8 + y$

$y = 6$

$x = 8 + y$

$y = 6$

$x = 8 + y$

Step 4

Replace all occurrences of $y$ with $6$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = 8 + y$ with $6$ .

$x = 8 + 6$

$y = 6$

Step 4.2

Simplify $x = 8 + 6$ .

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Step 4.2.1

Simplify the left side.

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Step 4.2.1.1

Remove parentheses.

$x = 8 + 6$

$y = 6$

$x = 8 + 6$

$y = 6$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Add $8$ and $6$ .

$x = 14$

$y = 6$

$x = 14$

$y = 6$

$x = 14$

$y = 6$

$x = 14$

$y = 6$

Step 5

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

$(14, 6)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(14, 6)$

Equation Form:

$x = 14, y = 6$

Step 7