Algebra Examples

$\frac{1}{3} y + 7 + \frac{1}{6} y = \frac{1}{2} y - 1$

Step 1

Simplify $\frac{1}{3} y + 7 + \frac{1}{6} y$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Combine $\frac{1}{3}$ and $y$ .

$\frac{y}{3} + 7 + \frac{1}{6} y = \frac{1}{2} y - 1$

Step 1.1.2

Combine $\frac{1}{6}$ and $y$ .

$\frac{y}{3} + 7 + \frac{y}{6} = \frac{1}{2} y - 1$

Step 1.2

To write $\frac{y}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{y}{3} \cdot \frac{2}{2} + \frac{y}{6} + 7 = \frac{1}{2} y - 1$

Step 1.3

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{y}{3}$ by $\frac{2}{2}$ .

$\frac{y \cdot 2}{3 \cdot 2} + \frac{y}{6} + 7 = \frac{1}{2} y - 1$

Step 1.3.2

Multiply $3$ by $2$ .

$\frac{y \cdot 2}{6} + \frac{y}{6} + 7 = \frac{1}{2} y - 1$

Step 1.4

Combine the numerators over the common denominator.

$\frac{y \cdot 2 + y}{6} + 7 = \frac{1}{2} y - 1$

Step 1.5

Add $y \cdot 2$ and $y$ .

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Step 1.5.1

Reorder $y$ and $2$ .

$\frac{2 \cdot y + y}{6} + 7 = \frac{1}{2} y - 1$

Step 1.5.2

Add $2 \cdot y$ and $y$ .

$\frac{3 \cdot y}{6} + 7 = \frac{1}{2} y - 1$

Step 1.6

Cancel the common factor of $3$ and $6$ .

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Step 1.6.1

Factor $3$ out of $3 \cdot y$ .

$\frac{3 (y)}{6} + 7 = \frac{1}{2} y - 1$

Step 1.6.2

Cancel the common factors.

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Step 1.6.2.1

Factor $3$ out of $6$ .

$\frac{3 y}{3 \cdot 2} + 7 = \frac{1}{2} y - 1$

Step 1.6.2.2

Cancel the common factor.

$\frac{3 y}{3 \cdot 2} + 7 = \frac{1}{2} y - 1$

Step 1.6.2.3

Rewrite the expression.

$\frac{y}{2} + 7 = \frac{1}{2} y - 1$

Step 2

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

$\frac{y}{2} + 7 = \frac{y}{2} - 1$

Step 3

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

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Step 3.1

Subtract $\frac{y}{2}$ from both sides of the equation.

$\frac{y}{2} + 7 - \frac{y}{2} = - 1$

Step 3.2

Combine the opposite terms in $\frac{y}{2} + 7 - \frac{y}{2}$ .

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Step 3.2.1

Subtract $\frac{y}{2}$ from $\frac{y}{2}$ .

$0 + 7 = - 1$

Step 3.2.2

Add $0$ and $7$ .

$7 = - 1$

Step 4

Since $7 \neq - 1$ , there are no solutions.

No solution