Pre-Algebra Examples

$\frac{7 x}{8} + 1 > \frac{7 x + 1}{8}$

Step 1

Multiply both sides by $8$ .

$(\frac{7 x}{8} + 1) \cdot 8 > \frac{7 x + 1}{8} \cdot 8$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $(\frac{7 x}{8} + 1) \cdot 8$ .

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Step 2.1.1.1

Apply the distributive property.

$\frac{7 x}{8} \cdot 8 + 1 \cdot 8 > \frac{7 x + 1}{8} \cdot 8$

Step 2.1.1.2

Cancel the common factor of $8$ .

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Step 2.1.1.2.1

Cancel the common factor.

$\frac{7 x}{8} \cdot 8 + 1 \cdot 8 > \frac{7 x + 1}{8} \cdot 8$

Step 2.1.1.2.2

Rewrite the expression.

$7 x + 1 \cdot 8 > \frac{7 x + 1}{8} \cdot 8$

Step 2.1.1.3

Multiply $8$ by $1$ .

$7 x + 8 > \frac{7 x + 1}{8} \cdot 8$

Step 2.2

Simplify the right side.

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Step 2.2.1

Cancel the common factor of $8$ .

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Step 2.2.1.1

Cancel the common factor.

$7 x + 8 > \frac{7 x + 1}{8} \cdot 8$

Step 2.2.1.2

Rewrite the expression.

$7 x + 8 > 7 x + 1$

Step 3

Solve for $x$ .

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Step 3.1

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

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Step 3.1.1

Subtract $7 x$ from both sides of the inequality.

$7 x + 8 - 7 x > 1$

Step 3.1.2

Combine the opposite terms in $7 x + 8 - 7 x$ .

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Step 3.1.2.1

Subtract $7 x$ from $7 x$ .

$0 + 8 > 1$

Step 3.1.2.2

Add $0$ and $8$ .

$8 > 1$

Step 3.2

Since $8 > 1$ , the inequality will always be true for any value of $x$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$