Solve for x 2x+3>(6x+7)/3

$2 x + 3 > \frac{6 x + 7}{3}$

Step 1

Multiply both sides by $3$ .

$(2 x + 3) \cdot 3 > \frac{6 x + 7}{3} \cdot 3$

Step 2

Simplify.

Simplify the left side.

Simplify $(2 x + 3) \cdot 3$ .

Apply the distributive property.

$2 x \cdot 3 + 3 \cdot 3 > \frac{6 x + 7}{3} \cdot 3$

Multiply.

Multiply $3$ by $2$ .

$6 x + 3 \cdot 3 > \frac{6 x + 7}{3} \cdot 3$

Multiply $3$ by $3$ .

$6 x + 9 > \frac{6 x + 7}{3} \cdot 3$

Simplify the right side.

Cancel the common factor of $3$ .

Cancel the common factor.

$6 x + 9 > \frac{6 x + 7}{3} \cdot 3$

Rewrite the expression.

$6 x + 9 > 6 x + 7$

Step 3

Solve for $x$ .

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

Subtract $6 x$ from both sides of the inequality.

$6 x + 9 - 6 x > 7$

Combine the opposite terms in $6 x + 9 - 6 x$ .

Subtract $6 x$ from $6 x$ .

$0 + 9 > 7$

Add $0$ and $9$ .

$9 > 7$

Since $9 > 7$ , 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:

$(- \infty, \infty)$