Solve for x 3x(5x-4)-7=x(15x-6)-2(3x+5)+3

$3 x (5 x - 4) - 7 = x (15 x - 6) - 2 (3 x + 5) + 3$

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x (15 x - 6) - 2 (3 x + 5) + 3 = 3 x (5 x - 4) - 7$

Simplify $x (15 x - 6) - 2 (3 x + 5) + 3$ .

Simplify each term.

Apply the distributive property.

$x (15 x) + x \cdot - 6 - 2 (3 x + 5) + 3 = 3 x (5 x - 4) - 7$

Rewrite using the commutative property of multiplication.

$15 x \cdot x + x \cdot - 6 - 2 (3 x + 5) + 3 = 3 x (5 x - 4) - 7$

Move $- 6$ to the left of $x$ .

$15 x \cdot x - 6 \cdot x - 2 (3 x + 5) + 3 = 3 x (5 x - 4) - 7$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$15 (x \cdot x) - 6 \cdot x - 2 (3 x + 5) + 3 = 3 x (5 x - 4) - 7$

Multiply $x$ by $x$ .

$15 x^{2} - 6 \cdot x - 2 (3 x + 5) + 3 = 3 x (5 x - 4) - 7$

$15 x^{2} - 6 x - 2 (3 x + 5) + 3 = 3 x (5 x - 4) - 7$

Apply the distributive property.

$15 x^{2} - 6 x - 2 (3 x) - 2 \cdot 5 + 3 = 3 x (5 x - 4) - 7$

Multiply $3$ by $- 2$ .

$15 x^{2} - 6 x - 6 x - 2 \cdot 5 + 3 = 3 x (5 x - 4) - 7$

Multiply $- 2$ by $5$ .

$15 x^{2} - 6 x - 6 x - 10 + 3 = 3 x (5 x - 4) - 7$

Simplify by adding terms.

Subtract $6 x$ from $- 6 x$ .

$15 x^{2} - 12 x - 10 + 3 = 3 x (5 x - 4) - 7$

Add $- 10$ and $3$ .

$15 x^{2} - 12 x - 7 = 3 x (5 x - 4) - 7$

Simplify each term.

Apply the distributive property.

$15 x^{2} - 12 x - 7 = 3 x (5 x) + 3 x \cdot - 4 - 7$

Rewrite using the commutative property of multiplication.

$15 x^{2} - 12 x - 7 = 3 \cdot 5 x \cdot x + 3 x \cdot - 4 - 7$

Multiply $- 4$ by $3$ .

$15 x^{2} - 12 x - 7 = 3 \cdot 5 x \cdot x - 12 x - 7$

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$15 x^{2} - 12 x - 7 = 3 \cdot 5 (x \cdot x) - 12 x - 7$

Multiply $x$ by $x$ .

$15 x^{2} - 12 x - 7 = 3 \cdot 5 x^{2} - 12 x - 7$

Multiply $3$ by $5$ .

$15 x^{2} - 12 x - 7 = 15 x^{2} - 12 x - 7$

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

Subtract $15 x^{2}$ from both sides of the equation.

$15 x^{2} - 12 x - 7 - 15 x^{2} = - 12 x - 7$

Add $12 x$ to both sides of the equation.

$15 x^{2} - 12 x - 7 - 15 x^{2} + 12 x = - 7$

Combine the opposite terms in $15 x^{2} - 12 x - 7 - 15 x^{2} + 12 x$ .

Subtract $15 x^{2}$ from $15 x^{2}$ .

$- 12 x - 7 + 0 + 12 x = - 7$

Add $- 12 x - 7$ and $0$ .

$- 12 x - 7 + 12 x = - 7$

Add $- 12 x$ and $12 x$ .

$0 - 7 = - 7$

Subtract $7$ from $0$ .

$- 7 = - 7$

Since $- 7 = - 7$ , the equation will always be true for any value of $x$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x 3x(5x-4)-7=x(15x-6)-2(3x+5)+3

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