2x(2x-3) >= 4x^(2)-3(2x+1)

Expand and Simplify: Expand both sides of the inequality. On the left side, apply the distributive property to 2x(2x-3). 2x(2x) - 2x(3) = 4x^2 - 6x On the right side, apply the distributive property to -3(2x+1). 4x^2 - 3(2x) - 3(1) = 4x^2 - 6x - 3 Now the inequality looks like this: 4x^2 - 6x >= 4x^2 - 6x - 3Subtract to Simplify: Subtract 4x^2 from both sides of the inequality to simplify. 4x^2 - 6x - 4x^2 >= 4x^2 - 6x - 3 - 4x^2 This simplifies to: -6x >= -6x - 3Isolate Constant Term: Subtract -6x from both sides of the inequality to isolate the constant term on one side. -6x + 6x >= -6x - 3 + 6x This simplifies to: 0 >= -3Analyze Simplified Inequality: Analyze the simplified inequality. The inequality 0 >= -3 is always true, because 0 is always greater than or equal to -3. This means that the original inequality holds for all values of x.

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$2 x(2 x-3) \geq 4 x^{2}-3(2 x+1)$

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Grade 8

Solve multi-step equations with fractional coefficients

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2 x(2 x-3) \geq 4 x^{2}-3(2 x+1)

Expand and Simplify: Expand both sides of the inequality. $\newline$ On the left side, apply the distributive property to $2x(2x-3)$ . $\newline$ $2x(2x) - 2x(3) = 4x^2 - 6x$ $\newline$ On the right side, apply the distributive property to $-3(2x+1)$ . $\newline$ $4x^2 - 3(2x) - 3(1) = 4x^2 - 6x - 3$ $\newline$ Now the inequality looks like this: $\newline$ $4x^2 - 6x \geq 4x^2 - 6x - 3$
Subtract to Simplify: Subtract $4x^2$ from both sides of the inequality to simplify. $\newline$ $4x^2 - 6x - 4x^2 \geq 4x^2 - 6x - 3 - 4x^2$ $\newline$ This simplifies to: $\newline$ $-6x \geq -6x - 3$
Isolate Constant Term: Subtract $-6x$ from both sides of the inequality to isolate the constant term on one side. $\newline$ $-6x + 6x \geq -6x - 3 + 6x$ $\newline$ This simplifies to: $\newline$ $0 \geq -3$
Analyze Simplified Inequality: Analyze the simplified inequality. $\newline$ The inequality $0 \geq -3$ is always true, because $0$ is always greater than or equal to $-3$ . $\newline$ This means that the original inequality holds for all values of $x$ .