Question

Solve each equation, if possible. $$ x(1+2 x)=(2 x-1)(x-2) $$

Short answer

\( x = \frac{1}{3} \)

Step-by-step solution

Expand both sides

First, expand both sides of the equation. On the left side, distribute the x: \[ x(1 + 2x) = x + 2x^2 \] On the right side, use the distributive property to expand: \[ (2x - 1)(x - 2) = 2x^2 - 4x - x + 2 = 2x^2 - 5x + 2 \]

Equate and Rearrange

Set the expanded expressions equal to each other and move all terms to one side to set the equation to zero: \[ x + 2x^2 = 2x^2 - 5x + 2 \] Subtract \(2x^2\) from both sides: \[ x = -5x + 2 \]

Solve for x

Isolate the x terms by adding 5x to both sides: \[ 6x = 2 \] Then, divide both sides by 6 to solve for x: \[ x = \frac{1}{3} \]

Verify the solution

Substitute \( x = \frac{1}{3} \) back into the original equation to verify: Left-hand side, \[ x(1 + 2x) = \frac{1}{3}(1 + 2 \cdot \frac{1}{3}) = \frac{1}{3}(\frac{5}{3}) = \frac{5}{9} \] Right-hand side, \[ (2x-1)(x-2) = (\frac{2}{3} - 1)(\frac{1}{3}-2) = -\frac{1}{3}(-\frac{5}{3}) = \frac{5}{9} \] Both sides are equal so the solution is verified. Therefore, the valid solution is \( x = \frac{1}{3} \).

Key concepts

Distributive Property

The distributive property is a fundamental algebraic principle often used in solving equations, especially those involving parentheses. It states that for any numbers or variables, \(a(b + c) = ab + ac\). This property allows you to expand expressions and simplify equations. For example, in our equation, we distribute the \(x\) on the left side: \[ x(1 + 2x) = x + 2x^2 \]. This action turns a single term multiplied by a binomial into a more manageable polynomial. The distributive property helps in the breaking down of complex mathematical expressions into simpler parts, making it easier to solve the equation. Always look for opportunities to use this property when dealing with parentheses.

Expanding Expressions

Expanding expressions involves applying the distributive property to multiply out factors and simplify the equation. In our example, on the right side of the equation, we had \( (2x-1)(x-2) \). We apply the distributive property step-by-step:

• First, distribute \( (2x) \) to both terms inside the second parentheses: \(2x \times x = 2x^2\) and \(2x \times -2 = -4x\).
• Next, distribute \( (-1) \) to both terms inside the second parentheses: \(-1 \times x = -x\) and \(-1 \times -2 = 2\).
• Combine all the terms: \[2x^2 - 4x - x + 2 = 2x^2 - 5x + 2\].

Expanding expressions is crucial in simplifying and making equations easier to manage. It transforms products into sums or differences, which are simpler to work with in further steps.

Rearranging Terms

After expanding the expressions, the next important step is rearranging them to set up the equation for solving. To do this, we usually move all terms to one side, forming a single, simplified expression equal to zero. Consider our example after expansion: \[ x + 2x^2 = 2x^2 - 5x + 2\]. To rearrange these terms, we subtract \(2x^2\) from both sides: \[x = -5x + 2\]. This process of rearranging helps in isolating the variable and identifying which terms need to be combined or eliminated. Simplifying equations step-by-step by rearranging terms is essential to clear the path towards finding the solution.

Isolating Variables

Isolating variables is the final and often most crucial step in solving equations. It means manipulating the equation to get the variable you are solving for on one side by itself. In our example, after rearranging terms to get \[x = -5x + 2\], we add \(5x\) to both sides: \[x + 5x = 2 \] which simplifies to \[6x = 2\]. To isolate \(x\), divide both sides by 6: \[ x = \frac{1}{3} \]. This step usually involves simplification through addition, subtraction, multiplication, or division until the variable stands alone. Isolating the variable is the step where the solution becomes clear. Always verify the solution by substituting it back into the original equation to ensure both sides are equal.