Problem 25 - Elementary Algebra

Question

Solve. $$2+2 x-3=3(x-1) x-3$$

Step-by-Step Solution

Answer

The solutions are \(x = -\frac{1}{3}\) and \(x = 2\).

Step 1: Expand Parentheses

Start by expanding the right-hand side of the equation to remove the parentheses. The given equation is:\[2 + 2x - 3 = 3(x-1)x - 3\]Expanding the right-hand side, you first multiply \(3(x-1)\), which is \(3x - 3\) and then multiply \((3x - 3)\) by \(x\) to get \(3x^2 - 3x\). Hence, the equation becomes:\[2 + 2x - 3 = 3x^2 - 3x - 3\]

Step 2: Simplify Both Sides

Next, simplify both sides of the equation. The left side simplifies to:\[2x - 1\]So now the equation is:\[2x - 1 = 3x^2 - 3x - 3\]

Step 3: Rearrange to Form a Quadratic Equation

Rearrange all terms to one side of the equation to form a standard quadratic equation (\(ax^2 + bx + c = 0\)). Subtract \(2x - 1\) from both sides:\[0 = 3x^2 - 3x - 3 - 2x + 1\]Simplifying gives:\[3x^2 - 5x - 2 = 0\]

Step 4: Factor the Quadratic Equation

Attempt to factor the quadratic equation \(3x^2 - 5x - 2 = 0\). Look for two numbers that multiply to \(3 imes -2 = -6\) and add to \(-5\). The numbers \(-6\) and \(1\) work.Rewrite the middle term using these numbers:\[3x^2 - 6x + x - 2 = 0\]Factor by grouping:\[3x(x - 2) + 1(x - 2) = 0\]This gives us the factored form:\[(3x + 1)(x - 2) = 0\]

Step 5: Solve for x

To find the solutions for \(x\), set each factor equal to zero:For \(3x + 1 = 0\):\[3x = -1 \ x = -\frac{1}{3}\]For \(x - 2 = 0\):\[x = 2\]Thus, the solutions are \(x = -\frac{1}{3}\) and \(x = 2\).

Key Concepts

Expanding Parentheses

Expanding parentheses is a crucial step when solving quadratic equations. It involves eliminating the parentheses by distributing the terms outside with the terms inside. This step is essential because it simplifies the equation to a form that is easier to work with.
In our original problem, the expression on the right side of the equation is \(3(x-1)x\).
To expand, start by distributing \(3\) over \((x - 1)\), which gives you \(3x - 3\).
Next, multiply this result by \(x\) to get \(3x^2 - 3x\).
By expanding parentheses, we rewrite the original equation in an expanded form, which allows for further simplification.

Simplifying Equations

Simplifying an equation involves combining like terms and making the equation as compact as possible. This ensures a clearer path toward solving it.
In the equation we started with after expanding, it was crucial to simplify both sides:\(2 + 2x - 3 = 3x^2 - 3x - 3\).
The left side simplifies to \(2x - 1\).
This step brings the equation closer to a standard form where quadratic techniques can be applied. The final simplified equation is \(2x - 1 = 3x^2 - 3x - 3\).
Once simplified, we're ready to rearrange it into a more standard quadratic format.

Factoring Quadratics

Once we have the quadratic equation in a simplified form like \(3x^2 - 5x - 2 = 0\), we need to factor it. Factoring involves expressing the quadratic as a product of two binomials.
To begin, we look for two numbers that multiply to the product of the leading coefficient and the constant term (\(3 \times -2 = -6\)) and add up to the middle coefficient, \(-5\).
The numbers \(-6\) and \(1\) suit these requirements.
Rewrite the equation by splitting the middle term using these two numbers: \(3x^2 - 6x + x - 2 = 0\).
Apply the grouping method to factor by grouping. First, take \(3x\) as a common factor in the first two terms giving \(3x(x - 2)\), and \(1\) as common in the last two terms, giving \(1(x - 2)\).
Unite them into the common factor: \((3x + 1)(x - 2) = 0\).
This form allows us to solve the equation by setting each factor equal to zero.

Solving Quadratic Equations

After factoring the quadratic equation into \((3x + 1)(x - 2) = 0\), solving it involves applying the zero product property. This property states if the product of two terms is zero, then at least one of the terms must be zero.
Set each factor equal to zero and solve for \(x\).

• For \(3x + 1 = 0\), solve by subtracting \(1\) from both sides to get \(3x = -1\), then divide by 3 to find \(x = -\frac{1}{3}\).
• For \(x - 2 = 0\), solve by adding 2 to both sides leading to \(x = 2\).

This results in two solutions: \(x = -\frac{1}{3}\) and \(x = 2\). Finding these solutions is the ultimate goal in solving quadratic equations, revealing the values of \(x\) that satisfy the original equation.