Problem 38 - Algebra A Combined Function

Question

Hint: For Exercises 33 through 38 , first divide the equation through by the coefficient of \(x^{2}\) (or \(\left.y^{2}\right)\). $$ 5(x-2)^{2}+5(y+1)=50 $$

Step-by-Step Solution

Answer

The simplified equation is \(x^2 - 4x + y = 5\).

Step 1: Distribute and Expand the Squared Term

Begin by expanding the squared term in the equation. The equation is given as:\[ 5(x-2)^2 + 5(y+1) = 50 \]First, expand the \((x-2)^2\) term:\[ (x-2)^2 = x^2 - 4x + 4 \]Substitute this back into the equation:\[ 5(x^2 - 4x + 4) + 5(y+1) = 50 \]

Step 2: Simplify the Equation

Now, distribute the 5 through the expanded terms:\[ 5x^2 - 20x + 20 + 5y + 5 = 50 \]Combine like terms:\[ 5x^2 - 20x + 5y + 25 = 50 \]

Step 3: Isolate the Constant and Variable Terms

Subtract 25 from both sides to start isolating the variable terms:\[ 5x^2 - 20x + 5y = 25 \]

Step 4: Divide Through by the Coefficient of \(x^2\)

Finally, divide every term in the equation by 5, which is the coefficient of \(x^2\):\[ x^2 - 4x + y = 5 \]This is the simplified form of the equation after dividing through by the coefficient of \(x^2\).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra, representing equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. These equations form a parabola when graphed, and the solution can be found using various methods, such as factoring, completing the square, or the quadratic formula. In our exercise, we begin with the function \( 5(x-2)^2 + 5(y+1) = 50 \). It's essential to recognize that any term like \((x-2)^2\) is a quadratic expression because it expands to involve \( x^2 \). Quadratic equations may have two, one, or no real solutions depending on the value of the discriminant; something that's determined as \( b^2 - 4ac \).
By mastering how to handle quadratic terms, solving equations becomes more achievable, laying the foundation for many topics in algebra and beyond.

Equation Simplification

Simplifying equations efficiently is a key skill in algebra that enables easier manipulation and solution of equations. In the given step-by-step solution, simplification involves expanding and combining terms: \((x-2)^2\) is expanded to \( x^2 - 4x + 4 \). These terms are then substituted back into the original equation. Simplifying further needed distributing the multiplier throughout each term.
This step leads to clearing out unnecessary complexity. The goal is to transform the equation into its simplest form so you can easily identify properties like coefficients and constants. Practicing simplification helps in recognizing patterns, thus enhancing problem-solving speed and accuracy.

Coefficient Division

Coefficient division is critical when you want to make equations more manageable by normalizing them. In this exercise, we divide all terms by 5, the coefficient of \(x^2\), simplifying \[5x^2 - 20x + 5y + 25 = 50\] to \[x^2 - 4x + y = 5\]. By performing this division, each term is scaled down, making the equation easier to interpret and solve. The benefit of normalizing the equation is you can focus on the relationships between the terms without the clutter of larger numbers.

• It allows for a clearer understanding of the structural parts of the equation.
• You can easily compare and contrast equations with different initial coefficients once they're standardized.

Learning how to divide by coefficients effectively is an invaluable strategy in algebraic manipulation, especially for complex equations with multiple terms.