Problem 88 - College Algebra Essentials

Question

Evaluate \(x^{2}-(x y-y)\) for \(x\) satisfying \(\frac{13 x-6}{4}=5 x+2\) and \(y\) satisfying \(5-y=7(y+4)+1\).

Step-by-Step Solution

Answer

The value of the expression \(x^{2}-(x y-y)\) for \(x\) satisfying \(\frac{13 x-6}{4}=5 x+2\) and \(y\) satisfying \(5-y=7(y+4)+1\) is -5.

Step 1: Solve the equation for x

We first need to solve the equation \(\frac{13 x-6}{4}=5 x+2\) for \(x\). Start by multiplying every term by 4 to get rid of the fraction: \(13x - 6 = 20x + 8\). Rearranging terms to have all x terms on one side and numerical values on the other gives \(7x = -14\). Solving for x, we find that \(x = -2\).

Step 2: Solve the equation for y

We then need to solve the equation \(5-y=7(y+4)+1\) for \(y\). Simplifying the equation gives \(5 - y = 7y + 28 + 1\). Moving every term involving y to one side and numerical values to the other gives \(8y = -24\). Solving for y, we find that \(y = -3\).

Step 3: Substitute x and y into the given expression

Substituting \(x = -2\) and \(y = -3\) into the expression \(x^{2}-(x y-y)\) gives \((-2)^{2}-((-2) * (-3) - (-3))\). Calculating each term carefully gives the result \(4 - (6 + 3) = -5\).

Key Concepts

Quadratic Expressions

Quadratic expressions are a form of algebraic expressions that include terms up to the second degree. Typically, these involve variables raised to the power of two, alongside linear and constant terms. In the context of our exercise, we analyze a quadratic expression in the form of \(x^2 - (xy - y)\). This expression is modified using the specific values derived for \(x\) and \(y\) from the prior steps.

Here is what a typical quadratic expression looks like:

• The standard form is \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.
• The highest power of the variable \(x\) is 2, hence making it a quadratic.

Quadratic expressions can often represent parabolas when graphed. They are foundational in solving quadratic equations, which can describe various real-world scenarios, from projectile motion to minimizing costs or maximizing profits in mathematical modeling.

Linear Equations

Proficiency in solving linear equations is essential for handling more complex algebraic processes, such as the one outlined in our problem. A linear equation typically describes a straight line when plotted on a graph. In our exercise, we encounter linear equations like \(\frac{13x-6}{4} = 5x + 2\) and \(5-y = 7(y+4)+1\).

Let's break down what linear equations involve:

• They are first-degree equations. Every term is either a constant or the product of a constant and the first power of a variable.
• The standard form is \(ax + b = 0\), where \(a\) and \(b\) are real numbers and \(a eq 0\).

The primary goal in solving linear equations is to find the value of variable(s) that make the equation true. By isolating the variable on one side, as done with \(x\) and \(y\) in the exercise, solutions are found. Once these solutions are obtained, they can be substituted back into other equations or expressions.

Substitution Method

The substitution method is a key strategy used for solving systems of equations, especially when dealing with one linear and one nonlinear equation. This method involves solving one of the equations for one variable and then substituting this expression into the other equation, simplifying it into something solvable.

Steps to using the Substitution Method effectively:

• Solve one of the equations for one variable. Preferably choose the equation that is simplest.
• Substitute the expression obtained into the other equation. This yields an equation with a single variable.
• Solve this new equation for the variable.
• Substitute back to find the value of the other variable.

In our exercise, the substitution method is implicit in how the known values for \(x = -2\) and \(y = -3\) are inserted into the quadratic expression \(x^2 - (xy - y)\). This makes it easier to evaluate complex expressions by replacing variables with numbers derived from prior solutions, simplifying the work involved.