Problem 87 - College Algebra Essentials

Question

Evaluate \(x^{2}-(x y-y)\) for \(x\) satisfying \(\frac{3(x+3)}{5}=2 x+6\) and \(y\) satisfying \(-2 y-10=5 y+18\).

Step-by-Step Solution

Answer

The result of evaluating \(x^{2}-(x y-y)\) for \(x\) satisfying \(\frac{3(x+3)}{5}=2 x+6\) and \(y\) satisfying \(-2 y-10=5 y+18\) is \(11 \frac{2}{7}\)

Step 1: Solving for x

First, rewrite the equation \(\frac{3(x+3)}{5}=2 x+6\) in a more manageable form by multiplying every term by 5 to get rid of the fraction: \(3x + 9 = 10x + 30\). Then, subtract \(3x\) and \(30\) from both sides to get \(7x = -21\). Finally, solve for \(x\) by dividing both sides by 7 to get \(x = -3\).

Step 2: Solving for y

Write the equation \(-2 y-10=5 y+18\) and rearrange it by adding \(2y\) and subtracting \(18\) from both sides to get \(7y = -8\). Finally, solve for \(y\) by dividing both sides by 7 to get \(y = -\frac{8}{7}\).

Step 3: Substituting x and y into the Expression

Substitute \(x = -3\) and \(y = -\frac{8}{7}\) into the expression \(x^{2}-(x y-y)\). This yields \((-3)^{2}-(( -3)(-\frac{8}{7}) -(-\frac{8}{7})) = 9 - (-\frac{24}{7}-(-\frac{8}{7})) = 9 - (-\frac{16}{7}) = 9 + \frac{16}{7} = \frac{63}{7} + \frac{16}{7} = \frac{79}{7}\).

Step 4: Simplifying the Final Result

You can simplify the final result a bit by converting it into a mixed number: \(\frac{79}{7} = 11 \frac{2}{7}\).

Key Concepts

Evaluating Algebraic Expressions

Understanding how to evaluate algebraic expressions is a fundamental skill in algebra. This involves substituting numerical values for variables and performing the operations indicated. Take the expression \(x^{2}-(xy-y)\), for example. To evaluate this, you must know the values of \(x\) and \(y\). This kind of task tests your ability to simplify expressions and execute operations in the correct order, following the rules of arithmetic and algebraic properties.

When evaluating, always remember to follow the order of operations - often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). Also, keep in mind to simplify expressions within parentheses first and work your way outwards, applying exponents before other operations, and finally, combining like terms to reach your simplified result.

Solving Linear Equations

Solving linear equations is another cornerstone of algebra. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Linear equations can often be written in the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants.

To solve for the unknown variable, you aim to isolate it on one side of the equation using inverse operations - subtracting the same number from both sides, adding the same number to both sides, dividing both sides by the same number, or multiplying both sides by the same number. In our textbook solution, two linear equations are solved step-by-step, with the aim of finding the values of \(x\) and \(y\) that make both equations true.

Substitution Method in Algebra

The substitution method is a powerful tool for solving systems of equations, where we have more than one equation working together. After finding the value of one variable, as seen in the solution to the earlier provided exercise, you can 'substitute' this value into another equation or expression. This is done by replacing the variable with its corresponding numerical value.

In our example, once you find the values of \(x\) and \(y\), you plug these values into the original algebraic expression \(x^{2}-(xy-y)\). This method is particularly useful as it allows us to reduce complex problems with multiple variables into simpler problems that can be solved with basic arithmetic.