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Chapter 1: Problem 79

Evaluate each algebraic expression for the given value of the variable. $$\frac{6 y-4 y^{2}}{y^{2}-15} ; y=5$$

Short Answer

Expert verified
Upon evaluating the given algebraic expression for \(y = 5\), we find that it is equal to -7

Step by step solution

01

Substitution

Substitute \(y = 5\) into the given algebraic expression. This gives us: \[\frac{6 \times 5 - 4 \times 5^2}{5^2 - 15}\]
02

Simplify the numerator

Perform the multiplication, then the subtraction in the numerator: \[6 \times 5 - 4 \times 5^2 = 30 - 100 = -70\]. The expression now becomes: \[\frac{-70}{5^2 - 15}\]
03

Simplify the denominator

Perform the multiplication, then the subtraction in the denominator : \[5^2 - 15 = 25 - 15 = 10\]. This simplifies the expression to: \[\frac{-70}{10}\]
04

Divide

Finally, divide the numerator by the denominator to get the value of the expression: \[-70 / 10 = -7\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution
Substitution is a fundamental process in algebra that allows us to replace a variable with a specific value. When dealing with algebraic expressions, it helps to effectively evaluate the expression and simplify it further. Let's break it down further using the original exercise.
  • The original expression is: \(\frac{6y-4y^{2}}{y^{2}-15}\).
  • Given \(y = 5\), substitution turns the expression into: \(\frac{6 \times 5 - 4 \times 5^2}{5^2 - 15}\).
  • This means that wherever we see the variable \(y\), we replace it with the number 5.
This operation sets the stage for simplification and helps in evaluating the whole expression accurately. Remember: the importance of substitution lies in its ability to transform the expression into a workable numeric form.
Numerator and Denominator Simplification
Simplifying the numerator and the denominator of a fraction is essential in making the math cleaner and more manageable. In our example:
  • The numerator \(6 \times 5 - 4 \times 5^2\) involves both multiplication and subtraction. It's simpler to first handle the multiplications: producing 30 and 100 respectively.
  • Then, we subtract: \(30 - 100 = -70\). This provides us with the simplified numerator of \(-70\).
  • The denominator \(5^2 - 15\) simplifies similarly: first calculating \(5^2 = 25\), and then performing the subtraction: \(25 - 15 = 10\).
After both the numerator and denominator are simplified to \(-70\) and \(10\) respectively, the fraction \(\frac{-70}{10}\) becomes ready for division.
Variable Evaluation
Evaluating a variable essentially means determining its contribution to the entire expression. In the given exercise, after substitution, each step builds on the previous to evaluate the impact of \(y = 5\):
  • The effect of \(y\) on the expression is observed post substitution, where calculations are simplified to real numbers.
  • Through simplifying the numerator and the denominator, the consequences of setting \(y = 5\) become clear and provide a pathway to the solution.
Variable evaluation assures that the algebraic expression is understood within the context of its defined variable values. By doing so, the process becomes more coherent as one translates abstract variables into concrete constants.
Division in Algebra
Division in algebra is a crucial step, especially after simplifying the numerator and the denominator of an expression. The aim is to find the simplest form of the expression by dividing these two results.
  • After simplification, we are left with \(\frac{-70}{10}\).
  • Perform the division: \(-70 / 10 = -7\).
  • The division reduces the expression to a single number, making it easier to interpret the final result as \(-7\).
Effective division ensures that any algebraic fraction is broken down to a meaningful solution. This step ties together the entire evaluation process, showing how prior developments built towards a simple and precise result.

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Most popular questions from this chapter

Without a calculator, you can add numbers using a number line, using absolute value, or using gains and losses. Which method do you find most helpful? Why is this so?

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$(24 \div 6) \div 2=24 \div(6 \div 2)$$

In each exercise, determine whether the given number is a solution of the equation. $$5 y+3-4 y-8=15 ; 20$$

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The algebraic expression \(\frac{6 x+6}{x+1}\) cannot have the same value when two different replacements are made for \(x\) such as \(x=-3\) and \(x=2\)

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. Irrational numbers cannot be negative.
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