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Chapter 1: Problem 16

Find all values of \(x\) satisfying the given conditions. \(y_{1}=2.5, y_{2}=2 x+1, y_{3}=x,\) and the difference between 2 times \(y_{1}\) and 3 times \(y_{2}\) is 8 less than 4 times \(y_{3}.\)

Short Answer

Expert verified
The value of \(x\) that satisfies the given conditions is \(x = 3\).

Step by step solution

01

Set up equation

From the given problem, we know that the difference between 2 times \(y_{1}\) and 3 times \(y_{2}\) is 8 less than 4 times \(y_{3}\). This can be written as: \(2y_{1} - 3y_{2} = 4y_{3} - 8\).
02

Substitute the given values of \(y_{1}\), \(y_{2}\) and \(y_{3}\)

Now we substitute the given values into the derived equation in Step 1. Substituting the value of \(y_{1}\) as 2.5, \(y_{2}\) as \(2x+1\), and \(y_{3}\) as \(x\) into the equation gives us: \(2(2.5) - 3(2x+1) = 4(x) - 8\).
03

Solve the equation

Solving the equation from Step 2 gives: \(5 - (6x+3) = 4x - 8\). Simplifying both sides gives: \(-2x = -6\). Dividing by -2 gives the value of \(x\).
04

Find the value of \(x\)

Finally, dividing by -2 gives us the value of \(x = 3\).

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Key Concepts

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

Equation Solving
Solving an equation is like cracking a code. It involves finding the value of the variable that makes the equation true. In our original exercise, the goal is to find the value of \(x\) that satisfies the given condition. Here鈥檚 how we did it:
  • The problem gives us a relationship between different expressions involving \(x\).
  • First, we set up an equation based on the problem statement: \(2y_{1} - 3y_{2} = 4y_{3} - 8\).
  • Next, we substitute the known values of \(y_{1}, y_{2},\) and \(y_{3}\) into this equation.
  • This substitution transformed our equation into a solvable format.
Once the equation was properly set up and values substituted, we used algebraic methods to solve for \(x\). It's important to carefully perform each operation to maintain the balance of the equation, ensuring the solution is correct.
Substitution Method
The substitution method is a powerful technique used to simplify and solve equations. It involves replacing variables in an equation with known values or expressions. Here鈥檚 how it was applied in our problem:
  • We identified the values of \(y_{1}, y_{2},\) and \(y_{3}\).
  • Then, we substituted \(y_{1} = 2.5\), \(y_{2} = 2x + 1\), and \(y_{3} = x\) into the main equation.
  • This substitution reduced the number of variables, making the equation easier to solve.
The main benefit of substitution is that it turns complex expressions into simpler ones, ultimately leading to the value of the variable we鈥檙e trying to find. This method is especially useful in systems of equations or when functions are given in terms of other variables.
Linear Equations
Linear equations are equations of the first degree, meaning they involve no exponents or powers higher than one. The standard form of a linear equation in one variable is \(ax + b = c\). In the original problem:
  • We ended up with a linear equation: \(-2x = -6\).
  • This was derived by combining like terms and moving all terms involving \(x\) to one side of the equation.
Solving linear equations involves straightforward arithmetic:
  • We need to isolate the variable \(x\).
  • This is achieved by performing the same operation on both sides of the equation.
  • By dividing both sides by \(-2\), we found \(x = 3\).
Linear equations are one of the simplest types of equations to solve, providing a straightforward pathway to finding the value of a variable.

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Most popular questions from this chapter

In Exercises \(115-122,\) find all values of \(x\) satisfying the given conditions. $$ \begin{aligned} &y_{1}=-x^{2}+4 x-2, y_{2}=-3 x^{2}+x-1, \text { and }\\\ &y_{1}-y_{2}=0 \end{aligned} $$

An isosceles right triangle has legs that are the same length and acute angles each measuring \(45^{\circ} .\) (GRAPH NOT COPY) a. Write an expression in terms of \(a\) that represents the length of the hypotenuse. b. Use your result from part (a) to write a sentence that describes the length of the hypotenuse of an isosceles right triangle in terms of the length of a leg.

In Exercises \(166-169\), determine whether each statement makes sense or does not make sense, and explain your reasoning. I obtained \(-17\) for the discriminant, so there are two imaginary irrational solutions.

Use the Pythagorean Theorem and the square root property to solve Exercises \(140-143 .\) Express answers in simplified radical form. Then find a decimal approximation to the nearest tenth. A baseball diamond is actually a square with 90 -foot sides. What is the distance from home plate to second base?

A tennis club offers two payment options. Members can pay a monthly fee of \(\$ 30\) plus \(\$ 5\) per hour for court rental time. The second option has no monthly fee, but court time costs \(\$ 7.50\) per hour. a. Write a mathematical model representing total monthly costs for each option for \(x\) hours of court rental time. b. Use a graphing utility to graph the two models in a \([0,15,1]\) by \([0,120,20]\) viewing rectangle. c. Use your utility's trace or intersection feature to determine where the two graphs intersect. Describe what the coordinates of this intersection point represent in practical terms. d. Verify part (c) using an algebraic approach by setting the two models equal to one another and determining how many hours one has to rent the court so that the two plans result in identical monthly costs.
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