In Exercises 61–66, find all values of x satisfying the given conditions. y1=x−35,y2=x−54,andy1−y2=1y_1 = \frac{x - 3}{5}, \quad $$y_2$$ = \frac{x - 5}{4}, \quad \text{and} \quad $$y_1$$ - $$y_2 = 1 y1$$=5x−3,y2=4x−5,andy1−y2=1

Ch. 1 - Equations and Inequalities

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Blitzer 8th Edition

Ch. 1 - Equations and Inequalities

Problem 63

Chapter 2, Problem 63

In Exercises 61–66, find all values of x satisfying the given conditions. $y_1 = $\frac{x - 3}{5}$, $\quad$ y_2 = $\frac{x - 5}{4}$, $\quad$ $\text{and}$ $\quad$ y_1 - y_2 = 1$

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Start by writing down the given equations: $ y_1 = \frac{x - 3}{5} $, $ y_2 = \frac{x - 5}{4} $, and the condition $ y_1 - y_2 = 1 $.

Substitute the expressions for $ y_1 $ and $ y_2 $ into the equation $ y_1 - y_2 = 1 $ to get: $ \frac{x - 3}{5} - \frac{x - 5}{4} = 1 $.

Find a common denominator for the fractions on the left side, which is 20, and rewrite the equation as: $ \frac{4(x - 3)}{20} - \frac{5(x - 5)}{20} = 1 $.

Combine the fractions over the common denominator: $ \frac{4(x - 3) - 5(x - 5)}{20} = 1 $.

Multiply both sides of the equation by 20 to eliminate the denominator, then simplify and solve the resulting linear equation for $ x $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Solving Linear Equations

Solving linear equations involves finding the value of the variable that makes the equation true. In this problem, you will combine expressions for y1 and y2 and solve for x by isolating the variable using algebraic operations like addition, subtraction, multiplication, or division.

Substitution Method

The substitution method involves replacing one variable with an equivalent expression from another equation. Here, y1 and y2 are given in terms of x, so you substitute these expressions into the equation y1 - y2 = 1 to form an equation with only x, simplifying the solving process.

Understanding Linear Functions

Linear functions are algebraic expressions where variables are to the first power and graph as straight lines. Recognizing that y1 and y2 are linear functions of x helps in setting up and solving equations involving their difference, as linearity ensures the operations and solutions remain straightforward.

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In Exercises 61–66, find all values of x satisfying the given conditions. y1=x−35,y2=x−54,andy1−y2=1y_1 = \(\frac{x - 3}{5}\), \(\quad\) y_2 = \(\frac{x - 5}{4}\), \(\quad\) \(\text{and}\) \(\quad\) y_1 - y_2 = 1y1=5x−3,y2=4x−5,andy1−y2=1

Verified video answer for a similar problem:

Key Concepts

Solving Linear Equations

Substitution Method

Understanding Linear Functions

In Exercises 61–66, find all values of x satisfying the given conditions. $y_1 = \(\frac{x - 3}{5}\), \(\quad\) y_2 = \(\frac{x - 5}{4}\), \(\quad\) \(\text{and}\) \(\quad\) y_1 - y_2 = 1$