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Find all values of x satisfying the given conditions. $$y_{1}=\frac{x-3}{5}, y_{2}=\frac{x-5}{4}, \text { and } y_{1}-y_{2}=1$$

Short Answer

Expert verified
After performing the steps, the value of \(x\) that satisfies all the conditions is \(x = -7\)

Step by step solution

01

Express \(y_{1}\) and \(y_{2}\) in terms of \(x\)

It is given that \(y_{1}=\frac{x-3}{5}\) and \(y_{2}=\frac{x-5}{4}\)
02

Substituting \(y_{1}\) and \(y_{2}\) into the equation \(y_{1}-y_{2}=1\)

Substitute \(y_{1}\) and \(y_{2}\) from first and second equations into \(y_{1}-y_{2}=1\). Therefore, \(\frac{x-3}{5}-\frac{x-5}{4}=1\)
03

Solve the equation for \(x\)

This results into a linear equation in one variable. To solve for \(x\), must first multiply all terms by the least common multiple of the denominators, which is 20, giving: \[20*\frac{x-3}{5} - 20*\frac{x-5}{4} = 20 \nonumber\] This simplifies to: \[4x - 12 - 5x + 25 = 20 \nonumber\] Combine like terms: \[-x + 13 = 20 \nonumber\] Subtract 13 from both sides to isolate x: \[-x = 20 - 13 \nonumber\] Divide through by -1 to find \(x = - (20 - 13) = -7\]

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Key Concepts

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

Linear Equation in One Variable
Imagine you're on a treasure hunt, following a mystery map. The clues lead you to a mysterious algebraic puzzle that needs solving. This puzzle is akin to a linear equation in one variable. It's a math sentence that explains how two things are equivalent, with an unknown quantity, typically represented by a variable like 'x' or 'y'. Like our treasure map, the goal is to uncover the unknown, which is the value of 'x'.

In our exercise, we are given an equation where the unknown variable 'x' only appears once, and it's not tangled up with any exponents or complicated functions. This means our puzzle has just one possible 'x' that solves it. To find this treasure, we simply perform operations that isolate 'x' on one side of the equation. It's like having a set of scales that need to be balanced; what you do to one side, you do to the other, and voila! You've found your 'x', and possibly the location of the hidden treasure.
Least Common Multiple
On our quest to master the linear equations, we come across a little trickster known as fractions. To tackle them, we must call upon a math magician – the least common multiple (LCM). Remember when you were a kid and tried to share candy equally? The LCM is like finding the best way to divide your treats so everyone gets the same amount.

The LCM of two or more numbers is the smallest number that is a multiple of each of them. It allows us to add, subtract, or compare fractions by giving them a common denominator, thus making it easier to perform operations with them. In the context of our linear equation, using the LCM to eliminate fractions is as satisfying as having a neat stack of candies, all the same size, ready for sharing. It simplifies our equation, making 'x' stand out like the last piece of chocolate – precious and clear!
Algebraic Expression Simplification
The final element of our quest is like the art of decluttering a room: algebraic expression simplification. It's all about transforming a crowded, messy expression into its most streamlined and efficient form. Picture this: you've got coins scattered all over the floor – pennies, nickels, dimes, and quarters. Simplifying is like grouping all those coins together by type, so they're easier to count.

In algebraic terms, simplification might involve combining like terms, reducing fractions, or clearing parentheses. By doing this, we tidy up the equation, and the solution becomes as clear as day. Like with our coins, we group the variables and constants separately; then adjust and simplify until we see the neat total – the single value of 'x'. In our exercise, we combine terms with 'x' and the numbers by themselves to unwrap the mystery – revealing the exact value of 'x', which is the key to unlocking the secrets of algebra!

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Most popular questions from this chapter

In Exercises \(103-104,\) use the graph of \(y=|4-x|\) to solve each inequality. $$ |4-x|<5 $$

In Exercises 122–133, use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. A company manufactures and sells blank audio cassette tapes. The weekly fixed cost is 10.000 dollar and it costs 0.40 dollar to produce each tape. The selling price is 2.00 dollar per tape. How many tapes must be produced and sold each week for the company to generate a profit?

A bank offers two checking account plans. Plan A has a base service charge of 4.00 dollar per month plus 10¢ per check. Plan B charges a base service charge of $2.00 per month plus 15¢ per check. a. Write models for the total monthly costs for each plan if x checks are written. b. Use a graphing utility to graph the models in the same [0, 50, 10] by [0, 10, 1] viewing rectangle. c. Use the graphs (and the intersection feature) to determine for what number of checks per month plan A will be better than plan B. d. Verify the result of part (c) algebraically by solving an inequality.

Explaining the Concepts. When graphing the solutions of an inequality, what does a parenthesis signify? What does a square bracket signify?

An isosceles right triangle has legs that are the same length and acute angles each measuring \(45^{\circ} .\) 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.

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