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Short Answer

Expert verified
The solution to the given equation \( 3x - 5 = 10 \) is \( x = 5 \).

Step by step solution

01

Add 5 to both sides of the equation.

By adding 5 to both sides of the equation, we can remove the constant term (-5) from the left side of the equation. This results in: \[ 3x - 5 + 5 = 10 + 5 \]
02

Simplify the equation.

After adding 5 to both sides of the equation, we can simplify the equation as follows: \[ 3x = 15 \]
03

Divide both sides of the equation by 3.

To isolate x, we need to remove the coefficient (3) on the left side of the equation. To do this, we divide both sides of the equation by 3: \[ \frac{3x}{3} = \frac{15}{3} \]
04

Simplify the equation and find the value of x.

After dividing both sides of the equation by 3, we simplify the equation to find the value of x: \[ x = 5 \] So the solution to the given equation \( 3x - 5 = 10 \) is \( x = 5 \) .

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Key Concepts

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

Understanding Algebra
Algebra is a branch of mathematics that deals with using symbols and letters to represent numbers and quantities in formulas and equations. It allows us to model real-world problems and abstract concepts mathematically. The ability to work with variables gives us a lot of flexibility.
When we use letters such as \( x \) or \( y \), these symbols become placeholders that can take on different values. In the given exercise, for example, the variable \( x \) represents a number that, once determined, makes the equation true.
Algebra is useful in various fields such as science, engineering, and technology because it allows us to solve equations and make predictions about unknown quantities.
  • Key components include variables, coefficients, and constants.
  • Expressions are combinations of variables, numbers, and operations.
  • Equations set two expressions equal to each other and include an equality sign \((=)\).
Solving Linear Equations
Solving linear equations involves finding the value of a variable that makes the equation true. Linear equations in one variable, like the one in the exercise \(3x - 5 = 10\), have a single solution.
The goal is to isolate the variable, \(x\), on one side of the equation. This process requires a systematic series of steps that use inverse operations to undo the operations applied to the variable.
The method often involves:
  • Adding or subtracting terms on both sides to remove constants from the side of the variable. Example: Adding 5 to both sides: \(3x - 5 + 5 = 10 + 5\).
  • Simplifying the equation to make it more manageable. Example: Simply the previous step to get: \(3x = 15\).
  • Dividing or multiplying to isolate the variable and find its specific value. Example: Divide both sides by 3: \(\frac{3x}{3} = \frac{15}{3}\).
Each step applies a basic arithmetic operation that progressively makes the equation simpler. This logical sequence guarantees that the original equation remains balanced, ultimately leading to the discovery of the solution, \(x = 5\).
Mathematical Operations in Equation Solving
Mathematical operations are fundamental tasks of arithmetic — addition, subtraction, multiplication, and division — that are used to manipulate equations. These operations allow us to change the appearance of equations without altering their equality, enabling us to solve for unknown variables effectively.
In the exercise, the operations were used specifically to simplify and solve the equation \(3x - 5 = 10\):
  • Addition: To counteract subtraction on the variable's side, we added 5 to both sides to eliminate \(-5\), leading to \(3x = 15\).
  • Division: By dividing both sides by 3, we counteracted multiplication, resulting in \(x = 5\).
Understanding how to appropriately apply these operations is crucial as they enable one to transition between different forms of an equation while keeping it balanced. Each operation has an inverse — addition with subtraction, multiplication with division — which makes it possible to "reverse" or undo operations in solving equations. This is a key technique in algebraic problem-solving that simplifies complex equations.

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Most popular questions from this chapter

Derive an equation analogous to the Henderson-Hasselbalch equation but relating pOH and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\)

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5} )\) and 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the pH after 20.0\(\%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the \(\mathrm{pH} :\) $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ d. Do your answers in parts \(b\) and \(c\) agree? Explain.

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 \(\mathrm{mL}\) of the weak acid solution has been added to 50.0 \(\mathrm{mL}\) of the 0.100 \(\mathrm{M}\) NaOH solution, the \(\mathrm{pH}\) of the resulting solution is 10.50 . Calculate the original concentration of the solution of weak acid.

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) with 0.100 \(\mathrm{M}\) \(\mathrm{HCl} .\)

Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.

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