Question

Solve each of these equations for \(x\) : (a) \(3.42 x=6.5\) (b) \(\frac{x}{12.3}=7.05\) (c) \(\frac{0.525}{x}=0.25\)

Short answer

Equation (a): \( x ≈ 1.90 \), Equation (b): \( x ≈ 86.715 \), Equation (c): \( x ≈ 2.1 \)

Step-by-step solution

Isolate the variable for equation (a)

Given the equation is \(3.42x = 6.5\). To isolate \(x\), divide both sides of the equation by 3.42. \[ x = \frac{6.5}{3.42} \]

Calculate the value of x for equation (a)

Perform the division to find \(x\). Using a calculator: \[ x ≈ 1.90 \]

Isolate the variable for equation (b)

Given the equation is \(\frac{x}{12.3} = 7.05\). To isolate \(x\), multiply both sides of the equation by 12.3. \[ x = 7.05 \times 12.3 \]

Calculate the value of x for equation (b)

Perform the multiplication to find \(x\). Using a calculator: \[ x ≈ 86.715 \]

Isolate the variable for equation (c)

Given the equation is \(\frac{0.525}{x} = 0.25\). To isolate \(x\), multiply both sides of the equation by \(x\) and then divide by 0.25. \[ 0.525 = 0.25x \] \[ x = \frac{0.525}{0.25} \]

Calculate the value of x for equation (c)

Perform the division to find \(x\). Using a calculator: \[ x ≈ 2.1 \]

Key concepts

algebra

Algebra is the branch of mathematics that involves variables and equations. When solving linear equations, your goal is to find the value of the unknown variable, often referred to as 'x'. Let's explore how algebra helps us to solve linear equations through a series of operations. These operations include addition, subtraction, multiplication, and division of both sides of the equation.
For example, consider the equation from the exercise:

(a) \(3.42x = 6.5\)

In algebra, to isolate 'x', we need to perform an operation that undoes the multiplication by 3.42. Thus, dividing both sides by 3.42 isolates 'x', giving us the solution:
\(x = \frac{6.5}{3.42} \ x ≈ 1.90\).

Understanding these basic operations and principles of algebra is key to solving any linear equation effectively.

isolation of variables

The isolation of variables is a fundamental concept in algebra. The idea is to manipulate the equation in such a way that the variable of interest stands alone on one side of the equation. Let’s delve into how we can achieve this.

For example, consider equation (b) from the exercise:

\(\frac{x}{12.3} = 7.05\)

To isolate 'x', we perform the inverse operation of division, which is multiplication. Multiplying both sides of the equation by 12.3 gives:
\(x = 7.05 \times 12.3\ x ≈ 86.715\).

By following the principle of inverse operations, you can isolate the variable in any linear equation. Tumultuous expressions become manageable as you perform these systematic operations.
Isolating the variable simplifies the equation, making it easier to find the value of the variable.

step-by-step solution

A step-by-step solution is a comprehensive method to solve equations. It involves breaking down the problem into smaller, more manageable steps. Each step leads you closer to the solution. Let's apply these steps to the equations from the exercise.

In equation (c):\(\frac{0.525}{x} = 0.25\)

We first want to remove 'x' from the denominator. To do this, multiply both sides by 'x':
\(0.525 = 0.25x\)

Next, isolate 'x' by dividing both sides by 0.25:
\(x = \frac{0.525}{0.25} \ \ \ \ x ≈ 2.1\).

Breaking the problem into these smaller steps ensures clarity and accuracy in the solution process. Whether dealing with straightforward or complex equations, a step-by-step approach makes the problem-solving procedure more accessible and comprehensible.