Question

Solve the following equations for \(x\). (a) \(14 x+16=44\) (b) \(\frac{1}{3} x+3=5\) (c) \(12 x=4(12-x)\) (d) \(3 x+15=4 x+12\)

Short answer

The solutions to the equations are: (a) \(x = 2\), (b) \(x = 6\), (c) \(x = 3\), and (d) \(x = 3\).

Step-by-step solution

Solving Equation (a)

Subtract 16 from both sides of the equation \(14x + 16 = 44\) to get \(14x = 44 - 16 = 28\). Then divide each side by 14 to solve for \(x\), giving \(x = 28 / 14 = 2\).

Solving Equation (b)

Subtract 3 from both sides of the equation \(\frac{1}{3}x + 3 = 5\) to get \(\frac{1}{3}x = 5 - 3 = 2\). Then multiply each side by 3 to solve for \(x\), giving \(x = 2 * 3 = 6\).

Solving Equation (c)

Expand the equation \(12x = 4(12 - x)\) to get \(12x = 48 - 4x\). Add \(4x\) to both sides of the equation to get \(16x = 48\). Then divide each side by 16 to solve for \(x\), which gives \(x = 48 / 16 = 3\).

Solving Equation (d)

Subtract \(3x\) from both sides of the equation \(3x + 15 = 4x + 12\) to get \(15 = x + 12\). Then subtract 12 from both sides to solve for \(x\), giving \(x = 15 - 12 = 3\).

Key concepts

Solving Equations

Solving equations is all about finding the value of the variable that makes the equation true. When we solve equations, we perform a series of mathematical operations to isolate the variable on one side of the equation. This means we are trying to get the variable by itself.
This process might involve:

• Adding or subtracting numbers on both sides.
• Multiplying or dividing numbers on both sides.
• Combining like terms, if necessary.

The key point is to maintain the balance of the equation. For instance, when solving the equation \(14x + 16 = 44\), we subtract 16 from both sides to keep the equation balanced, resulting in \(14x = 28\). Then, dividing both sides by 14 gives \(x = 2\).
Remember, whatever operation you do on one side, you must do the same on the other.This step-by-step process ensures that you maintain a valid equation all the way to the solution.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to make them easier to solve. This might include expanding expressions, as well as adding, subtracting, or dividing terms.
Let’s take the equation \(12x = 4(12 - x)\). First, we expand the right side: \(4 \times 12\) and \(4 \times -x\), which gives us \(48 - 4x\).
Then, we rearrange the terms by adding \(4x\) to both sides to consolidate the x-terms: \(12x + 4x = 48\), simplifying to \(16x = 48\). Finally, dividing both sides by 16 isolates \(x\), resulting in \(x = 3\).
These techniques require practice, but they are powerful tools for simplifying complex equations into something more manageable.

Equations in One Variable

Equations in one variable are equations where we solve for a single unknown quantity, usually denoted as \(x\). They appear in the form of linear equations like those in the exercise, where all terms involve just \(x\) and constant numbers.
This simplicity makes them a great starting point for studying algebra. For example, take the equation \(3x + 15 = 4x + 12\). Here, we are working with a single variable, \(x\). Our goal is to find its specific value.
First, subtract \(3x\) from both sides to have \(15 = x + 12\). Next, subtract 12 from both sides to isolate \(x\), leading to \(x = 3\). These equations help form the foundation for understanding more complex mathematical concepts in algebra.