Problem 5 - Basic Engineering Mathematics

Question

Without plotting graphs, determine the gradient and \(y\)-axis intercept values of the following equations: (a) \(y=7 x-3\) (b) \(3 y=-6 x+2\) (c) \(y-2=4 x+9\) (d) \(\frac{y}{3}=\frac{x}{3}-\frac{1}{5}\) (e) \(2 x+9 y+1=0\)

Step-by-Step Solution

Answer

(a) Slope: 7, y-intercept: -3; (b) Slope: -2, y-intercept: 2/3; (c) Slope: 4, y-intercept: 11; (d) Slope: 1, y-intercept: -3/5; (e) Slope: -2/9, y-intercept: -1/9.

Step 1: Identify Slope-Intercept Form

The slope-intercept form of a line is given by the equation \(y = mx + b\), where \(m\) represents the slope (gradient) and \(b\) represents the y-intercept. We will convert each equation into this form to identify the slope and y-intercept.

Step 2: Solve Part (a)

The equation given is \(y = 7x - 3\). This is in the slope-intercept form \(y = mx + b\), where \(m = 7\) and \(b = -3\). Thus, the slope is 7 and the y-intercept is -3.

Step 3: Solve Part (b)

The equation given is \(3y = -6x + 2\). Divide each term by 3 to put it in slope-intercept form: \(y = -2x + \frac{2}{3}\). The slope \(m = -2\) and the y-intercept \(b = \frac{2}{3}\).

Step 4: Solve Part (c)

The equation is \(y - 2 = 4x + 9\). Add 2 to both sides to isolate \(y\): \(y = 4x + 11\). The slope \(m = 4\) and the y-intercept \(b = 11\).

Step 5: Solve Part (d)

The equation given is \(\frac{y}{3} = \frac{x}{3} - \frac{1}{5}\). Multiply through by 3: \(y = x - \frac{3}{5}\). The slope \(m = 1\) and the y-intercept \(b = -\frac{3}{5}\).

Step 6: Solve Part (e)

The equation given is \(2x + 9y + 1 = 0\). Rearrange to express \(y\) explicitly: \(9y = -2x - 1\). Divide through by 9: \(y = -\frac{2}{9}x - \frac{1}{9}\). The slope \(m = -\frac{2}{9}\) and the y-intercept \(b = -\frac{1}{9}\).

Key Concepts

Linear Equations

Linear equations are a type of algebraic equation where the highest power of the variable is one. These equations describe a straight line when graphed on a coordinate plane. The standard form of a linear equation is usually expressed as \(y = mx + b\), where:

• \(y\) represents the dependent variable
• \(x\) represents the independent variable
• \(m\) is the slope or gradient
• \(b\) is the y-intercept

In the context of a real-world situation, linear equations can model relationships where there is a constant rate of change between two quantities. This makes them extremely useful for predicting outcomes and making decisions.

Gradient

The gradient, or slope, of a line is a measure of its steepness. It indicates how much the \(y\) value changes for a given change in the \(x\) value. In slope-intercept form, \(y = mx + b\), \(m\) represents the gradient.

• If \(m > 0\), the line slopes upwards.
• If \(m < 0\), the line slopes downwards.
• If \(m = 0\), the line is horizontal, indicating no change in \(y\) as \(x\) changes.

The gradient is calculated as \(m = \frac{rise}{run} = \frac{\Delta y}{\Delta x}\). It reflects how rapidly or slowly the \(y\) value changes as \(x\) increases or decreases.

Y-Intercept

The y-intercept is a critical point where a line crosses the y-axis. In a linear equation of the form \(y = mx + b\), \(b\) represents the y-intercept.This point is significant because it shows where the line intersects when the \(x\) value is zero. For example, in the equation \(y = 7x - 3\), the line meets the y-axis at -3. This means, when \(x = 0\), \(y = -3\).Understanding the y-intercept allows you to easily sketch or understand the position of the line on a graph, giving insight into initial values or starting points in real-world contexts.

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. It is a method of representing real-world problems in a form that can be solved more easily through mathematical processes. In the context of linear equations, algebra involves:

• Solving equations to find the value of unknown variables
• Re-arranging equations to solve for a particular variable
• Using properties of equality to transform equations into simpler forms

Working with algebra allows students to develop strong problem-solving skills, which are applicable to a wide range of practical situations beyond mathematics.