Problem 48 - Elementary Algebra

Question

Write the equation of the line that satisfies the given conditions. Express final equations in standard form. \(x\) intercept of 5 and slope of \(-\frac{3}{10}\)

Step-by-Step Solution

Answer

The standard form of the equation is \(3x + 10y = 15\).

Step 1: Identify the Point-Slope Form

To start, use the x-intercept and slope to find an equation of the line. Since the x-intercept is 5, you have the point (5, 0). The slope is given as \(-\frac{3}{10}\). The point-slope form of a line is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope.

Step 2: Substitute Given Values into Point-Slope Form

Substitute the known values into the point-slope form equation. With point (5, 0) and slope \(-\frac{3}{10}\), the equation becomes \(y - 0 = -\frac{3}{10}(x - 5)\). This simplifies to \(y = -\frac{3}{10}(x - 5)\).

Step 3: Expand the Equation

Distribute the slope \(-\frac{3}{10}\) across \(x - 5\). This gives you \(y = -\frac{3}{10}x + \frac{15}{10}\). Simplifying the constant term, the equation becomes \(y = -\frac{3}{10}x + \frac{3}{2}\).

Step 4: Convert to Standard Form

To express the equation in standard form, move all terms to one side, resulting in \(\frac{3}{10}x + y = \frac{3}{2}\). Multiply through by 10 to eliminate the fractions: \(3x + 10y = 15\). This is the equation in standard form \(Ax + By = C\) where \(A = 3\), \(B = 10\), and \(C = 15\).

Step 5: Verify Standard Form

Check that the final standard form equation \(3x + 10y = 15\) satisfies the original conditions. With a slope of \(-\frac{3}{10}\) and x-intercept of 5, substituting \(x = 5\) should yield \(y = 0\).

Key Concepts

x-intercept

The x-intercept is where a line crosses the x-axis, meaning the y-value is zero. It's a crucial part of a linear equation because it helps to graph the line and understand its behavior. For the given problem, the x-intercept is 5, which means the point (5, 0) lies on the line. This tells us that when x equals 5, y equals 0. To determine the x-intercept from an equation like the one derived in the step-by-step solution, you can set y to zero and solve for x. This is useful when verifying the x-intercept from a derived equation to ensure it matches your initial condition or to find the x-intercept from any linear equation. With linear equations, remember:

• The x-intercept shows where the line will hit the x-axis.
• It's used to derive the point-slope form of the equation.
• In practice, it's easy to visually spot on a graph where y is zero.

point-slope form

The point-slope form of a line is a way to write the equation of a line when you know a point on the line and the slope. The formula is given by \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope.In the problem, we have a point (5, 0) and a slope of \( -\frac{3}{10} \). Using this information, the point-slope form becomes \( y - 0 = -\frac{3}{10}(x - 5) \). This form is useful because it highlights the slope and a specific point, making transitions to other forms easier.Some key advantages include:

• It directly incorporates both a known point and the slope.
• From here, you can easily manipulate the equation to other forms like slope-intercept or standard form.

This form is particularly effective for constructing a line when starting with a specific known point and given slope, effectively guiding us step-by-step through the derivation process.

standard form

Standard form is a way of writing linear equations as \( Ax + By = C \), where \( A, B, \) and \( C \) are integers, and \( A \) should be a non-negative integer. Converting a line from point-slope or slope-intercept form to standard form typically involves algebraic manipulation to eliminate fractions and move all terms to one side of the equation.In this exercise, the line's equation was derived to end in standard form as \( 3x + 10y = 15 \). This transformation required removing fractions by multiplying through by 10, ensuring each term was adequately arranged to satisfy the conditions of standard form. The result is an easy-to-understand format that simplifies determining intercepts and evaluating points.Understanding standard form:

• It's best for calculation-oriented tasks, especially involving intersections of lines.
• Shows clear integer coefficients and an integer constant.
• Easily reveals the relationship between the variables by showing both x and y on the same side of the equation.

Recognizing the utility and conversion steps to standard form can greatly enhance the flexibility in handling linear equations, making it a valuable tool in many mathematical contexts.