Problem 6
Question
Determine whether the equation is in slope-intercept form. $$ x+23 y=-15 $$
Step-by-Step Solution
Verified Answer
The equation in slope-intercept form is \(y = -\frac{1}{23}x - \frac{15}{23}\).
1Step 1: Identify the given equation
The given equation is \(x+23y=-15\).
2Step 2: Rearrange the equation into slope-intercept form
To get the equation into slope-intercept form, isolate the term involving \(y\) on one side of the equation. This involves subtracting \(x\) from both sides of the equation, which results in \(23y = -x - 15\).
3Step 3: Simplify the equation
Finally, divide every term by 23 to solve for \(y\). This gives the equation in slope-intercept form: \(y = -\frac{1}{23}x - \frac{15}{23}\).
Key Concepts
Linear EquationsAlgebraic ManipulationEquation Rearrangement
Linear Equations
Linear equations are foundational in algebra and represent straight lines on a graph. They consist of variables and constants, and the highest power of the variable is never greater than one. In its simplest form, a linear equation in two variables, like 'x' and 'y', can be expressed as \(y = mx + b\), where 'm' represents the slope of the line and 'b' represents the y-intercept, which is the point where the line crosses the y-axis. The slope 'm' defines the steepness and the direction of the line: a positive slope means the line goes upwards from left to right, while a negative slope means it goes downwards. When the equation does not look like the standard form, algebraic manipulation is often necessary to rearrange it.
Algebraic Manipulation
Algebraic manipulation involves using mathematical operations to reshape equations without changing their solutions or the relationships between variables. This skill is paramount in solving and graphing linear equations. Reaching the slope-intercept form from any linear equation requires such manipulations, which can include adding, subtracting, multiplying, or dividing both sides of the equation by the same number. Operations also have to 'undo' each other properly; for example, if there is a term subtracting from 'y', you add it to both sides to keep the equation balanced.
- Addition/Subtraction: Used to move terms from one side of the equation to the other.
- Multiplication/Division: Used to isolate the variable and solve for it.
Equation Rearrangement
Equation rearrangement is the process of shaping an equation into a more useful or recognizable form. For linear equations, one common goal is to rearrange them into the slope-intercept form. This form is highly useful for graphing the equation swiftly and understanding its characteristics with ease. The slope-intercept form makes it instantly clear what the slope and y-intercept of the line are.
In the provided exercise, rearrangement required subtracting 'x' and then dividing every term by 23 to isolate 'y'. This alters the form of the equation but retains the same graph and solutions. The final form, \(y = -\frac{1}{23}x - \frac{15}{23}\), is in the desired slope-intercept format and clearly shows the slope as \( -\frac{1}{23} \) and the y-intercept as \( -\frac{15}{23} \), thus fully preparing us to graph the line or understand its behavior in relation to the coordinate system.
In the provided exercise, rearrangement required subtracting 'x' and then dividing every term by 23 to isolate 'y'. This alters the form of the equation but retains the same graph and solutions. The final form, \(y = -\frac{1}{23}x - \frac{15}{23}\), is in the desired slope-intercept format and clearly shows the slope as \( -\frac{1}{23} \) and the y-intercept as \( -\frac{15}{23} \), thus fully preparing us to graph the line or understand its behavior in relation to the coordinate system.
Other exercises in this chapter
Problem 6
Determine whether the lines are perpendicular. $$ y=3, x=4 $$
View solution Problem 6
Write in standard form an equation of the line that passes through the given point and has the given slope. Use integer coefficients. \((3,4), m=-4\)
View solution Problem 6
Write in slope-intercept form the equation of the line that passes through the given points. $$ (-1,1) \text { and }(2,5) $$
View solution Problem 7
In Exercises \(7-11\), a mountain climber is scaling a 400 -foot cliff. The climber starts at the bottom at \(t=0\) and climbs at a constant rate of 124 feet pe
View solution