Problem 70
Question
Use intercepts to graph each equation. \(3 x+5 y+15-0\)
Step-by-Step Solution
Verified Answer
The x-intercept is (-5,0) and the y-intercept is (0, -3). The graph is a straight line passing through these intercepts.
1Step 1: Find the x-intercept
The x-intercept is found by setting y=0 and then solving for x. In the given equation, \(3 x+5 (0)+15=0\), simplifies to \(3x = -15\). So, \(x = -15 / 3\) which is \(x = -5\). The x-intercept is thus, (-5,0).
2Step 2: Find the y-intercept
The y-intercept is found by setting x=0 and then solving for y. In the given equation, \(3 (0)+5y+15=0\), simplifies to \(5y = -15\). So, \(y = -15 / 5\) which is \(y = -3\). The y-intercept is thus, (0,-3).
3Step 3: Graph the line
Plot these intercepts on the graph and then draw a line that passes through these two points. This line represents the solution of the given linear equation.
Key Concepts
Understanding the X-InterceptUnderstanding the Y-InterceptExploring Linear Equations
Understanding the X-Intercept
The x-intercept of a linear equation is the point where the line crosses the x-axis on a graph. At this point, the value of y is always 0. To find the x-intercept of an equation, you set y to 0 and solve for x.
For example, in the equation given in the exercise, we have:
For example, in the equation given in the exercise, we have:
- Start with the equation: \(3x + 5y + 15 = 0\).
- Set \(y = 0\): \(3x + 5(0) + 15 = 0\).
- This simplifies to \(3x + 15 = 0\).
- Solving for x gives \(3x = -15\).
- Divide both sides by 3 to find \(x = -5\).
Understanding the Y-Intercept
The y-intercept of a linear equation is the point where the line crosses the y-axis. At the y-intercept, the value of x is always 0. To find the y-intercept, you set x to 0 and then solve for y.
Here's how it works using the equation from the exercise as an example:
Here's how it works using the equation from the exercise as an example:
- Start with the equation: \(3x + 5y + 15 = 0\).
- Set \(x = 0\): \(3(0) + 5y + 15 = 0\).
- This simplifies to \(5y + 15 = 0\).
- Solve for y: \(5y = -15\).
- Divide both sides by 5 to find \(y = -3\).
Exploring Linear Equations
Linear equations represent straight lines on a graph, and they come in the form \(Ax + By + C = 0\). These equations are about finding a relationship between two variables, typically x and y.
To graph a linear equation, you need at least two points to draw the line. This is where intercepts become very useful:
Linear equations are foundational in algebra as they describe direct relationships between variables. This means when one variable changes, the other changes at a consistent rate. Understanding how to work with linear equations is crucial for solving many real-world problems, such as calculating distances, budgeting expenses, and predicting trends.
To graph a linear equation, you need at least two points to draw the line. This is where intercepts become very useful:
- The x-intercept gives one point on the graph where y is zero.
- The y-intercept provides another point where x is zero.
Linear equations are foundational in algebra as they describe direct relationships between variables. This means when one variable changes, the other changes at a consistent rate. Understanding how to work with linear equations is crucial for solving many real-world problems, such as calculating distances, budgeting expenses, and predicting trends.
Other exercises in this chapter
Problem 70
Graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations. $$
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Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$ for the given function. $$f(x)=-3 x^{2}+x-1$$
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Find a. \((f \circ g)(x)\) b. the domain of \(f^{\circ}\) g. $$ f(x)-\frac{x}{x+5}, g(x)=\frac{6}{x} $$
View solution Problem 71
Describe how to find the inverse of a one-to-one function.
View solution