Problem 52
Question
For the linear function \(f(x)=m x+b, f(-3)=23\) and \(f(2)=-7 .\) Find \(m\) and \(b\).
Step-by-Step Solution
Verified Answer
The values of \(m\) and \(b\) are -30 and -67, respectively.
1Step 1: Setting up the equations
Start by plugging in the values of \(x\) into the equation to create two new equations. This gives: \\[\begin{align*} f(-3)=23 & \Rightarrow -3m+b=23 \\ f(2)=-7 & \Rightarrow 2m+b=-7 \end{align*}\\] So, the problem now reduces to finding \(m\) and \(b\) in this system of two equations.
2Step 2: Subtract the second equation from the first
By subtracting equation 2 from equation 1, the variable \(b\) will eliminate out. The resulting equation will only involve \(m\). Carry out the subtraction: \\[\begin{align*} (-3m+b)-(2m+b) &= 23-(-7) \\ -3m+2m&=23+7 \\ -m&=30 \end{align*}\\] Solving for \(m\), it is found that \(m=-30\).
3Step 3: Substituting \(m\) into the first equation
Substitute \(m=-30\) into equation 1 to find \(b\): \\[\begin{align*} -3m+b&=23 \\ -3(-30)+b&=23 \\ 90+b&=23 \\ b&=23-90 \end{align*}\\] Solving for \(b\), it is found that \(b=-67\).
Key Concepts
Solving Linear EquationsSystems of Linear EquationsAlgebraic ManipulationFinding Slope and Y-intercept
Solving Linear Equations
Understanding how to solve linear equations is crucial for tackling a variety of problems in algebra. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. When solving a linear equation, the goal is to find the value of the unknown variable that makes the equation true.
For instance, when given the linear function in the form of \( f(x) = mx + b \), and two values of the function, such as \( f(-3) = 23 \) and \( f(2) = -7 \), we aim to determine the values of \( m \) (slope) and \( b \) (y-intercept). The solution begins by substituting the x-values and f(x)-values into the linear equation to create two new equations based on the given function values. These equations form the foundation upon which the values of \( m \) and \( b \) can be derived through further steps that include algebraic manipulation.
For instance, when given the linear function in the form of \( f(x) = mx + b \), and two values of the function, such as \( f(-3) = 23 \) and \( f(2) = -7 \), we aim to determine the values of \( m \) (slope) and \( b \) (y-intercept). The solution begins by substituting the x-values and f(x)-values into the linear equation to create two new equations based on the given function values. These equations form the foundation upon which the values of \( m \) and \( b \) can be derived through further steps that include algebraic manipulation.
Systems of Linear Equations
When you have two or more linear equations using the same variables, this is called a system of linear equations. Solving such a system means finding the values of the variables that satisfy all equations simultaneously. One common method of solving systems of linear equations is by elimination or substitution.
As in the problem from the exercise, we first obtain two equations from the linear function by plugging in the given values. Then, by manipulating these equations, we aim to eliminate one of the unknowns. For instance, subtracting the second equation from the first one eliminates \( b \), allowing us to solve for \( m \). This method simplifies complex problems and is essential for understanding how different equations interact with each other.
As in the problem from the exercise, we first obtain two equations from the linear function by plugging in the given values. Then, by manipulating these equations, we aim to eliminate one of the unknowns. For instance, subtracting the second equation from the first one eliminates \( b \), allowing us to solve for \( m \). This method simplifies complex problems and is essential for understanding how different equations interact with each other.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying equations to solve for an unknown. It requires a good grasp of algebraic rules, such as the distributive property, combining like terms, and inverse operations.
In our example, after establishing two equations, we used these concepts to subtract one equation from the other. This subtraction, which also illustrates combining like terms, led to an equation with only one variable, \( m \), which we then solved. Understanding the techniques of algebraic manipulation is integral to solving not only linear equations but a wide range of mathematical problems.
In our example, after establishing two equations, we used these concepts to subtract one equation from the other. This subtraction, which also illustrates combining like terms, led to an equation with only one variable, \( m \), which we then solved. Understanding the techniques of algebraic manipulation is integral to solving not only linear equations but a wide range of mathematical problems.
Finding Slope and Y-intercept
Every linear function can be expressed in slope-intercept form, which is \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. The slope measures the steepness of the line and is calculated as the rise over the run between any two points on the line. In contrast, the y-intercept is the point where the line crosses the y-axis.
In the given exercise, once the value for \( m \) is found using algebraic manipulation, one of the original equations is used to solve for \( b \), the y-intercept. This involves substituting the slope back into either of the original equations and solving for \( b \). Mastery of finding the slope and y-intercept empowers students to graph linear functions and understand the graphical implications of these parameters.
In the given exercise, once the value for \( m \) is found using algebraic manipulation, one of the original equations is used to solve for \( b \), the y-intercept. This involves substituting the slope back into either of the original equations and solving for \( b \). Mastery of finding the slope and y-intercept empowers students to graph linear functions and understand the graphical implications of these parameters.
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Problem 51
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