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Complete the following tables using algebraic methods. a. \(y=2 x-10\) CAN'T COPY THE GRAPH b. \(y=20+0.5 x\) CAN'T COPY THE GRAPH c. \(y=-3 x+15\) CAN'T COPY THE GRAPH d. \(y=12-\frac{3}{4} x\) CAN'T COPY THE GRAPH

Short Answer

Expert verified
Complete tables: a) \((0,-10), (5,0), (10,10)\), b) \((0,20), (10,25), (20,30)\), c) \((0,15), (3,6), (5,0)\), d) \((0,12), (4,9), (8,6)\).

Step by step solution

01

Understand the Problem

We are given four linear equations, and we need to complete tables of values for each equation. This involves finding the corresponding values of \(y\) for different values of \(x\).
02

Identify Points for Equation a

For the equation \(y = 2x - 10\), choose values for \(x\). Common choices are \(x = 0\), \(x = 5\), and \(x = 10\). Calculate \(y\) for each of those \(x\) values.
03

Calculate Values for Equation a

Substitute each \(x\) value into \(y = 2x - 10\):- For \(x = 0\): \(y = 2(0) - 10 = -10\)- For \(x = 5\): \(y = 2(5) - 10 = 0\)- For \(x = 10\): \(y = 2(10) - 10 = 10\)Table for (a) is \( (0, -10), (5, 0), (10, 10) \).
04

Identify Points for Equation b

For the equation \(y = 20 + 0.5x\), choose values such as \(x = 0\), \(x = 10\), and \(x = 20\).
05

Calculate Values for Equation b

Plug each \(x\) into \(y = 20 + 0.5x\):- For \(x = 0\): \(y = 20 + 0.5(0) = 20\)- For \(x = 10\): \(y = 20 + 0.5(10) = 25\)- For \(x = 20\): \(y = 20 + 0.5(20) = 30\)Table for (b) is \( (0, 20), (10, 25), (20, 30) \).
06

Identify Points for Equation c

For \(y = -3x + 15\), select points like \(x = 0\), \(x = 3\), and \(x = 5\).
07

Calculate Values for Equation c

Insert each \(x\) into \(y = -3x + 15\):- For \(x = 0\): \(y = -3(0) + 15 = 15\)- For \(x = 3\): \(y = -3(3) + 15 = 6\)- For \(x = 5\): \(y = -3(5) + 15 = 0\)Table for (c) is \( (0, 15), (3, 6), (5, 0) \).
08

Identify Points for Equation d

For \(y = 12 - \frac{3}{4}x\), choose points such as \(x = 0\), \(x = 4\), and \(x = 8\).
09

Calculate Values for Equation d

Substitute \(x\) into \(y = 12 - \frac{3}{4}x\):- For \(x = 0\): \(y = 12 - \frac{3}{4}(0) = 12\)- For \(x = 4\): \(y = 12 - \frac{3}{4}(4) = 12 - 3 = 9\)- For \(x = 8\): \(y = 12 - \frac{3}{4}(8) = 12 - 6 = 6\)Table for (d) is \( (0, 12), (4, 9), (8, 6) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebraic Methods
Algebraic methods are essential for solving linear equations, which are equations of the first degree. These equations form a straight line when plotted on a graph. In the context of a linear equation like \( y = mx + b \), \( m \) represents the slope, and \( b \) is the y-intercept. The y-intercept is where the line intersects the y-axis when \( x = 0 \).

These methods involve substituting chosen values of \( x \) into the equation to solve for \( y \). This substitution helps find specific points that lie on the line. Each substitution reveals a pairing of \( x \) and \( y \) values that are solutions to the equation, like
  • for \( y = 2x - 10 \), inserting \( x = 5 \) results in \( y = 0 \).
This process is straightforward and only requires basic arithmetic operations. Algebraic methods are not only practical but also pivotal for understanding more complex mathematical concepts.
Table of Values
Creating a table of values is a simple and effective way to understand and solve linear equations. For any given equation, you choose a set of x-values. Then, you substitute these into the equation to calculate the corresponding y-values. This results in ordered pairs \((x, y)\).

For instance, using equation \( y = 20 + 0.5x \), you might choose the x-values 0, 10, and 20:
  • When \( x = 0 \), \( y = 20 \)
  • When \( x = 10 \), \( y = 25 \)
  • When \( x = 20 \), \( y = 30 \)
These pairs \((0, 20), (10, 25), (20, 30)\) visually represent points on the line of the equation. Such tables are essential as they provide a structured approach to exploring and understanding the linear relationship described by the equation, making complex concepts much more accessible.
Graphing Linear Equations
Graphing linear equations involves plotting points from a table of values on a coordinate plane and drawing a straight line through them. It's a great way to visualize algebraic relationships.

For example, in the equation \( y = -3x + 15 \), once you have calculated the pairs \((0, 15), (3, 6), (5, 0)\), you can plot these points on a graph:
  • Plots the point \((0, 15)\) on the y-axis
  • Plots the point \((3, 6)\) further right on the graph
  • Plots \((5, 0)\), which is the x-intercept
Align these points and connect them to form a straight line, representing all solutions to the equation. This visual method can help you quickly determine the range of solutions for different values of \( x \). It's excellent for developing an intuitive understanding of the equation's behavior.
Substitution Method
The substitution method in solving linear equations involves replacing one variable, typically \( y \), with an expression from another equation. When using this method, you solve one equation for a variable, then substitute this expression into another equation.

Consider the equation \( y = 12 - \frac{3}{4}x \). You might substitute chosen values like 0, 4, 8 for \( x \) to find \( y \). For example:
  • For \( x = 0 \): \( y = 12 \)
  • For \( x = 4 \): \( y = 9 \)
  • For \( x = 8 \): \( y = 6 \)
This method helps manage complex problems by reducing the number of variables and simplifying calculations. It enhances consistency and accuracy in deriving solutions for various equation forms.

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Most popular questions from this chapter

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