How Do You Find the X-Coordinates of Points A and B on the Line y = (x/2) + 1?

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Discussion Overview

The discussion revolves around finding the x-coordinates of points A and B on the line defined by the equation y = (x/2) + 1. Participants explore methods for substituting given y-values into the equation to solve for the corresponding x-values, focusing on the mathematical reasoning behind the process.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant suggests substituting the y-values of points A and B into the linear equation to find the x-coordinates.
  • Another participant proposes rewriting the equation as a function of y to solve for x, then substituting the y-values.
  • A participant provides detailed calculations showing the steps to find x for both points A and B, arriving at A(1, 1.5) and B(3, 2.5).
  • Another participant presents the solution in a mathematical notation format, confirming the x-coordinates as x_A = 1 and x_B = 3.
  • One participant comments on the efficiency of solving for x directly from the equation.

Areas of Agreement / Disagreement

Participants generally agree on the method of substituting y-values to find x-coordinates, but there are variations in the approaches discussed. No explicit consensus on a single method is established.

Contextual Notes

The discussion assumes familiarity with algebraic manipulation and does not address potential limitations or alternative methods for solving the equation.

mathdad
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The equation y = (x/2) + 1 is given. It forms a straight line going through the points A(x, 1.5) and B(x, 2.5). Find the x-coordinates of points A and B.

Do I substitute the value of y for each point into the given equation and solve for x individually?
 
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Since we are given y's and asked for the corresponding x's, I would write the given linear equation as a function of y (solve for x), and then plug in the y's to get the corresponding x's. :D
 
Not too bad.
 
What do you get when you solve for x?
 
For A(x, 1.5) we have

1.5 = (x/2) + 1

1.5 - 1 = x/2

0.5 = x/2

0.5(2) = x

1 = x

A(1, 1.5)

For B(x, 2.5) we have

2.5 = (x/2) + 1

2.5 - 1 = x/2

1.5 = x/2

1.5(2) = x

3 = x

B(3, 2.5)
 
$$x(y)=2(y-1)$$

$$x_A=x\left(y_A\right)=2(1.5-1)=1$$

$$x_B=x\left(y_B\right)=2(2.5-1)=3$$
 
By solving the given equation for x, it is faster.
 

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