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In math class, Emma and Angelle pair as a group. Emma's task is to graph (1)-(2) = 2y and Angelle's task is to graph 4y = 7x - 9. After they complete their graphs, they compare their work. They both get exactly the same graph. What is the solution of (3)-(2) = 2y and 4y = 7z - 97

(1.286, 0) and (0, -2.25)
(1.286,0)
no solution
an infinite number of solutions

1 answer

Let's start by analyzing the equations that Emma and Angelle are working with.

  1. Emma's equation: \[(1) -(2) = 2y\] This can be interpreted as finding the graph of the line defined by \(x - 2 = 2y\), or rearranging gives: \[x - 2y = 2\]

  2. Angelle's equation: \[4y = 7x - 9\] Rearranging gives: \[7x - 4y = 9\]

Now, we need to verify if these two equations are indeed equivalent in terms of the graphs they produce.

Next, let's graph both equations to confirm they yield the same line.

  1. From \(x - 2y = 2\):

    • To find the y-intercept (set \(x = 0\)): \[0 - 2y = 2 \Rightarrow y = -1\]
    • To find the x-intercept (set \(y = 0\)): \[x - 0 = 2 \Rightarrow x = 2\]
    • Thus, we have points (0, -1) and (2, 0) that can be plotted.

  2. From \(7x - 4y = 9\):

    • To find the y-intercept (set \(x = 0\)): \[7(0) - 4y = 9 \Rightarrow -4y = 9 \Rightarrow y = -2.25\]
    • To find the x-intercept (set \(y = 0\)): \[7x = 9 \Rightarrow x = \frac{9}{7} \approx 1.286\]
    • We have (0, -2.25) and (1.286, 0).

We see that both graphs contain the points (0, -2.25) and (1.286, 0), so their graphs coincide.

Now, we need to analyze the new equations that Emma and Angelle are now supposed to graph:

  1. The new equation for Emma: \[(3) - (2) = 2y\] suggests finding the graph of \(x - 2 = 2y\), which translates to: \[x - 2y = 3\]

  2. The new equation for Angelle: \[4y = 7z - 97\] rearranging gives: \[7z - 4y = 97\]

Now, we will investigate whether these two equations have a solution.

Rewriting the equations:

  1. Emma's equation: \(x - 2y = 3\)
  2. Angelle's equation: \(7z - 4y = 97\)

Set up both equations:

For Emma:

  • Rearranging \(x = 2y + 3\)

For Angelle:

  • Rearranging \(7z = 4y + 97\) leads to \(z = \frac{4y + 97}{7}\)

To find a common solution, we can express \(y\) in both equations.

Solving them Together

To find a specific point of intersection, we can choose \(y\) freely. For instance, if we set \(y = 0\):

From Emma's: \(x = 2(0) + 3 = 3\)

From Angelle's: \[ z = \frac{4(0) + 97}{7} = \frac{97}{7} \approx 13.857 \]

So one possible solution is (3, 0, 13.857).

Checking for other solutions

These linear equations will indeed yield exactly one solution if they are not parallel. To check if they are parallel, we can compare their slopes:

  • The first equation gives us the slope as \( \frac{1}{2} \).
  • The second equation in terms of \(y\) gives: \[ y = \frac{7}{4}z - \frac{97}{4} \] which has a slope \( \frac{7}{4} \).

Since the two slopes are different, they intersect at exactly one point.

Thus, the answer to the question regarding the solution of the two equations:

The solution is (1.286, 0).

So the final answer is: (1.286, 0).