/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 Think About It In Exercises 67 a... [FREE SOLUTION] | 91Ó°ÊÓ

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Think About It In Exercises 67 and \(68,\) the graphs of the two equations appear to be parallel. Yet, when you solve the system algebraically, you find that the system does have a solution. Find the solution and explain why it does not appear on the portion of the graph shown. $$ \left\\{\begin{array}{rr}{100 y-x=} & {200} \\ {99 y-x=} & {-198}\end{array}\right. $$

Short Answer

Expert verified
The solution to the system of equations is (400, 6).

Step by step solution

01

Rearrange the equations

The first step is to organize both equations to be in the form y = mx + b. The equations become: \(y = \frac{1}{100}x + 2\) and \(y = \frac{1}{99}x - 2\)
02

Set the two equations equal to each other

Since both expressions are equal to y, they can be set equal to each other to find the value of x. So, \(\frac{1}{100}x + 2 = \frac{1}{99}x - 2\).
03

Solve for x

Rearrange the equation to bring like terms together and solve for x. So, \((1/100 - 1/99)x = -4 \Rightarrow x = (99 - 100)(-400) = 400\).
04

Substitute x into one of the original equations

In order to find the corresponding y-coordinate, substitute this x-value into the first original equation: \(100y - 400 = 200 \Rightarrow y = 6\)

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Key Concepts

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

Parallel Lines
Parallel lines in the context of linear equations have the same slope but different intercepts when graphed on a coordinate plane. This is why they appear never to intersect and usually suggest that a system of equations is inconsistent, meaning it has no solution.
However, sometimes appearances can be deceptive. In the given exercise, the graphs of the equations might look parallel, but a deeper algebraic analysis reveals the true picture. The slight difference in slopes between the lines, even if visually negligible, can lead to them intersecting - albeit sometimes outside the visible section of the graph.
The misinterpretation usually stems from a certain segment of the graph being viewable, which gives the illusion of parallelism due to scale or viewing limitations. Thus, it is crucial to rely not just on graphical representations but also on algebraic computations to verify the nature of a set of linear equations.
Algebraic Solution
Finding an algebraic solution often involves transforming the given system of equations into a simpler form which can easily be solved. Here, the equations are restructured into the form of \(y = mx + b\) known as slope-intercept form.
This aids in setting the equations equal because now, both equations express \(y\) in terms of \(x\). When you solve \((\frac{1}{100}x + 2 = \frac{1}{99}x - 2)\), you work to isolate \(x\).
You combine like terms and solve for \(x\) by removing fractions typically through multiplication, leading to \(x = 400\). It's like peeling an onion of equations layer by layer until you find the core value of \(x\), which provides a solid solution.
After finding \(x\), the next task is to substitute it back into either of the original equations to find \(y\), completing the solution with \(x = 400\) and \(y = 6\).
Graphing Linear Equations
When graphing linear equations, the line represents all possible solutions of that equation, with its slope \(m\) indicating the angle of the line and \(b\) the y-intercept where it crosses the y-axis.
For the exercise, transforming the given equations into a form easier to interpret visually can often make understanding the solution clearer. This means converting equations into \(y = mx + b\) form, allowing for simple plotting and slope comparisons.
Here, note the differences in slopes: \(\frac{1}{100}\) vs \(\frac{1}{99}\). Even a small difference implies that the lines must eventually intersect if extended far enough. This is crucial because, in many graphing exercises, the domain of the graph might hide where intersections actually occur.
Therefore, graphing equations helps interpret solutions graphically, but it's essential to cross-verify with algebraic methods, ensuring all dimensions of the problem are understood. So always expand the view if necessary, as graphical limitations might obscure reality.

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Most popular questions from this chapter

The _____ function of a linear programming problem gives the quantity to be maximized or minimized.

Finding Minimum and Maximum Values, find the minimum and maximum values of the objective function and where they occur, subject to the constraints \(x \geq 0, y \geq 0, x+4 y \leq 20\) \(x+y \leq 18,\) and \(2 x+2 y \leq 21 .\) $$ z=2 x+4 y $$

Data Analysis: Wildlife A wildlife management team studied the reproductive rates of deer in three tracts of a wildlife preserve. Each tract contained 5 acres. In each tract, the number of females \(x ,\) and the percent of females \(y\) that had offspring the following year were recorded. The table shows the results.$$ \begin{array} { | c | c | c | c | } \hline \text { Number, } x & { 100 } & { 120 } & { 140 } \\ \hline \text { Percent, y } & { 75 } & { 68 } & { 55 } \\\ \hline \end{array} $$ (a) Use the data to create a system of linear equations. Then find the least squares regression parabola for the data by solving the system. (b) Use a graphing utility to graph the parabola and the data in the same viewing window. (c) Use the model to estimate the percent of females that had offspring when there were 170 females. (d) Use the model to estimate the number of females when 40\(\%\) of the females had offspring.

Advanced Applications In Exercises 69 and 70 , solve the system of equations for \(u\) and \(v .\) While solving for these variables, consider the transcendental functions as constants. (Systems of this type appear in a course in differential equations.) $$ \left\\{\begin{aligned} u \cos 2 x+& v \sin 2 x=& 0 \\ u(-2 \sin 2 x)+v(2 \cos 2 x) &=\csc 2 x \end{aligned}\right. $$

Fitting a Line to Data To find the least squares regression line \(y=a x+b\) for a set of points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) you can solve the following system for \(a\) and \(b\) . $$ \left\\{\begin{array}{c}{n b+\left(\sum_{i=1}^{n} x_{i}\right) a=\left(\sum_{i=1}^{n} y_{i}\right)} \\ {\left(\sum_{i=1}^{n} x_{i}\right) b+\left(\sum_{i=1}^{n} x_{i}^{2}\right) a=\left(\sum_{i=1}^{n} x_{i} y_{i}\right)}\end{array}\right. $$ In Exercises 55 and \(56,\) the sums have been evaluated. Solve the given system for \(a\) and \(b\) to find the least squares regression line for the points. Use a graphing utility to confirm the result. $$ \left\\{\begin{aligned} 5 b+10 a &=20.2 \\ 10 b+30 a &=50.1 \end{aligned}\right. $$

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