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Chapter 4: Problem 10

Solve each system of linear equations by graphing. See Examples 3 through 6 $$ \left\\{\begin{array}{l} {x+y=3} \\ {x-y=5} \end{array}\right. $$

Short Answer

Expert verified
The solution is the point \( (4, -1) \).

Step by step solution

01

Convert Equations to Slope-Intercept Form

To graph, first convert each equation into slope-intercept form, \( y = mx + b \).For the first equation, \( x + y = 3 \):Subtract \( x \) from both sides to get: \( y = -x + 3 \).For the second equation, \( x - y = 5 \):Add \( y \) to both sides and subtract 5 from both sides to get: \( y = x - 5 \).
02

Graph the First Line

Now, graph \( y = -x + 3 \):1. Start at the y-intercept \( (0, 3) \).2. Use the slope \( -1 \) (which is \( -1/1 \)) to move 1 unit down and 1 unit right.3. Plot another point by moving down one and right one, then draw the line through the points.
03

Graph the Second Line

Now, graph \( y = x - 5 \):1. Start at the y-intercept \( (0, -5) \).2. Use the slope \( 1 \) (which is \( 1/1 \)) to move 1 unit up and 1 unit right.3. Plot another point by moving up one and right one, then draw the line through the points.
04

Find the Intersection

Locate the point where the two lines intersect on the graph. This is the solution to the system of equations. Upon graphing, we find the intersection occurs at the point \( (4, -1) \).

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Key Concepts

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

Graphing Linear Equations
Graphing linear equations is a powerful visual technique that helps you understand the relationships between variables in a system of equations. On a coordinate plane, each linear equation represents a straight line.
By plotting these lines, you can easily see how they interact with one another.Here's how you graph a linear equation step-by-step:
  • Convert the equation to slope-intercept form, if it isn鈥檛 already. This makes it easy to identify the slope and y-intercept.
  • Identify the y-intercept, which is where the line crosses the y-axis.
  • From the y-intercept, use the slope to find the next point on the line. For a slope expressed as a fraction \( \frac{rise}{run} \), the "rise" indicates how many units you move up or down, and the "run" indicates how many units you move to the right.
  • Draw a straight line through these points.
Once both equations are graphed, look for the point where they intersect, as this point is key to solving the system of equations graphically.
Slope-Intercept Form
The slope-intercept form of a linear equation is one of the most common and useful formats. It is expressed as \( y = mx + b \). This form highlights two critical aspects of the line: the slope \( m \) and the y-intercept \( b \).
  • The slope \( m \) measures the steepness of the line. A positive slope means the line goes up as you move right, and a negative slope indicates the line goes down.
  • The y-intercept \( b \) is where the line crosses the y-axis. It's the value of \( y \) when \( x \) is zero.
This format makes it straightforward to graph the line and predict any point's coordinates on that line efficiently. Converting equations into this form is often the first step when tackling systems of linear equations graphically.
Intersection Point
The intersection point is crucial when dealing with systems of linear equations. It represents the solution to the system, showing that the x and y values satisfy both equations simultaneously.
Finding this point graphically involves plotting both equations on the same set of axes. Once both lines are drawn, the intersection point becomes evident, as it's where the lines cross each other.To ensure accuracy when identifying the intersection point:
  • Carefully draw the lines based on the slope and y-intercept of each equation.
  • Use graph paper or a digital graphing tool for precision.
  • The coordinates of the intersection give you the solution in the form \( (x, y) \).
In cases where lines do not intersect (are parallel), the system has no solution. If the lines coincide, the system has infinite solutions, as every point is an intersection.
Solving Algebraically
Beyond graphing, systems of equations can also be solved algebraically. This is particularly useful when precise values are needed and the graphing approach lacks accuracy due to limitations in scale or tools.
Two popular methods to solve algebraically include:
  • Substitution: Solve one equation for one variable and substitute this expression in the other equation. Simplify to find the values of both variables.
  • Elimination: Add or subtract equations to eliminate one variable, making it possible to solve for the other. Then, substitute back to find the eliminated variable.
The algebraic solution provides an exact answer and confirms the graphical solution by giving the precise coordinates of the intersection point.

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Most popular questions from this chapter

The annual U.S. per capita consumption of whole milk has decreased since 1980 , while the per capita consumption of lower fat milk has increased. For the years \(1980-2005\), the function \(y=-0.40 x+15.9\) approximates the annual U.S. per capita consumption of whole milk in gallons, and the function \(y=0.14 x+11.9\) approximates the annual U.S. per capita consumption of lower fat milk in gallons. Determine the year in which the per capita consumption of whole milk equaled the per capita consumption of lower fat milk. (Source: Economic Research Service: U.S.D.A.) (IMAGE CANNOT COPY)

Solve each system by substitution. When necessary, round answers to the nearest hundredth. \(\left\\{\begin{array}{l}{y=5.1 x+14.56} \\ {y=-2 x-3.9}\end{array}\right.\)

Solve each system of linear equations by graphing. See Examples 3 through 6 $$ \left\\{\begin{array}{l} {x=3} \\ {y=-1} \end{array}\right. $$

Macadamia nuts cost an astounding 16.50 dollars per pound, but research by an independent firm says that mixed nuts sell better if macadamias are included. The standard mix costs 9.25 dollars per pound. Find how many pounds of macadamias and how many pounds of the standard mix should be combined to produce 40 pounds that will cost 10 dollars per pound. Find the amounts to the nearest tenth of a pound.

Elise Everly is preparing 15 liters of a \(25 \%\) saline solution. Elise has two other saline solutions with strengths of \(40 \%\) and \(10 \% .\) Find the amount of \(40 \%\) solution and the amount of \(10 \%\) solution she should mix to get 15 liters of a \(25 \%\) solution. $$ \begin{array}{|c|c|c|} \hline \text { Concentration } & {\text {Liters of }} & {\text {Liters of }} \\\ {\text { Rate }} & {\text { Solution }} & {\text { Pure Salt }} \\ \hline 0.40 & {x} & {0.40 x} \\ \hline 0.10 & {y} & {?} \\ \hline 0.25 & {15} & {?} \\ \hline \end{array} $$
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