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Chapter 4: Problem 18

Solve each system of linear equations by graphing. See Examples 3 through \(6 .\) \(\left\\{\begin{array}{l}2 x+y=1 \\ 3 x+y=0\end{array}\right.\)

Short Answer

Expert verified
The solution to the system is \((-1, 3)\).

Step by step solution

01

Convert Equations to Slope-Intercept Form

To graph the equations, we need them in slope-intercept form, which is \(y = mx + b\). For the first equation \(2x + y = 1\), we rearrange to get \(y = -2x + 1\). For the second equation \(3x + y = 0\), we rearrange to get \(y = -3x\).
02

Graph the First Equation

Plot the line for the equation \(y = -2x + 1\). Start by plotting the y-intercept \((0, 1)\). From there, use the slope \(-2\), meaning go down 2 units and right 1 unit from the y-intercept to mark another point \( (1, -1) \). Draw a line through these points.
03

Graph the Second Equation

Plot the line for the equation \(y = -3x\). Start by plotting the y-intercept \((0, 0)\). From there, use the slope \(-3\), meaning go down 3 units and right 1 unit to mark another point \((1, -3)\). Draw a line through these points.
04

Identify the Point of Intersection

Find where the two lines intersect - this point is the solution to the system of equations. The lines intersect at \((-1, 3)\).
05

Verify the Solution

Substitute \((-1, 3)\) into the original equations to verify it works for both: - For \(2x + y = 1\): substitute to get \(2(-1) + 3 = 1\), which is true. - For \(3x + y = 0\): substitute to get \(3(-1) + 3 = 0\), which is also true.

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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 visual way to display relationships between variables. It involves plotting equations on a coordinate grid so that their solutions can be identified easily. Each equation corresponds to a straight line on this graph. To begin, you need to express each equation in slope-intercept form, which helps in easily identifying both the slope and the y-intercept. The y-intercept is the point where the line crosses the y-axis.
  • Select two key points - the y-intercept and another point determined by the slope.
  • Plot these points on the graph and draw a straight line through them.
With both equations graphed, the system of equations can be solved by locating the point where the two lines cross, known as the point of intersection.
Slope-Intercept Form
The slope-intercept form of a linear equation is written as: \[ y = mx + b \]where \( m \) represents the slope and \( b \) represents the y-intercept. This form is beneficial for quickly drawing graphs, as it provides a straightforward way to locate both the starting point and direction of the line.
  • The slope \( m \) describes the steepness of the line. It tells you how many units to move up or down for each unit you move to the right.
  • The y-intercept \( b \) is where the line crosses the y-axis.
By rearranging equations into this form, you make graphing more intuitive and less error-prone, facilitating the solution of the system of equations.
Point of Intersection
The point of intersection of two lines on a graph represents the solution to a system of linear equations. This is the precise point where both equations have the same values for \( x \) and \( y \).
  • To find it graphically, carefully plot both equations on the same set of axes.
  • Look for the point where the two lines intersect, which can often be seen as a clear crossing point.
In the example given, the lines intersect at \[ (-1, 3) \]indicating that this is the set of \( x \) and \( y \) values that satisfy both equations simultaneously. This graphical method provides a visual confirmation of the solution.
Verifying Solutions
Once you have a proposed solution from the graph, it is essential to verify it by plugging the point of intersection back into the original equations.
  • Substitute the \( x \) and \( y \) values into each equation separately.
  • Ensure that each equation holds true with these substituted values.
In our original example, substituting the point \[ (-1, 3) \]into both equations should yield true statements.
  • For the first equation: \[ 2(-1) + 3 = 1 \]which simplifies accurately.
  • For the second equation: \[ 3(-1) + 3 = 0 \]which confirms the correctness as well.
Verification is vital because it confirms that the intersection has been correctly identified and is not a graphical error.

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Most popular questions from this chapter

Solve each system of equations by the substitution method. $$ \left\\{\begin{array}{l} \frac{1}{3} x-y=2 \\ x-3 y=6 \end{array}\right. $$

Solve each system of equations by the addition method. If a system contains fractions or decimals, you may want to first clear each equation of fractions or decimals.$$ \left\\{\begin{array}{l} 3 x+y=4 \\ 9 x+3 y=6 \end{array}\right. $$

Solve each system by the addition method. $$ \left\\{\begin{array}{l} 2 x+3 y=14 \\ 3 x-4 y=-69.1 \end{array}\right. $$

Solve each system of equations by the addition method. If a system contains fractions or decimals, you may want to first clear each equation of fractions or decimals.$$ \left\\{\begin{array}{c} 10 x+3 y=-12 \\ 5 x=-4 y-16 \end{array}\right. $$

Solve each system of equations by the addition method. $$ \left\\{\begin{array}{l} x-2 y=8 \\ -x+5 y=-17 \end{array}\right. $$
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