Problem 27
Question
In Exercises 27-36, solve the system by graphing. $$ \left\\{\begin{aligned} x+7 y &=-5 \\ 3 x-2 y &=8 \end{aligned}\right. $$
Step-by-Step Solution
Verified Answer
The solution would be the coordinates of the point of intersection of the two plotted lines.
1Step 1: Convert both equations to slope-intercept form
First, we need to rewrite each equation into slope-intercept form. \[y = mx + b\]For \(x+7y=-5\), by subtracting 'x' from both sides, and dividing by 7, we get: \[y = -\frac{1}{7}x - \frac{5}{7}\]For \(3x-2y=8\), by subtracting '3x' from both sides, and then dividing by -2, we get: \[y = \frac{3}{2}x - 4\]
2Step 2: Graph the equations
Using the slope-intercept forms obtained in the previous step, graph each line on the same graph. For \(y=-\frac{1}{7}x - \frac{5}{7}\), plot the y-intercept at -5/7, and then use the slope (-1/7) to find additional points. For \(y=\frac{3}{2}x - 4\), plot the y-intercept at -4, and then use the slope (3/2) to find additional points.
3Step 3: Determine the solution
The solution to the system is the point where the two graphs intersect. If the graphs intersect at a point, then find the (x, y) coordinates. It would be the solution to the system.
Key Concepts
Graphing Linear EquationsSlope-Intercept FormSolving Equations by GraphingIntersection of Lines
Graphing Linear Equations
Graphing linear equations is like translating a math problem into a visual picture. When you graph a line, you're plotting points on a grid where the line passes through. Each linear equation produces a straight line when graphed. To graph any linear equation, you usually start by finding several points that satisfy the equation and then plot those points on a coordinate plane. Connect the dots, and ta-da! You have a line.
- Each point on the line is a solution to the equation.
- Graphs can help visualize and solve systems of equations.
Slope-Intercept Form
The slope-intercept form is one of the most common ways to express a linear equation. It looks like this: \[ y = mx + b \] where \( m \) is the slope and \( b \) is the y-intercept. Understanding this form allows you to quickly sketch a graph. Let's break it down:
- The "slope" \( m \) represents how steep the line is. It tells you how much \( y \) changes for a change in \( x \). A positive slope goes up, and a negative slope goes down. For example, a slope of \(-\frac{1}{7}\) means the line gradually goes down.
- The "y-intercept" \( b \) is where your line crosses the y-axis. It's like the starting point on the y-axis. In our example, one line intercepts at \(-\frac{5}{7}\), so it starts just below zero.
Solving Equations by Graphing
Solving equations by graphing involves plotting both equations on the same coordinate grid and looking for their point of intersection. This method can be visually satisfying as you directly see the solution.
- First, you need both equations in slope-intercept form to easily identify slopes and y-intercepts.
- Graph each line carefully: start at the y-intercept and use the slope to find as many points as needed. For instance, from a y-intercept of \(-\frac{5}{7}\) and a slope of \(-\frac{1}{7}\), you can plot a new point by going down 1 unit and right 7 units.
- Finally, observe where the lines meet. The intersection point gives the solution \((x, y)\) of your system of equations.
Intersection of Lines
The intersection of lines is a crucial concept when solving systems of linear equations graphically.
- When two lines intersect, the point of intersection represents a common solution for the equations.
- If the lines cross at exactly one point, the system has a unique solution. In our exercise, after graphing, you would look for this shared meeting point.
- It's also possible for lines to be parallel (no intersection, hence no solution) or coincident (the same line overlapping, infinite solutions).
Other exercises in this chapter
Problem 27
In Exercises 25-28, solve the system by the method of elimination. $$ \left\\{\begin{array}{l} 0.4 a+0.7 b=3 \\ 0.8 a+1.4 b=7 \end{array}\right. $$
View solution Problem 27
A total of \(\$ 15,000\) is invested in two funds paying \(5 \%\) and \(8 \%\) simple interest. (There is more risk in the \(8 \%\) fund.) The combined annual i
View solution Problem 28
In Exercises 23-28, sketch the graph of the system of linear inequalities. $$ \left\\{\begin{aligned} 4 x-y & \leq 13 \\ -x+2 y & \leq 22 \\ x & \geq 0 \\ y & \
View solution Problem 28
In Exercises 25-28, solve the system by the method of elimination. $$ \left\\{\begin{array}{r} 0.2 u-0.1 v=1 \\ -0.8 u+0.4 v=3 \end{array}\right. $$
View solution