What is the solution of [tex]$3 x+5=2 x-7 ?$[/tex]
A. the [tex]$x$[/tex]-coordinates of the intersection point of the lines [tex]$y=3 x+5$[/tex] and [tex]$y=2 x-7$[/tex]
B. the [tex]$x$[/tex]-coordinates of the [tex]$x$[/tex]-intercepts of the lines [tex]$y=3 x+5$[/tex] and [tex]$y=2 x-7$[/tex]
C. the [tex]$y$[/tex]-coordinate of the intersection point of the lines [tex]$y=3 x+5$[/tex] and [tex]$y=2 x-7$[/tex]
D. the [tex]$y$[/tex]-coordinates of the [tex]$y$[/tex]-intercepts of the lines [tex]$y=3 x+5$[/tex] and [tex]$y=2 x-7$[/tex]
Answer (1)
Solve the equation 3 x + 5 = 2 x − 7 to find x = − 12 .
Determine the x-intercepts by setting y = 0 for both lines, resulting in x = − 3 5 and x = 2 7 .
Calculate the y-coordinate of the intersection point by substituting x = − 12 into either line equation, yielding y = − 31 .
Find the y-intercepts by setting x = 0 for both lines, giving y = 5 and y = − 7 .
The solution to the equation is x = − 12 , the x-intercepts are x = − 3 5 and x = 2 7 , the y-coordinate of the intersection point is y = − 31 , and the y-intercepts are y = 5 and y = − 7 .
− 12
Explanation
Problem Analysis We are given the equation 3 x + 5 = 2 x − 7 . Our goal is to find the value of x that satisfies this equation. We also need to find the x and y coordinates of the intersection of the lines y = 3 x + 5 and y = 2 x − 7 , the x-intercepts of these lines, and the y-intercepts of these lines.
Solving for x To solve the equation 3 x + 5 = 2 x − 7 , we first subtract 2 x from both sides: 3 x − 2 x + 5 = 2 x − 2 x − 7 x + 5 = − 7 Next, we subtract 5 from both sides: x + 5 − 5 = − 7 − 5 x = − 12 So, the solution to the equation is x = − 12 .
Finding the Intersection Point's x-coordinate The x-coordinate of the intersection point of the lines y = 3 x + 5 and y = 2 x − 7 is the solution to the equation 3 x + 5 = 2 x − 7 , which we found to be x = − 12 .
Finding the x-intercept of y = 3x + 5 To find the x-intercept of the line y = 3 x + 5 , we set y = 0 and solve for x :
0 = 3 x + 5 Subtract 5 from both sides: − 5 = 3 x Divide by 3: x = 3 − 5 So, the x-intercept of the line y = 3 x + 5 is x = − 3 5 .
Finding the x-intercept of y = 2x - 7 To find the x-intercept of the line y = 2 x − 7 , we set y = 0 and solve for x :
0 = 2 x − 7 Add 7 to both sides: 7 = 2 x Divide by 2: x = 2 7 So, the x-intercept of the line y = 2 x − 7 is x = 2 7 .
Finding the Intersection Point's y-coordinate To find the y-coordinate of the intersection point, we substitute x = − 12 into either y = 3 x + 5 or y = 2 x − 7 . Let's use y = 3 x + 5 :
y = 3 ( − 12 ) + 5 y = − 36 + 5 y = − 31 So, the y-coordinate of the intersection point is y = − 31 .
Finding the y-intercept of y = 3x + 5 To find the y-intercept of the line y = 3 x + 5 , we set x = 0 :
y = 3 ( 0 ) + 5 y = 5 So, the y-intercept of the line y = 3 x + 5 is y = 5 .
Finding the y-intercept of y = 2x - 7 To find the y-intercept of the line y = 2 x − 7 , we set x = 0 :
y = 2 ( 0 ) − 7 y = − 7 So, the y-intercept of the line y = 2 x − 7 is y = − 7 .
Final Answer In summary:
The solution to the equation 3 x + 5 = 2 x − 7 is x = − 12 .
The x-coordinate of the intersection point of the lines y = 3 x + 5 and y = 2 x − 7 is x = − 12 .
The x-intercept of the line y = 3 x + 5 is x = − 3 5 .
The x-intercept of the line y = 2 x − 7 is x = 2 7 .
The y-coordinate of the intersection point of the lines y = 3 x + 5 and y = 2 x − 7 is y = − 31 .
The y-intercept of the line y = 3 x + 5 is y = 5 .
The y-intercept of the line y = 2 x − 7 is y = − 7 .
Examples
Understanding linear equations and their intersections is crucial in many real-world applications. For instance, in economics, the intersection of supply and demand curves determines the market equilibrium price and quantity. Similarly, in physics, analyzing the motion of objects often involves solving linear equations to find when and where objects meet. These concepts are also used in computer graphics to calculate intersections of lines and planes, which is essential for rendering 3D scenes. By mastering these fundamental algebraic skills, you'll be well-equipped to tackle a wide range of practical problems across various disciplines.
