Which graph represents the solution to the given system?
-4x+3y= -12 and -2x+3y=-18
again, the canned answer is wrong
this cannot be answered, since no graph is shown
however, the first line has a slope of 4/5 and a y-intercept of -4
the second would have a slope of 2/3 and a y-intercept of -6
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Anonymous
Nov 3, 2021
#lien your heart out
Oh boy, let me put on my graphing wig! Alright, so let's solve this system.
First, let's rewrite the equations in slope-intercept form (y = mx + b).
For -4x + 3y = -12, we'll isolate y:
3y = 4x - 12
y = (4/3)x - 4
For -2x + 3y = -18, isolating y gives us:
3y = 2x - 18
y = (2/3)x - 6
Now, let's plot these lines on a graph.
(Drawing a graph)
Okie dokie, we have two lines. Now, which graph represents the solution? Well, since the lines have different slopes, they are not parallel, which means they will intersect at a single point.
So the solution to this system is a single point of intersection.
On the graph, if you see two lines intersecting at a point, that's your solution! Voila!
To find the graph that represents the solution to the given system of equations, we need to solve the system first. Let's start by using the method of elimination:
Given system:
-4x + 3y = -12 (Equation 1)
-2x + 3y = -18 (Equation 2)
To eliminate the y variable, we can multiply Equation 2 by 2:
-4x + 3y = -12 (Equation 1)
-4x + 6y = -36 (Equation 3) (Multiplying Equation 2 by 2)
Now, we can subtract Equation 1 from Equation 3 to eliminate the x variable:
(-4x + 6y) - (-4x + 3y) = (-36) - (-12)
-4x + 6y + 4x - 3y = -36 + 12
3y = -24
Divide both sides of the equation by 3:
y = -8
Now that we have the value of y, we can substitute it back into either Equation 1 or Equation 2. Let's use Equation 1:
-4x + 3(-8) = -12
-4x - 24 = -12
-4x = -12 + 24
-4x = 12
x = -3
Therefore, the solution to the system is x = -3 and y = -8. To graph this solution, plot the point (-3, -8) on a coordinate plane.
The answer is that the graph representing the solution to the given system is a single point (-3, -8) on the coordinate plane.