Circle D is below what is the arc measure of BC in degrees
1 answer:
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Answer: For the system given, there is no solution.
The slopes are the same for both equations, so the lines are parallel. With no intersection, there is no solution.
It may be easier to graph the system if you change the equations to slope intercept form y = mx + b
That locates the y-intercept, <em><u>b</u></em> and you can use the slope, <em><u>m</u></em> to plot additional points to draw the lines.
The attachment shows the graph of this system of equations.
Step-by-step explanation: Isolate y on the left
<u>First equation:</u>
4x + 6y = -12 Subtract 4x from both sides
6y = -4x -12 Divide both sides by 6
y = -4x/6 -12/6 Simplify
y = -2x/3 - 2 The y-intercept is -2, <u>the slope is -2/3</u>
<u>Second equation:</u>
2x + 3y = 6 Subtract 2x from both sides
3y = –2x + 6 . Divide both sides by 3
y = –2x/3 + 6/3 Simplify
y = –2x/3 +2 The y-intercept is -2, <u>the slope is -2/3 </u>
At this point we see that the slopes are the same, so the lines are parallel.
