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Sloan [31]
3 years ago
9

Both lines pass through the point (-3,-4). one has a slope of 4 and the other has a slope of -1/4

Mathematics
1 answer:
Alchen [17]3 years ago
3 0
If the question is to find the slope-intercept form of both lines, here's the answer:
Both lines pass through the point (-3,-4), so we can use these coordinates in both equations. The slope-intercept form is represented by y=mx+b, with m the slope, b the intersection of the line with Y'Y for x=0, y and x the coordinates of a point.
Let's first apply all these for the first line, with a slope of 4.
y = mx + b
y=-3; x=-4; m=4. All we need to do is find b.
-3 = 4(-4) + b
-3 = -16 + b
b=13
So the equation of the first line is y= 4x + 13.

Now, we'll do the same thing but for the second line:
y=-3; x=-4; m=-1/4, and we need to find b.
-3 = (-1/4)(-4) + b
-3 = 1 + b
b= -4
So the equation of the second line is y=(-1/4)x - 4

Hope this Helps! :)
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lisov135 [29]

Answer:

(a) The system of the equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. has no solution.

(b) The system of the equations \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. has many solutions y=\frac{2x}{3}-\frac{5}{3}

Step-by-step explanation:

(a) To find the solutions of the following system of equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. you must:

Multiply 2x-3y=3 by 2:

\begin{bmatrix}4x-6y=6\\ 4x-6y=3\end{bmatrix}

Subtract the equations

4x-6y=3\\-\\4x-6y=6\\------\\0=-3

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(b) To find the solutions of the system \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. you must:

Isolate x for 4x-6y=10

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Substitute x=\frac{5+3y}{2} into the second equation

16\cdot \frac{5+3y}{2}-24y=40\\8\left(3y+5\right)-24y=40\\24y+40-24y=40\\40=40

The system has many solutions.

Isolate y for 4x-6y=10

y=\frac{2x}{3}-\frac{5}{3}

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Answer:

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r=c/2(3.14)

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r=7

Step-by-step explanation:

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