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Flura [38]
3 years ago
10

Determine the slope and y-intercept of the line represented by the linear equation: 1/2y + 2 = 0. Graph the line using the slope

and y-intercept. Describe, in complete sentences, what type of linear relationship the solutions (points) of the line have with each other. Choose two points on the line and prove that they are solutions of the linear equation. Complete your work in the space provided or upload a file that can display math symbols if your work requires it.

Mathematics
1 answer:
yarga [219]3 years ago
5 0

Answer:

1) Attached please find the graph of the equation 1/2·y + 2 = 0

2) The values of the y-coordinates for the points are the same, while the values of the x-coordinate of the points increases as we move from left to right

3) The result of substituting the x values of two points on the line in the equation 1/2·y + 2 = 0, gives y = -4, which proves that they are solutions of the linear equation

Step-by-step explanation:

1) The given equation is 1/2·y + 2 = 0

Therefore, the given equation in slope and intercept form, y = m·x + c, can be presented as follows;

1/2·y + 2 = 0  

1/2·y = 0 - 2 = -2

y = -2 × 2 = -4

y = -4

Therefore, by comparing the above equation with the equation for a straight line, y = m·x + c, we have;

m = 0, c = -4

Therefore, the graph is an horizontal line with y-intercept at (0, -4)

2) The relationship of the solutions (points) on of the line have with each other are;

a) The values of the y-coordinates for the points are the same, while the values of the x-coordinate of the points increases as we move from left to right

3) Two points on the line are the points (3, -4) and (33, -4)

With the slope = 0, the linear equation is given as follows;

y = 0·x - 4

For the point (3, -4) we have;

y = 0×3 - 4 = -4

y = -4

For the point (33, -4) we have;

y = 0×33 - 4 = -4

y = -4

Therefore, the two points are solutions of the linear equation 1/2·y + 2 = 0, where y = -4.

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X-y=6<br> 2x-3z=16<br> 2y+z=4<br><br> Solve the systems of equations.
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Answer:

Given System of equation:

x-y =6                                  .....,[1]

2x-3z = 16                           ......[2]  

2y+z = 4                              .......[3]

Rewrite the equation [1] as

y = x - 6                                .......[4]

Substitute the value of [4] in [3], we get

2(x-6)+z = 4

Using distributive property on LHS ( i.e,  a \cdot (b+c) =a \cdot b+ b \cdot c )

then, we have

2x - 12 +z =4

Add 12 to both sides of an equation:

2x-12+z+12=4+12

Simplify:

2x +z = 16                         .......[5]

On substituting equation [2] in [5] we get;

2x+z=2x -3z

or

z = -3z

Add 3z both sides of an equation:

z+3z = -3z+3z

4z = 0

Simplify:

z = 0

Substitute the value of z = 0 in [2] to solve for x;

2x-3(0) = 16

or

2x = 16

Divide by 2 both sides of an equation:

\frac{2x}{2} =\frac{16}{2}

Simplify:

x= 8

Substitute the value of x =8 in equation [4] to solve for y;

y = 8-6 = 2

or

y = 2

Therefore, the solution for the given system of equation is;  x = 8 , y = 2 and z =0

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Step-by-step explanation:

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