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labwork [276]
3 years ago
8

Solve these simultaneous equations:

Mathematics
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

1) We have the system:

5*x - 3*y = 15

4*x + 3*y = 6

To solve this, we first need to isolate one of the variables in one of the equations, let's isolate x in the first equation:

x = 15/5 + (3/5)*y = 3 + (3/5)*y

Now we can replace this in the other equation to get:

4*( 3 + (3/5)*y) + 3*y = 6

and solve this for y.

12 + (12/5)*y  + 3*y = 6

(12/5 + 3)*y = 6 - 12 = -6

(12/5 + 15/5)*y = -6

(27/5)*y = -6

y = -6*(5/27) = 1.11

Now we can replace this in the equation:

x = 3 + (3/5)*y

To get the value of x.

x = 3 + (3/5)*1.11 = 3.67

Then the solution of this system is the point (3.67, 1.11)

2) Now we have the system:

2*x + 5*y = 26

4*x + 3*y = 24

The solution method is the same as before:

x = 26/2 - (5/2)*y = 13 - (5/2)*y

Now we replace this in the other equation:

4*( 13 - (5/2)*y) + 3*y = 24

52 - 10*y + 3*y = 24

52 - 7*y = 24

52 - 24 = 7*y

28 = 7*y

28/7 = y

4 = y

now we replace this in the equatio:

x = 13 - (5/2)*y

x = 13 - (5/2)*4 = 13 - 10 = 3

The solution of this sytem is (3, 4)

3) Now we have the system:

3*x + 3*y = 39

2*x - 3*y = -2

first we isolate x in the first equation:

x = 39/3 - 3*y/3 = 13 - y

Now we can replace this in the other equation:

2*(13 - y) - 3*y = -2

26 - 2*y - 3*y = -2

26 - 5*y = -2

26 + 2 = 5*y

28 = 5*y

28/5 = y = 5.6

Now we can replace this in the equation:

x = 13 - y

To get the x-value

x = 13 - 5.6 = 7.4

Then the solution for this system is (7.4, 5.6)

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Answer:

(A)Segment EF, segment FG, segment GH, and segment EH are congruent

Step-by-step explanation:

<u>Step 1</u>

Quadrilateral EFGH with points E(-2,3), F(1,6), G(4,3), H(1,0)

<u>Step 2</u>

Using the distance formula

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

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Given G(4,3), H(1,0)

|GH|=\sqrt{(0-3)^2+(1-4)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt{2}

Given E (−2, 3), H (1, 0)

|EH|=\sqrt{(0-3)^2+(1-(-2))^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{18}=3\sqrt{2}

<u>Step 3</u>

Segment EF ,E (−2, 3), F (1, 6)

Slope of |EF|=\frac{6-3}{1+2} =\frac{3}{3}=1

Segment GH, G (4, 3), H (1, 0)

Slope of |GH|= \frac{0-3}{1-4} =\frac{-3}{-3}=1

<u>Step 4</u>

Segment EH, E(−2, 3), H (1, 0)

Slope of |EH|= \frac{0-3}{1+2} =\frac{-3}{3}=-1

Segment FG, F (1, 6,) G (4, 3)

Slope of |EH| =\frac{3-6}{4-1} =\frac{-3}{3}=-1

<u>Step 5</u>

Segment EF and segment GH are perpendicular to segment FG.

The slope of segment EF and segment GH is 1. The slope of segment FG is −1.

<u>Step 6</u>

<u>Segment EF, segment FG, segment GH, and segment EH are congruent. </u>

The slope of segment FG and segment EH is −1. The slope of segment GH is 1.

<u>Step 7</u>

All sides are congruent, opposite sides are parallel, and adjacent sides are perpendicular. Quadrilateral EFGH is a square

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