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Step2247 [10]
3 years ago
11

Sole by substitution. -3x+2y=16 X-2y=-12

Mathematics
2 answers:
leva [86]3 years ago
7 0

Answer:

For the given equations the values of x and y by using substitution method are  x=1  and y=\frac{13}{2}

Step-by-step explanation:

Given system of equations are

-3x+2y=16\hfill (1)

x-2y=-12\hfill (2)

Solve the given equations by substitution method:

Now (2) implies

x-2y=-12

x=-12+2y\hfill (3)

Now substitute the x value in equation (1)

-3x+2y=16

-3(-12+2y)+2y=16

36-6y+2y=16

36-4y=16

-4y=16-36

y=-(\frac{-26}{4})

y=\frac{26}{4}

Therefore  y=\frac{13}{2}

Now substitute the y value in equation (3)

x=-12+2y

x=-12+2\times \frac{13}{2}

x=-12+13

x=1

Therefore x=1  and y=\frac{13}{2}

bekas [8.4K]3 years ago
6 0
Answer: (-2,5)

Step 1: Rewrite equations

-3x+2y=16

x-2y=-12

Step 2: Changing the second equation

With substitution, you have to already have the equation to solve for the specific variable (ex: x=y+1;y=x+1). Since the equation doesn’t give us one, we can flip the second equation to give us this.

Let’s add 2y on both sides.

x-2y=-12
+2y +2y
_________
x=2y-12

Now we have the equation to solve for x!

Step 3: Substituting

Now we need to substitute x into the first equation. Let’s do this now.

-3(2y-12)+2y=16

Step 4: Solving for y

-3(2y-12)+2y=16

*distribute*

-6y+36+2y=16

*combine like terms*

-4y+36=16

*subtract 36 on both sides*

-4y=-20

*divide both numbers by -4*

y=5

This is y! Now we need to solve for x.

Step 5: Solving for x

To find x, let’s just substitute y into the equation and solve

x=2y-12

x=2(5)-12

x=10-12

x=-2

Step 6: Ordered pair

(X,y) —> (-2,5)


This is your answer! Hope this helps comment below for more questions :)
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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian
Alla [95]

Answer: mass (m) = 4 kg

              center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

<u>Points (0,0) and (2,1):</u>

y = \frac{1-0}{2-0}(x-0)

y = \frac{x}{2}

<u>Points (2,1) and (0,3):</u>

y = \frac{3-1}{0-2}(x-0) + 3

y = -x + 3

Now, find total mass, which is given by the formula:

m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }

Calculating for the limits above:

m = \int\limits^2_0 {\int\limits^a_\frac{x}{2}  {2(x+y)} \, dy \, dx  }

where a = -x+3

m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2}  {(xy+\frac{y^{2}}{2} )} \, dx  }

m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx  }

m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx  }

m = 2.(\frac{-3.2^{2}}{2}+3.2-0)

m = 2(-4+6)

m = 4

<u>Mass of the lamina that occupies region D is 4.</u>

<u />

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }

M_{y} = \int\limits^a_b {\int\limits^a_b {x.\rho(x,y)} \, dA }

M_{x} and M_{y} are moments of the lamina about x-axis and y-axis, respectively.

Calculating moments:

For moment about x-axis:

M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }

M_{x} = \int\limits^2_0 {\int\limits^a_\frac{x}{2}  {2.y.(x+y)} \, dy\, dx }

M_{x} = 2\int\limits^2_0 {\int\limits^a_\frac{x}{2}  {y.x+y^{2}} \, dy\, dx }

M_{x} = 2\int\limits^2_0 { ({\frac{y^{2}x}{2}+\frac{y^{3}}{3})}\, dx }

M_{x} = 2\int\limits^2_0 { ({\frac{x(-x+3)^{2}}{2}+\frac{(-x+3)^{3}}{3} -\frac{x^{3}}{8}-\frac{x^{3}}{24}  )}\, dx }

M_{x} = 2.(\frac{-9.x^{2}}{4}+9x)

M_{x} = 2.(\frac{-9.2^{2}}{4}+9.2)

M_{x} = 18

Now to find the x-coordinate:

x = \frac{M_{y}}{m}

x = \frac{63}{4}

x = 15.75

For moment about the y-axis:

M_{y} = \int\limits^2_0 {\int\limits^a_\frac{x}{2}  {2x.(x+y))} \, dy\,dx }

M_{y} = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2}  {x^{2}+yx} \, dy\,dx }

M_{y} = 2.\int\limits^2_0 {y.x^{2}+x.{\frac{y^{2}}{2} } } \,dx }

M_{y} = 2.\int\limits^2_0 {x^{2}.(-x+3)+\frac{x.(-x+3)^{2}}{2} - {\frac{x^{3}}{2}-\frac{x^{3}}{8}  } } \,dx }

M_{y} = 2.\int\limits^2_0 {\frac{-9x^3}{8}+\frac{9x}{2}   } \,dx }

M_{y} = 2.({\frac{-9x^4}{32}+9x^{2})

M_{y} = 2.({\frac{-9.2^4}{32}+9.2^{2}-0)

M{y} = 63

To find y-coordinate:

y = \frac{M_{x}}{m}

y = \frac{18}{4}

y = 4.5

<u>Center mass coordinates for the lamina are (15.75,4.5)</u>

3 0
3 years ago
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