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liubo4ka [24]
4 years ago
10

What is the value of y in the solution to the system of equations?

Mathematics
1 answer:
LUCKY_DIMON [66]4 years ago
8 0

To solve for y you have to simplify both sides of the equation, then isolating the variable which equals to...

y = -4x/3 + 4

Next, you have to move all terms that don't contain y to the right side and solve.

y = 10 + 2x/3

____

Now we solve the first variable in one of the equations, then substitute the result into the other equation.

Point Form: (-3,8)

Equation Form: x = -3, y = 8

____

Leaving it to be the only choice which is 'D.) 8', I hope this helps, as always. I wish you the best of luck and I hope you have a nice day..

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Mrs. Stevenson wants to split her class into 4 teams: Team A, Team B, Team C, and Team D. There are 28 students in her class. To
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The probability that both of the students are on Team D is about 5.56%

There are 4 teams, so the first person has a 1/4 chance of being on Team D.

However, for the second person, there are only 6 spots left out of 27 to be on Team D.

To find the combined probability, we need to multiply the two fractions.

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Step-by-step explanation:

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How many solutions does the system have?
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Answer:

y = 3/2 , x = 3/2

Step-by-step explanation:

Solve the following system:

{x + y = 3 | (equation 1)

10 x = 15 | (equation 2)

Divide equation 2 by 5:

{y + x = 3 | (equation 1)

0 y+2 x = 3 | (equation 2)

Divide equation 2 by 2:

{y + x = 3 | (equation 1)

0 y+x = 3/2 | (equation 2)

Subtract equation 2 from equation 1:

{y+0 x = 3/2 | (equation 1)

0 y+x = 3/2 | (equation 2)

Collect results:

Answer: {y = 3/2 , x = 3/2

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Answer/Step-by-step explanation:

 All of the solutions to the equation 3x^2 - 12 = 0 are x = 12 and x = -2

Answer: False

Explanation:

3x^2-12+12=0+12

3x^2=12

\frac{3x^2}{3}=\frac{12}{3}

x^2=4

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{4},\:x=-\sqrt{4}

x=2,\:x=-2

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There are two unique solutions to the equations (x-3)^2 = 16

Note: Each variable in the matrix can have only one possible value, and this is how you know that this matrix has one unique solution.

Answer: True

Explanation:

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=7,\:x=-1

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Ths solutions for the equation 2(x-3)^3 - 18 = 0 are x = 6 and x = 0

Answer: False

Explanation:

2\left(x-3\right)^3-18+18=0+18

2\left(x-3\right)^3=18

\frac{2\left(x-3\right)^3}{2}=\frac{18}{2}

\left(x-3\right)^3=9

\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}

=\sqrt[3]{9}+3,\:x=\frac{6-\sqrt[3]{9}}{2}+i\frac{\sqrt[3]{9}\sqrt{3}}{2},\:x=\frac{6-\sqrt[3]{9}}{2}-i\frac{\sqrt[3]{9}\sqrt{3}}{2}

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The solutions for the equation 2(x-5)^2-8=0 are x = 7 and x = -7

Answer: False

Explanation:

2\left(x-5\right)^2-8+8=0+8

2\left(x-5\right)^2=8

\frac{2\left(x-5\right)^2}{2}=\frac{8}{2}

\left(x-5\right)^2=4

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=7,\:x=3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation (x + 3)^2-25 = -8 are x = 2 and x = -8

Answer: False

Explanation:

\left(x+3\right)^2-25+25=-8+25

\left(x+3\right)^2=17

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{17}-3,\:x=-\sqrt{17}-3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 2(2x-1)^2=18 are x = 5 and x = -4

Answer: False

Explanation:

\frac{2\left(2x-1\right)^2}{2}=\frac{18}{2}

\left(2x-1\right)^2=9

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=2,x=-1

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The only solution for equation (2x-1)^2-49=0 is x = 4

Answer: False
Explanation:

\left(2x-1\right)^2-49+49=0+49

\left(2x-1\right)^2=49

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=4,\:x=-3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 3(x+2)^2 - 3 = 0 are x = -3 and x = -1

Answer: True
Explanation:

3\left(x+2\right)^2-3+3=0+3

3\left(x+2\right)^2=3

\frac{3\left(x+2\right)^2}{3}=\frac{3}{3}

\left(x+2\right)^2=1

\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=-1,\:x=-3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 5x^2 - 180 = 0 are x = 6 and x  = -6

Answer: True

Explanation:

5x^2-180+180=0+180

5x^2=180

\frac{5x^2}{5}=\frac{180}{5}

x^2=36

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{36},\:x=-\sqrt{36}

x=6,\:x=-6

<u><em>~Lenvy~</em></u>

7 0
3 years ago
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