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2 years ago

the system of equations are y=-2x=3 and y=x is graphed what is the soultions to the system of equations

Mathematics

2 answers:

bearhunter [10]2 years ago

6 0

Answer:

x=-1, y=-1

Step-by-step explanation:

Because -2x-3 and x are both equal to y, we set them equal to each other and solve for x:

-2x-3=x

-3=3x

-1=x

x=-1

So, because y=x, then y=-1 as well

kumpel [21]2 years ago

4 0

Answer:

y=1

Step-by-step explanation:

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nika2105 [10]

Answer:

You should have use this.

Step-by-step explanation:

Papa maths or domes

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3 years ago

On Monday, Jayla raised $30 and Carlos raised $50.

gayaneshka [121]

Answer:

84 days

Step-by-step explanation:

You would use the LCM to solve this.

12 and 7 don't share any common factors other than 1,

so you would multiply 12 x 7, which gets you 84.

After 84 days they would have raised the same amount.

8 0

3 years ago

The output is eleven more then the input

enot [183]

The input is known as x and the output is known as y in math.
So you’re stating that we’re making the equation equal the output, y, and we’re adding based off of the word “more”. Now that we know we’re adding, we have to find what we are adding together. We are adding x and 11 is that it equals y. Note that x and y could be many different variables and values.
Y= x+11

5 0

3 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

5 0

3 years ago

Read 2 more answers

Hey need help with some homework my teacher did not explain this very well so could you help me answer it and explain how iis do

Diano4ka-milaya [45]

y in (-oo:+oo)

(3*y+6)/11 = -9 // + 9

(3*y+6)/11+9 = 0

(3*y+6)/11+(9*11)/11 = 0

3*y+9*11+6 = 0

3*y+105 = 0

(3*y+105)/11 = 0

(3*y+105)/11 = 0 // * 11

3*y+105 = 0

3*y+105 = 0 // - 105

3*y = -105 // : 3

y = -105/3

y = -35

y = -35

7 0

3 years ago

Read 2 more answers