The first step in determining the solution to the systems of equations y=x^2-4x-3 and y=2x+5 algebraically is to set the two equations equal as -x2-4x-3=2x+5 what is the first step

x=1, y=8

Step-by-step explanation:

You start with the equations:

and

Then you say y = y, then

now you want to put all your numbers in one side of the equation and combine like terms in that side of the equation.

And now you have a quadratic equation, where you can determine the value of x using the Bhaskara method.

Then your solutions are x = -1 and x = -5.

Combine like terms onto one side of the equation

Combine like terms onto one side of the equation

Step-by-step explanation:

You have

-x^2  -4x  - 3  = 2x + 5

It seems that we should combine like terms to make this equations smaller.

-x^2 - 4x - 3  - 2x  - 5 = 0

-x^2  -6x  - 8 = 0

- (x + 2)(x + 4) = 0

x = -2  or  x  = -4

D

Step-by-step explanation:

Ok so for part a we're going to use substitution, since both equations have y alone.

Step 1: Use substitution

y = x - 1    y = -x + 3

x - 1 = -x + 3

If you think about it, this makes sense. You're looking for the point where they intersect, and you know that the y coordinate for that point will be the same in both equations. Therefore, we can set the rest of the equations equal to each other. Now we just need to simplify.

Step 2: add x to each side

x - 1 + x = -x + 3 + x

The x's on the right cancel out so we get:

2x - 1 = 3

Step 4: Add 1 to each side

2x - 1 + 1 = 3 + 1

2x = 4

Step 5: Divide both sides by 2

2x/2 = 4/2

x = 2

Step 6: Plug this x value back into one of the original equations to find the y coordinate

y = x - 1

y = 2 - 1

y = 1

Therefore your answer to part a is (2,1)

I've graphed the solution and attached it. It shouldn't be too hard to replicate - just draw both lines and they should intersect at (2,1)

Hope that helps! If you need clarification or anything please feel free to ask :)

Option 4 is correct that is Combine like terms onto one side of the equation.

Step-by-step explanation:

Given System of Equations are

y = -2x² - 4x - 3   and y = 2x + 5

To solve the above system of equation algebraically.

First step already taken which is Both equations set equal.

we get,

-2x² - 4x - 3 = 2x + 5

Now to further solve it we have to move term from either of side to other side or combine them onto one side of the equation.

we get,

-2x² - 4x - 3 - 2x - 5 = 0

Therefore, Option 4 is correct that is Combine like terms onto one side of the equation

Step-by-step explanation:

The answer would be D- Combine like terms onto one side of the equation.

The next step is to put all terms into the left side of the equation and group the like trems

Step-by-step explanation:

The first step in determining the solution to the system of equations, and , algebraically is to set the two equations equal as

The next step is to put all terms into the left side of the equation and group the like trems:

Now you can multiply this equation by -1:

and solve it using quadratic formula:

You can start by subtracting different equations from each other.

3x + 2y + 3z = 1

subtract

3x + 2y + z = 7

2z = -6

divide by 2

z = -3

add the following two expressions together:

3x + 2y + z = 7

3x + 2y + 3z =1

6x + 4y + 4z = 8

subtract the following two expressions:

6x + 4y + 4z = 8

5x + 5y + 4z = 3

x - y = 5

^multiply the whole equation above by 3

3x - 3y = 15

subtract the following two expressions:

3x - 3y = 15

3x + 2y = 10

-5y = 5

divide each side by -5

y=-1

take the following expression from earlier:

x - y = 5

substitute y value into above equation

x - - 1 = 5

2 negatives make a positive

x + 1 = 5

subtract 1 from each side

x = 4

Therefore x = 4, y = -1, z = -3

I checked these with all 3 equations and they worked :)

(it's quite complicated, comment if you don't understand anything) :)