Question

2. Describe the solution to this system of equation below. Justify your thinking.

3. Look at the system of equation.Solve the system algebraically. Show all of your work.

Answer 1

To solve a system of equations algebraically, identify the unknowns and known values, choose an equation, plug in the known values, and solve for the unknowns.

Step-by-step explanation:

1. Identify the unknowns in the system of equations.
2. Identify the known values in the system of equations.
3. Choose an equation, plug in the known values, and solve for the unknown.

For example, if the given system of equations is:

Equation 1: 3x + 2y = 7

Equation 2: 2x - y = 4

We can identify the unknowns as x and y, and the known values as 3, 2, 7, and 4. From there, we can choose one of the equations, substitute the known values, and solve for the unknown. Let's choose Equation 2 and plug in the values:

2(3) - y = 4

6 - y = 4

-y = 4 - 6

-y = -2

y = 2

Now, we can substitute the value of y back into Equation 1:

3x + 2(2) = 7

3x + 4 = 7

3x = 7 - 4

3x = 3

x = 1

Therefore, the solution to the given system of equations is x = 1 and y = 2.

Answer 2

Answers
#2 = infinite solutions.

if you simplify the 2nd equation by dividing each term by 2, the equations are the same and they will be on the same line.

EQ #1 x - 2y = 4
EQ #2 2x - 4y = 8 Divide each term by 2

2/2 x -4/2 y = 8/2
Simplify
x - 2y = 4

Now we have

EQ #1 x - 2y = 4
EQ #2 x - 2y = 4

So every point on the line is a solution to the system. The system has an infinite number of solutions, the two equations are just different forms of the same equation.

#3 ( -3, 3 ) or x = -3, y = 3

I will use elimination method to solve

EQ#1 -8x -4y = 12
EQ#2 x = 3 -2y
Change equation 2 to standard form

EQ#1 -8x -4y = 12
EQ#2 x + 2y = 3

Multiply all terms in EQ#2 by 2
2 ( x + 2y = 3) = 2x + 4y = 6

So you can see when we add these two together the -4y and +4y zero each other out and you can solve for x.

EQ#1 -8x -4y = 12
EQ#2 2x + 4y = 6
Add. _________

-6x = 18
Divide both sides by -6
-6/-6 x = 18/-6
x = -3

Substitute value of x = -3 into the 2nd equation to solve for y

x = 3 -2y
-3 = 3 - 2y
Subtract 3 from both sides to isolate y
-3 -3 = 3 -3 - 2y
Simplify
-6 = -2y
Divide both sides by -2 to solve for y

-6/-2 = -2/-2 y

y = 3

Check your work using equation #2 by substituting x and y values

x = 3 -2y
-3 = 3 - 2(3)
-3 = 3 - 6
-3 = -3

Solution is correct
Problem solved!

I showed the math on the attachment but it might be hard to figure it out