Question 3 - open up to see pic

Mathematics

1 answer:

Maru [420]3 years ago

4 0

(12, 11):

2((12) + 3) - (11) = 7 + (12)

2 • 15 - 11 = 19

2 • 4 = 19

8 $\neq$ 19

(1, 0):

2((1) + 3) - (0) = 7 + (1)

2 • 4 = 8

8 = 8

(-5, 1):

2((-5) + 3) - (1) = 7 + (-5)

2 • -2 - 1 = 2

2 • -3 = 2

-6 $\neq$ 2

Hope this helps.

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Which classification best describes the following system of equations? 3x+6y-12z=36 x=2y-4z=12 4x+8y-16z=48 inconsistent and dep

Viktor [21]

Answer:

Consistent and dependent

Step-by-step explanation:

We are given the following equation:

1. $3x+6y-12z=36$

2. $x+2y-4z=12$

3. $4x+8y-16z=48$

For equation 1 and 3, if we take out the common factor (3 and 4 respectively) out of it then we are left with $x+2y-4z=12$ which is the same as the equation number 2.

There is at least one set of the values for the unknowns that satisfies every equation in the system and since there is one solution for each of these equations, this system of equations is consistent and dependent.

6 0

3 years ago

I need help 6x-2y=10 x-2y=-5 solve by elimination

dem82 [27]

<h3>Explanation</h3>

Given the system of equations.

$\begin{cases} 6x - 2y = 10 \\ x - 2y = - 5 \end{cases}$

Solve the system of equations by eliminating either x-term or y-term. We will eliminate the y-term as it is faster to solve the equation.

To eliminate the y-term, we have to multiply the negative in either the first or second equation so we can get rid of the y-term. I will multiply negative in the second equation.

$\begin{cases} 6x - 2y = 10 \\ - x + 2y = 5 \end{cases}$

There as we can get rid of the y-term by adding both equations.

$(6x - x) + ( - 2y + 2y) = 10 + 5 \\ 5x + 0 = 15 \\ 5x = 15 \\ x = \frac{15}{5} \longrightarrow \frac{ \cancel{15}}{ \cancel{5}} = \frac{3}{1} \\ x = 3$

Hence, the value of x is 3. But we are not finished yet because we need to find the value of y as well. Therefore, we substitute the value of x in any given equations. I will substitute the value of x in the second equation.

$x - 2y = - 5 \\ 3 - 2y = - 5 \\ 3 + 5 = 2y \\ 8 = 2y \\ \frac{8}{2} = y \\ y = \frac{8}{2} \longrightarrow \frac{ \cancel{8}}{ \cancel{2}} = \frac{4}{1} \\ y = 4$