SOMEONE PLEASE HELP
(06.01)
If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalent to x from the first equation is substituted into the second equation.

x + 4y = −9
2x + 5y = −6 (6 points)

2x + 5(4y − 9) = −6
2x + 5(−4y − 9) = −6
2(4y − 9) + 5y = −6
2(−4y − 9) + 5y = −6

(06.03)
How can one half x − 5 = one third x + 6 be set up as a system of equations? (6 points)

A) 2y + x = −10
3y + x = 18
B)2y + 2x = −10
3y + 3x = 18
C) 2y − x = −10
3y − x = 18
D)2y − 2x = −10
3y − 3x = 18

(2) x = -9 -4y (at this point we are moving the 4y to the other side of the equation and since its positive in one side, it goes to the other with a minus on it.)

(3) x = -4y - 9 (all we do here is rearrenge the factors on the right side)

Then we should replace the x in the second equation with (3)

(4) 2x + 5y = −6

(5) 2(−4y − 9) + 5y = −6

And we find the answer thats listed.

(06.03)

First we need to write the equation given.

(1) [tex]\frac{1}{2} x - 5 = \frac{1}{3} x + 6[/tex]

Now we work with each side for separate. Since both sides are equal to eachother we can make them equal to y to both sides. And we get this two equations

(2) [tex]y = \frac{1}{2} x - 5[/tex]

(3) [tex]y = \frac{1}{3} x + 6\\[/tex]

We multiply (2) for 2 in both sides of the equation, to remove the fraction. And we will multiply (3) for 3 in both sides of the equation, to remove the fraction.

And we have this new system of equations.

(4) 2y = x - 10

(5) 3y = x + 18

Now we need to rearrenge the factors, and we will move the x to the left side in both equations, and since both are positive, they pass with a minus sign on it.

So we get.

2y − x = −10

3y − x = 18

Thus the answer is C.