Me with this question
what is the equation of a line that joins the point of intersection of 5x-2y+3=0 and 4x-3y+1=0 to the point of intersection of x=y and x=3y+4?

 y = x

Step-by-step explanation:

First, find the point where 5x - 2y + 3 and 4x - 3y + 1 intersect using the Elimination Method.

5x - 2y + 3 = 0  →  3(5x - 2y + 3 = 0)  →  15x - 6y + 9 = 0

4x - 3y + 1 = 0   → -2(4x - 3y + 1 = 0)  →   -8x + 6y - 2 = 0

                                                                 7x         + 7 = 0

                                                                 7x               = -7

                                                                   x               = -1

5x - 2y + 3 = 0

5(-1) - 2y + 3 = 0

-5 - 2y + 3 = 0

    -2y  - 2 = 0

    -2y        = 2

        y        = -1

(-1, -1)

Next, find the point where x = y and x = 3y + 4 intersect using the Substitution Method.

x = y

x = 3y + 4    →    y = 3y + 4

                      -2y = 4

                         y = -2

x = y

x = -2

(-2, -2)

Now, find the line that passes through (-1, -1) and (-2, -2) using the Point-Slope formula.  (x₁, y₁) = (-1, -1)  and m = 1

y - y₁ = m(x - x₁)

y + 1 = 1(x + 1)

    y = x

mnkjn

step-by-step explanation:

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb