Can someone help me do these?

Question

1. 3(m+4)^2-8=2(m+4)

2. 2y-5+3y=4-(y+2)

Answer 1

1)

3(m + 4)^2 - 8 = 2(m + 4)

First, move everything to the left hand side.

3(m + 4)^2 - 2(m + 4) - 8 = 0

Now, factor this quadratic; this is no different than factoring 3y^2 - 2y - 8 = 0.

[3(m + 4) + 4] [(m + 4) - 2] = 0

Simplify within the brackets.

[3m + 12 + 4] [m + 4 - 2] = 0
[3m + 16] [m - 2] = 0

Equate each factor to 0.

3m + 16 = 0
m - 2 = 0

m = {-16/3, 2}

Answer 2

Substitute x = m+4 to get
3x² - 2x - 8 and solve for the roots. Then substitute m = x -4 to get the m solution values.

2y - 5 + 3y = 4 - (y + 2)
2y - 5 + 3y = 4 -y - 2
6y = 7
y = 7/6

HTH

Doug

Answer 3

1. solve as a quatratic where x=(m+4)

3(m+4)^2 - 2(m+4) - 8 = 0

m= (2 +- sqrt(4+96))/6 - 4
m= 12/6 - 4 or -8/6 - 4
m = -2 or -5 1/3

2. first simplify

5y - 5 = -4y - 8
9y - 5 = -8
9y = -3
y = -1/3

Answer 4

1. 3(m+4)^2-8=2(m+4)
3(m^2+8m+16)-8=2m+8
3m^2+24m+48-8=2m+8
3m^2+22m+32=0
(3m+16)(m+2)=0
3m+16=0
m=-16/3
m+2=0
m=-2
m=-16/3, -2

2. 2y-5+3y=4-(y+2)
5y-5=4-y-2
6y=7
y=7/6

Answer 5

Well this is basic standerd math. (I think) For number two you have to add all the ys so you have 5y then left over you have the equation:
5y - 5=(y+) then you take 5y and minus it from 5 and you get: 0=(y+2) Then you go in the brackets and do that equation which in the end equals
: 2y and that your answer

Answer 6

1.
Let m+4=t;
3t^2-2t-8=0;
3t^2+4t-6t-8=0;
t(3t+4)-2(3t+4)=0;
(t-2)*(3t+4)=0;
t-2=0 or 3t+4=0;
t=2 or t=-4/3
Case 1
m+4=2;
m=-2.
Case 2
m+4=-4/3;
m=-4/3-4
=-16/3
Hence, m=-2 or m=-16/3.

2
5y-5=2-y;
6y=7
;y=7/6

Answer 7

you cheater cheater pumpkin eater, do your math homework by yourself, or ask your math teacher, or your parents, dont ask for help in the internet, remember there are child molesters in the internet