Can anyone help me solve these two math equations?

Question

[1.]Solve for q: p²q-3q=14



[2.]Solve for z: y²z+7z=y

Answer 1

You need to re-arrange the equations.

(1)

p²q-3q=14

therefore
q (p²-3)=14

then divide both sides by (p²-3)

and (2)
y²z+7z=y

therefore
z(y²+7)=y

and then divide both sides by (y²+7)

undoubtedly some kind soul will provide you with the final answers thereby doing your homework for you.

Answer 2

I cannot read the exponent, but assuming its a 2, I would suggest the following.

Since you have two variables and only one equation, you cannot get a numerical answer. First factor the q from each term on the left. q(p^2 - 3) = 14 Then divide the equation by (p^2 - 3) and get q = 14 / ( p^2 - 3).

This same process with the second equation gives z = y / (y^2 + 7).

Hope this answers your question, Mike R

Answer 3

Solve for q: p²q-3q=14
q is common . Take it out.
q( p²-3)=14

q=14/(p^2-3)

2)
y²z+7z=y
z(y^2+7)=y

z=y/(y^2+7)

Answer 4

1.) q =14 / (p^2-3)

2.) z = y / (y^2 + 7)

Answer 5

[1]... first factor the q out of both terms
q(p^2 - 3) = 14
now divide by (p^2 - 3) to get
q = (14) / (p^2 - 3)

[2] ... first factor out the z.
z(y^2 + 7) = y
z = y / (y^2 + 7)

Answer 6

the respond to the 1st question is C)y<-14 because of the fact -6-8=-14 no longer -sixteen. the respond to the 2nd question is D)2x-5. because of the fact the equation does equivalent 2x-10 and all the others are extremely restating this.