can anyone figure how to solve for z, given x and y?

[Sabraroo]

Hey guys!! Can anyone figure these problems out and tell me in an EASY way how to do them? lol thanks. The directions are "Solve the positive values of 'z' for the equations."

. . .1) 3x(1 + y) = 9 + z where y = 4, x = 2

. . .2) (2zy)/y² = 14/y

. . .3) 2zx²y = 18 where x = 1, z = y

 

[Concorde]

Hello Sab,

There are a couple of ways to approach this problem.

First you can simplify symbolically and plug in the values afterwards or you can plug the numbers in from the get go and work from there. I will work the first one both ways for you because it illustrates an important concept in math which is the symbols are just placeholders for numbers and you should not get hung up on the fact a symbol is present as opposed to a number.

Original equation

3x(1+y)=9+z

Used Distributive property to simplify the equation

3x(1)+3x(y)=9+z

Subtracted 9 from both sides (I flipped the equation too)

z=3x(1)+3x(y)-9

Plugged in numerical values for x and y

z=(3)(2)(1)+3(2)(4)-9

Simplification again

z=21



Original Equation

3x(1+y)=9+z

Flipped equation again, subtracted 9 from both sides

z=3x(1+y)-9

Inserted known values for x and y

z=3(2)(1+4)-9

Simplification; note how this value and the value I got the
other way match!

z=21



Original equation

(2zy)/(y^2)=(14)/(y)

Multiplied both sides by y^2, the 14 is times by y^2 and
this gives us (14y^2)/y so one of the y's is cancelled out

2zy=14y

Divide both sides by 2y

z=(14y)/(2y)

Simplification: 14/2=7 and y/y=1, 7*1=7

z=7


I'll leave the last problem for you to practice on.

 

[Denis]

Re: can anyone figure these out and tell how?

Hey guys!! Can anyone figure these problems out and tell me in an EASY way how to do them? lol thanks. The directions are Solve the positive values of 'z' for the equations.
3x(1+y)=9+z where y=4, x=2

Sabraroo, where did you get this: "Solve the positive values of 'z' for the equations"?
Makes no sense...

Should appear something like:
3x(1+y)=9+z
If x=2 and y=4, solve for z.

Now (as you were shown by Concorde), simply substitute:
9 + z = 3(2)(1 + 4)
9 + z = 6(5)
9 + z = 30
z = 30 - 9
z = 21

If you don't follow that, you're in need of listening in class :shock:

 

[Concorde]

Dennis, I believe the sitpulation of solving for only positive values of z is because of the last of the three equations he has posted.