Literal Equations. Solve Each Equation For The Indicated Variable.
1) g=6x, for x
2) u=2x-2, for x
3) z=m-x, for x
4) g=ca, for a
5) u=x-k, for x
6) g=c+x, for x
7) u=k/a, for a
8) g=xc, for x
9) 12am=4, for a
10) -3x+2c=-3, for x

1. g/6=x

2. (u+2)/2=x

3. m/z=x

4. g/c=a

5. u+k=x

6. g-c = x

7. k/u = a

8. g/c =x

9. 4/(12m)=a or 1/(3m)=a this is the simplified version

10. (-3-2c)/-3=x   x=1+2c/3 Teachers usually prefer the first one

Step-by-step explanation:

so for this, they aren't looking for exact answers they just want you to get the variable they are looking for by itself

1. you want to get x by itself so you divide both sides by 6

g/6=x

2. same thing here you want x by itself. First, you add 2 to both sides

(u+2)=2x

x is still not by itself so divide both sides by 2

(u+2)/2=x

3. you want x by itself so there are two ways to go about this

You could add x to both sides

z+x=m and then subtract z from both sides leaving you with x=m-z or

subtract m from both sides

z-m= -x divide both sides by -1 which you will then get -z+m=x and with some rearranging that is still equal to m-z=x

4. you want a by itself so divide both sides by c

g/c=a

5. add k to both sides to get x by itself

u+k=x

6. subtract c from both sides to get x by itself

g-c=x

7. you want a by itself so to get rid of the fraction multiply both sides by a

ua=k and then divide both sides by u to get a by itself a=k/u

8. want x by itself so divide both sides by c this is very similar to problem 4

g/c=x

9. you want a by itself so you divide both sides by 12

am=4/12  this is equivalent to am=1/3 then divide by m cause a is still not alone a=1/3 divide by m. you have to do the reciprocal of m because you are dividing by a fraction so it will be like 1/3 * 1/m  =a which is 1/(3m)=a

10. subtract 2c from both sides since we are trying to get x by itself

-3x= -3 - 2c   X is still not by itself so divide by -3 on both sides

x=(-3-2c)/-3 = 1+2c/3 these are equivalent usually teachers will prefer the first one

y=mx+

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations: