c=3d-274d+10c=120 solving using substitution

Question

This is algebra solving using substituion

 

c=3d - 27

4d + 10c = 120

 

 

Answer 1

c=3d - 27
4d + 10c = 120

 

Since we know what 'C' is equal to (Given c=3d-27), we can plug that into the second equation, giving us:

4d+10(3d-27)=120

 

We multiple the parenthetical out with the factor in front of it

4d+30d-270=120

 

Combine like terms

34d-270=120390

 

Add 270 from both sides:

(34d-270)+270=(120)+270

34d=390

 

Then we divide both sides by 34:

(34d)/34=(-390)/34

 

d= - 390/34

 

We can simplify that to:

 

d= 195/17

 

Make this a proper fraction:

 

d=11 8/17 

 

Now we know the value of D!

 

So we plug D back into the first equation

 

c=3d-27

 

c=3(11 8/17) -27

 

Multiply the parenthetical:

c=34 7/17 - 27

 

and then combine the terms:

 

c =7 7/17

 

Hope that helped!!!

Answer 2

Because c = 3d - 27, we can replace "c" with "3d - 27"

So in the equation 4d + 10c = 120 we can replace c (as stated above)

to get the equation 4d + 10(3d - 27) = 120

Now we have an equation with one variable and can solve for d.

 

4d + 10(3d - 27) = 120    [distribute]

4d + 30d - 270 = 120    [combine like terms]

34d - 270 = 120    [add 270]

34d = 390    [divide by 34]

d = 390/34    [simplify]

d = 195/17

 

Now to solve for c, pick either original equation, plug in the value of d, and solve for c. (The one where c is isolated is easier to use.)

c = 3d - 27

c = 3(195/17) - 27

c = 585/17 - 27

c = 585/17 - 459/17

c = 126/17