Search 75,755 tutors
0 0

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

This is algebra solving using substituion

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

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!!!
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