how do you solve c=3d-27 ; 4d+10c=120
(1) c = 3d - 27
(2) 4d + 10c = 120
there are several ways to solve this problem (system of linear equations),, but in this case, we substitution method since in equation (1), variable c is expressed in terms of the other variable d,, we then substitute c from equation (1) to equation (2):
4d + 10(3d-27) = 120
4d + 30d - 270 = 120
34d = 390
d = 195/17
*substituting this value of d to either equations (in this case, to (1)):
c = 3d - 27
c = 3(195/17) - 27
c = 585/17 - 27
c = (585 - 459)/17
c = 126/17
so there,, :)
How do you solve this problem? V = Bh
Solve each formula for the variable shown in RED. In this problem the variable in RED is the letter B.
To solve the given system of equations:
1. Start by isolating one of the variables in one of the equations.
Let's solve equation (1) for d:
c = 3d - 27
Add 27 to both sides:
c + 27 = 3d
Divide both sides by 3:
(c + 27)/3 = d
2. Substitute the expression for d in equation (2) with the value found in step 1.
4d + 10c = 120
Replace d with (c + 27)/3:
4((c + 27)/3) + 10c = 120
3. Simplify the equation by multiplying through by 3 to remove the fraction:
4(c + 27) + 30c = 360
Distribute:
4c + 108 + 30c = 360
Combine like terms:
34c + 108 = 360
Subtract 108 from both sides:
34c = 252
Divide both sides by 34:
c = 252/34
Simplify the fraction:
c ≈ 7.41
4. Substitute the value of c back into equation (1) to find d:
c = 3d - 27
Replace c with 7.41:
7.41 = 3d - 27
Add 27 to both sides:
7.41 + 27 = 3d
Simplify:
34.41 = 3d
Divide both sides by 3:
34.41/3 = d
d ≈ 11.47
So, the solution to the system of equations is approximately d ≈ 11.47 and c ≈ 7.41.