system of equations
c-3d=27
4d+10c=120
systems of linear equations by substitution. I need help, I am stuck
x=4
y=5
so if the equations involve c and d, what are x and y?
The first equation says
c-3d = 27
so,
c = 3d+27
Use that in the other equation:
4d+10c=120
first, divide by 2 so you can work with smaller numbers:
2d+5c = 60
Now, since c = 3d+27, that means
2d+5(3d+27) = 60
2d+15d+135 = 60
17d = -75
d = -75/17
c = 3d+27 = 3(-75/17) + 27 = 234/17
yucky answers, but hey - they're just numbers, eh?
To solve this system of equations:
Equation 1: c - 3d = 27
Equation 2: 4d + 10c = 120
We can use a common method like substitution or elimination. Let's solve it using elimination.
Step 1: Multiply both sides of Equation 1 by 10 to make the coefficients of c equal.
10(c - 3d) = 10(27)
10c - 30d = 270
Now, we have:
Equation 3: 10c - 30d = 270
Equation 4: 4d + 10c = 120
Step 2: Add Equation 3 to Equation 4 to eliminate c.
(10c - 30d) + (4d + 10c) = 270 + 120
14c - 26d = 390
Now, we have:
Equation 5: 14c - 26d = 390
Step 3: Solve Equation 5 for c.
14c = 390 + 26d
c = (390 + 26d) / 14
c = (195 + 13d) / 7
Now, we have the value of c in terms of d.
Step 4: Substitute this expression of c into Equation 1 to solve for d.
(195 + 13d) / 7 - 3d = 27
Simplify the equation:
195 + 13d - 21d = 189
Combine like terms:
-8d = -6
Divide both sides by -8:
d = -6 / -8
d = 3/4
Step 5: Substitute the value of d back into the expression for c to find its value.
c = (195 + 13(3/4)) / 7
c = (195 + 39/4) / 7
c = (780 + 39) / 28
c = 819 / 28
c ≈ 29.25
Therefore, the solution to the system of equations is c ≈ 29.25 and d = 3/4.