Solve each system using substitution. check your solution.
a = 2/3b -3
a = 2b - 18
c = 3d -27
4d + 10c = 120
Ah, sometimes we do not see a question immediately. Also I am wondering what you have done so far yourself.
to find d you have to plug c = 3d -27 into 4d + 10c = 120. then maybe to find a you have to make 2/3b -3 =2b - 18. Not sure if that's right.
Sure Miley, those will work. However I told him that when he posted them earlier.
2 +2
2
-
6
To solve the first system of equations using substitution, we need to find the value of one variable in terms of the other from one equation and substitute it into the other equation. Let's begin:
Given:
a = (2/3)b - 3 ----(Equation 1)
a = 2b - 18 ----(Equation 2)
From Equation 1, let's solve for b:
a = (2/3)b - 3
Multiply through by 3 to eliminate the fraction:
3a = 2b - 9
Now, we have the value of b in terms of a. We can substitute this into Equation 2:
2b - 18 = 3a ----(Equation 2)
2(2/3)b - 18 = 3a (Substituting the value of b from Equation 1)
(4/3)b - 18 = 3a
Multiply through by 3/4 to eliminate the fraction:
(3/4)(4/3)b - (3/4)(18) = (3/4)(3a)
b - 13.5 = (9/4)a
Now, we have b in terms of a. We can plug this into Equation 1 to solve for a:
a = (2/3)b - 3
a = (2/3)(b - 13.5) - 3 (Substituting the value of b from above)
a = (2/3)b - 9 - 3
a = (2/3)b - 12
We have obtained the value of a in terms of b. So, the solution to the first system is:
a = (2/3)b - 12
b = (9/4)a + 13.5
To check the solution, substitute the values of a and b into the original equations and see if they satisfy both equations.
Similarly, let's solve the second system using the method of substitution:
Given:
c = 3d - 27 ----(Equation 3)
4d + 10c = 120 ----(Equation 4)
From Equation 3, let's solve for c:
c = 3d - 27
Now, substitute this value into Equation 4:
4d + 10c = 120
4d + 10(3d - 27) = 120 (Substituting the value of c from Equation 3)
4d + 30d - 270 = 120
34d - 270 = 120
34d = 120 + 270
34d = 390
d = 390/34
d = 11.47 (rounded to two decimal places)
Now that we have the value of d, we can substitute it back into Equation 3 to find the value of c:
c = 3d - 27
c = 3(11.47) - 27
c = 34.41 - 27
c = 7.41 (rounded to two decimal places)
Therefore, the solution to the second system of equations is:
c = 7.41
d = 11.47
To check the solution, substitute the values of c and d into the original equations and see if they satisfy both equations.