Solving simulataneous equations?

Question

7c + 3d = 29

5c - 4d = 33

Im stuck, please help :(

Answer 1

take the first equation, and solve for d:

7c + 3d = 29

subtract 7c from both sides

3d = 29-7c

divide both sides by 3

d = (29-7c)/3

Now, substitute in for d in the second equation:

5c - 4d = 33

5c - 4(29-7c)/3 = 33

multiply both sides by 3

15c - 4(29-7c) = 99

15c - 116 +28c = 99

add 116 to both sides

15c +28c = 215

43c = 215

c = 215/43

c = 5

Now, put that into the first equation:

7c + 3d = 29

7*5 + 3d = 29

35 + 3d = 29

subtract 35 from both sides

3d = -6

divide both sides by 3

d = -2

Now put both numbers into the second equation to double check our answers:

5c - 4d = 33

5 *5 - 4*(-2) = 33

25 + 8 = 33

33 = 33, so our answers are correct.

c = 5, d = -2

Answer 2

The easiest way to solve these problems is to eliminate one of the variables by adding the two equations together, then solve for the remaining variable, then use that answer to get the other variable.

In this case, let's try to eliminate d first.

First multiply all of the items in the first equation by 4  and all of the items in the 2nd equation by 3 to get:

(4)7c + (4)3d = (4)29

(3)5c - (3)4d = (3)33

or further simplifying:

28c + 12d = 116

15c - 12d = 99

Now if we add the two equations together, you'll notice that the d variable disappears.  I chose the 3 and 4 multipliers to make this happen.

After adding the equations together you get:

43c +0d =115

now you can solve for c to get

c=115/43.

Now you can substitute this value of into the 2nd equation and solve for d as follows:

5(115/43) - 4d = 33

4d = 33-5(115/43)

d=[33-5(115/43)]/4

you could also solve for d as you did for c earlier by eliminating the c from the equations, then subtracting the equations, and solving for d:

(5)7c + (5)3d = 5(29)

(7)5c - (7)4d = (7)33

simplifying further:

35c + 15d = 145

35c - 28d = 231

Subtracting the 2nd from the first gives:

0c +43d = 86

solve for d to get:

d= 86/43

so the answers are

c=115/43 and d=86/43

These answers are a bit more complicated than I expected for a problem.  The neat thing you can do though is check your work by substituting these values into the starting equations to make sure they work.

Hope this helps

Answer 3

multiply top by 4 , bottom by 3

28c + 12d = 116

15c -  12d = 99

add them

43c  = 215

c= 5

7(5) + 3d = 29

35 + 3d = 29

3d = -6

d= -2

Answer 4

7c + 3d = 29                                      5c - 4d = 33

multiply  above by -5                             multiply above by 7

-35c - 15d = -145                               35c -28d = 231

Subtract the one problem from the other to elimate the c and solve for d.  Then substitute your answer from the d in one original equation to find c.

Answer 5

there are several ways to solve this  but by far the easiest is to arrange the equations so that one of the unknowns will zero out and solve for the other..

EXP   in the above equations multiply the top equation by 4 an the bottom equation by 3 to get

             28C + 12d =116

             15C  -12d   = 99  adding the two equations  the ds add to zero and one gets

43 C = 215    c = 5  now substitute the vakue of c in either ofthe above equations  5(5) -4d = 33  d = -2