Number operation review 10 pot

c  2d 13

c  2(11  2c) 13

c  22  4c 13

22  3c 13

22 13  3c

9 3c

c 3

Now substitute this answer into either original equation for c to find d.

2c  d 11

2(3)  d 11

6  d 11

d 5

Thus, c 3 and d 5.

Linear Combination

Linear combination involves writing one equation over another and then adding or subtracting the like terms so

that one letter is eliminated.

Example

x  7 3y and x  5 6y

First rewrite each equation in the same form.

x  7 3y becomes x  3y 7

x  5 6y becomes x  6y 5.

Now subtract the two equations so that the x terms are eliminated, leaving only one variable:

x  3y 7

 (x  6y 5)

(x  x)  ( 3y  6y) 7  (5)

3y 12

y 4 is the answer.

Now substitute 4 for y in one of the original equations and solve for x.

x  7 3y

x  7 3(4)

x  7 12

x  7  7 12  7

x 19

Therefore, the solution to the system of equations is y 4 and x 19.

–ALGEBRA REVIEW–

89