Basic Engineering Mathematics, Fifth Edition

Solving simultaneous equations 91

(b) By elimination

x+ 2 y=−1(1)

4 x− 3 y= 18 (2)

If equation (1) is multiplied throughout by 4, the

coefficient ofxwill be the same as in equation (2),

giving

4 x+ 8 y=−4(3)

Subtracting equation (3) from equation (2) gives

4 x− 3 y= 18 (2)

4 x+ 8 y=− 4 (3)

0 − 11 y= 22

Hence, y=

22

− 11

=− 2

(Note: in the above subtraction,

18 −− 4 = 18 + 4 = 22 .)

Substitutingy=−2 into either equation (1) or equa-
tion (2) will givex= 3 as in method (a). The solution
x= 3 ,y=− 2 is the only pair of values that satisfies
both of the original equations.

Problem 2. Solve, by a substitution method, the

simultaneous equations

3 x− 2 y= 12 (1)

x+ 3 y=−7(2)

From equation (2),x=− 7 − 3 y

Substituting forxin equation (1) gives

3 (− 7 − 3 y)− 2 y= 12

i.e. − 21 − 9 y− 2 y= 12

− 11 y= 12 + 21 = 33

Hence,y=

33

− 11

=− 3

Substitutingy=−3 in equation (2) gives

x+ 3 (− 3 )=− 7

i.e. x− 9 =− 7

Hence x=− 7 + 9 = 2

Thus,x= 2 ,y=− 3 is the solutionof thesimultaneous
equations. (Such solutions should always be checked

by substituting values intoeach of the original two

equations.)

Problem 3. Use an elimination method to solve

the following simultaneous equations

3 x+ 4 y=5(1)

2 x− 5 y=− 12 (2)

If equation (1) is multiplied throughout by 2 and equa-

tion (2) by 3, the coefficient ofxwill be the same in the

newly formed equations. Thus,

2 ×equation (1) gives 6x+ 8 y= 10 (3)

3 ×equation (2) gives 6x− 15 y=− 36 (4)

Equation (3) – equation (4) gives

0 + 23 y= 46

i.e. y=^46

23

= 2

(Note+ 8 y−− 15 y= 8 y+ 15 y= 23 yand 10−− 36 =

10 + 36 =46.)

Substitutingy=2 in equation (1) gives

3 x+ 4 ( 2 )= 5

from which 3 x= 5 − 8 =− 3

and x=− 1

Checking, bysubstitutingx=−1andy=2inequation

(2), gives

LHS= 2 (− 1 )− 5 ( 2 )=− 2 − 10 =− 12 =RHS

Hence,x=− 1 andy= 2 is the solution of the simul-

taneous equations.

The elimination method is the most common method of

solving simultaneous equations.

Problem 4. Solve

7 x− 2 y= 26 (1)

6 x+ 5 y= 29 (2)

When equation (1) is multiplied by 5 and equation (2)

by 2, the coefficients ofyin each equation are numeri-

cally the same, i.e. 10, but are of opposite sign.