Page 99 - Mathematics

Example 5.32 Substitute x = into equation (ii)
2
to obtain the value of y.
Solve the following linear
simultaneous equations by From 2x + 5y = − 1, it follows that,
elimination method. 2(2) 5y+ = − 1 Tanzania Institute of Education
3x −  2y = 8
 45+ = − 1
 2x + 5y = − 1
5y = − 5
Solution y = − 1
Label the two equations as follows.
2
3x −  2y = 8 (i) Therefore, x − and y = − 1is
 the solution of the simultaneous
 2x + 5y = − 1 (ii)
equation.

Eliminate y by multiplying equation Example 5.33
(i) by 5 and equation (ii) by 2 as
follows: Solve the following simultaneous
5(3x −  2 ) 8 5y = × equations by elimination method.
  x y
1 2
 2(2x + 5 )y = − ×  2 + 5 = 4

15x −  10y = 40 (iii)  x y
  + = 6
 4x + 10y = − 2 (iv)   4 2

Since the coefficients of y in equations Solution
(iii) and (iv) are equal with opposite Label the two equations.
signs, add the two equations. That is,
15x −  10y = 40  x + y = 4 (i)
+   2 5

 4x + 10y = − 2  x y
19x = 38  + = 6 (ii)
  4 2
Solve for the value of x in the
equation of one unknown. That is, For simplicity, transform the two
19x = 38 equations, (i) and (ii) into equations
with integer coefficients of the
38 variables. In order to remove the
x = 19 fractions, multiply the equations by Mathematics Form One

= 2 the LCM of the denominators.

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Mathematics form 1.indd 93 25/10/2024 09:51:31
Mathematics form 1.indd 93