Page 98 - Mathematics

It follows that, Multiply equation (i) by –2 to get a
3x = 6 new equation.
x = 2
 − 2(2x + y ) 10= ×− 2
Substitute the value of x into one  3x − 2y = 1

of the equations to obtain the value
of y. − 4x − 2y = − 20 (iii)
2
Substitute x = into the equation Subtract equation (ii) from equation
(ii) to obtain, (iii) as follows:
y
Tanzania Institute of Education
3(2) += 9 −  4x − 2y = − 20
6 y+= 9 −  3x − 2y = 1

y = 3 − 7x =− 21
3
Therefore, x = 2 and y = is the
solution of the given simultaneous Solve for the value of x in the
equations. equation of one unknown. That is,

− 7x = − 21
Example 5.31 21
x = 7
Solve the following simultaneous
equations by elimination method. = 3

 2x y+= 10 Substitute the value of x into (i) or
 (ii) or (iii) to find the value of y.
 3x − 2y = 1
From equation (i), substituting the
Solution value of x gives

Label the given simultaneous 2(3) += 10
y
equations.
y
 2x y+= 10 (i) 6 += 10

 3x − 2y = 1 (ii) y = 10 6− 4.
Mathematics Form One eliminated. Eliminate y to form Therefore, x = 3 and y =
4
=
Choose the variable to be
an equation with variable x only.

The coefficient of y in equation (i)
is 1 and in equation (ii) is 2.−

92

25/10/2024 09:51:30
Mathematics form 1.indd 92 25/10/2024 09:51:30
Mathematics form 1.indd 92