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Simultaneous Equations
By the end of the lesson you will be able to :
• Solve simultaneous equations by the elimination method.
Solve by the substitution method
As you see it demands a considerable work due to the 
fact that the unknowns are always accompanied by 
coefficients different from 1.
Elimination Method
3 x  +  y  =  9
5 x  ­  y   =  7
5 x  +  3 y  =  29
   x  +  3 y   = 13
x = 4
y = 3
3 x  ­ 2  y  =  8
2 x  ­ 2 y   =  4
x = 4
y = 2
2 x  ­  8 y  =  2
2 x  ­  3y   =  7
x = 5
y = 1
3 x  +  2 y  =  22
   x  +     y   =  9
x = 4
y = 5
2 x  +      y  =  7
3 x  +  2 y   = 12
x = 2
y = 3
Let's re­do the first exercise we solved in this 
class, but now by using elimination method.
3 x  + 4 y  =  18
5 x  ­  3 y   = 1
x =2
y = 3
The sum of two numbers is 12 and their difference 
is 4.8. Find the numbers. 
x = 8.4
y = 3.6
Solve from Book  pages 110­111

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