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b ab a abx
u u
u u
u ?^1 x y^1
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or 2 2 2 2 2 2
1
ab ab a b
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ab abx
or
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aba bx
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ab(a b) (a b)(a b)
x
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, or
a bab
a b a baba b
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ab(a b) (a b)(a b)
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(x,y) ,Exercise 12⋅ 2
Solve by the method of substitution (1 - 3) :
- 7 x 3 y 31
9 x 5 y 41
2. 1
2 3
x y1
3 2
x y- 2
b
y
axaxby a^2 b^2Solve by the method of elimination (4 - 6) :
- 7 x 3 y 31 5. 7 x 8 y 9 6. axby c
9 x 5 y 41 5 x 4 y 3 a^2 xb^2 y c^2
Solve by the method of cross-multiplication (7 - 15) : - 2 x 3 y 5 0 8. 3 x 5 y 9 0 9. x 2 y 7
4 x 7 y 6 0 5 x 3 y 1 0 2 x 3 y 0 - 4 x 3 y 12 11. 7 x 8 y 9 12. 3 xy 7 0 2 xy 3
2 x 5 5 x 4 y 3 - axby a^2 b^2 14. y( 3 x) x( 6 y)
2 bxay ab^3 ( 3 x) 5 (y 1 )^ - (x 7 )(y 3 ) 7 (y 3 )(x 1 ) 5