Find a so that each of the following equations may have x=1 and y=1 as a solution.
(i) 3x+ay=6
(ii) ax-2y=10
(iii) 99x+12ay=63
(iv) 5x+2ay=36
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3 (1)+a (1)=6
3+a=6
a=6-3
a=3
2) a (1)-2 (1)=10
a-2=10
a=12
3] 99 (1)+12a (1)=63
99+12a =63
12a =-36
a =-3
4] 5 (1)+2a(1)=36
5+2a =36
2a=31
a=31÷2
3+a=6
a=6-3
a=3
2) a (1)-2 (1)=10
a-2=10
a=12
3] 99 (1)+12a (1)=63
99+12a =63
12a =-36
a =-3
4] 5 (1)+2a(1)=36
5+2a =36
2a=31
a=31÷2
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2
(1)3.
(2)12
(3)-3
(4)31/2
(2)12
(3)-3
(4)31/2
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