Potatoo's answer to Clara's Secondary 2 Maths Singapore question.
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3x-5=y-1
x=(y+4)/3 -(1)
2x-4=y-3
2x-y=1 -(2)
Substitute eqn 1 into eqn 2
2[(y+4)/3]-y=1
[(2y+8)/3]-y=1
2y+8-3y=3
y=5
When y=5,x=(5+4)/3
x=3
Length = 3(3)-5=4cm
Breadth = 2(3)-4=2cm
Area = length x breadth
= 4cm x 2cm
= 8cm^2
x=(y+4)/3 -(1)
2x-4=y-3
2x-y=1 -(2)
Substitute eqn 1 into eqn 2
2[(y+4)/3]-y=1
[(2y+8)/3]-y=1
2y+8-3y=3
y=5
When y=5,x=(5+4)/3
x=3
Length = 3(3)-5=4cm
Breadth = 2(3)-4=2cm
Area = length x breadth
= 4cm x 2cm
= 8cm^2
Date Posted:
3 years ago
simultaneous equation, hope this helps !
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