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Your answers in (a) are correct.
For (b), we notice that if we change “x” in (a) to “x/2”, we would end up with the equation in (b).
For instance, the 1st term is (a) is 2x³. If we substitute “x” with “x/2”, we get:
2(x/2)³ = 2(x³/2³) = 2x³/8 = 1/4 x³
(Not sure if you know what I mean coz I didn’t write it out but only typed it out.)
The same goes to the 2nd term and 3rd term. That’s also why the 4th term remains unchanged, since it’s just a constant without any “x”.
Hence, the solution to the new equation in (b) would be similar to (a), except that we also need to change “x” in the final step to “x/2”.
Therefore,
x/2 = -2, 1/2, 3
Multiply both sides by 2 (Take note that all 3 answers must times 2):
x = -4, 1, 6
For (b), we notice that if we change “x” in (a) to “x/2”, we would end up with the equation in (b).
For instance, the 1st term is (a) is 2x³. If we substitute “x” with “x/2”, we get:
2(x/2)³ = 2(x³/2³) = 2x³/8 = 1/4 x³
(Not sure if you know what I mean coz I didn’t write it out but only typed it out.)
The same goes to the 2nd term and 3rd term. That’s also why the 4th term remains unchanged, since it’s just a constant without any “x”.
Hence, the solution to the new equation in (b) would be similar to (a), except that we also need to change “x” in the final step to “x/2”.
Therefore,
x/2 = -2, 1/2, 3
Multiply both sides by 2 (Take note that all 3 answers must times 2):
x = -4, 1, 6
Ohhhhh icic. I understand now. Tysm!!!
No prob. I just noticed that I forgot to write “=0” behind “+6” (before the “…”).
This is just one way to present the solution. Your school teacher may have another way to present it.
This is just one way to present the solution. Your school teacher may have another way to present it.
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This is another way to present the working, but personally I find it harder to understand.
Date Posted:
3 years ago
I think you meant how to present your solution? I’ll take a photo of the working and post it again.