sstrike's answer to Calisto Yeow's Primary 6 Maths Algebra Singapore question.
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Date Posted:
3 years ago
A → 3p
S → 5u
⅓ × 3p = p
⅖ × 5u = 2u
p+2u = 45 ⇨
① p = 45-2u
5u×$5 + 3p×$9 = $751
② 25u + 27p = 751
① → ②
25u + 27(45-2u) = 751
u=(27×45-751)÷(2×27-25) = 16
p=45-2u=13
5u+3p
= 5×16+3×13
= 119 (Ans)
S → 5u
⅓ × 3p = p
⅖ × 5u = 2u
p+2u = 45 ⇨
① p = 45-2u
5u×$5 + 3p×$9 = $751
② 25u + 27p = 751
① → ②
25u + 27(45-2u) = 751
u=(27×45-751)÷(2×27-25) = 16
p=45-2u=13
5u+3p
= 5×16+3×13
= 119 (Ans)
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