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secondary 3 | A Maths
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Cheryl
Cheryl

secondary 3 chevron_right A Maths chevron_right Singapore

Can someone help me with question 13, thank you !!!

Date Posted: 3 years ago
Views: 186

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Given : x/2 + y/5 = 5 ①
x/2 = 5 - y/5
Multiply both sides by 10,
5x = 50 - 2y
x = (50 - 2y) / 5 ②
Given : 2/x + 5/y = 5/6 ③
Sub ② into ③,
2 / ((50 - 2y)/5) + 5/y = 5/6
2 × 5/(50-2y) + 5/y = 5/6
5 / (25 - y) + 5/y = 5/6
1 / (25 - y) + 1/y = 1/6
Multiply both sides by 6(25 - y)(y),
6y + 6(25 - y) = (25 - y)y
6y + 150 - 6y = 25y - y²
y² - 25y + 150 = 0
(y - 15)(y - 10) = 0
y - 15 = 0 or y - 10 = 0
y = 15 or y = 10

So, x = (50 - 2(15)) / 5 or x = (50 - 2(10)) / 5
x = 20/5 or x = 30/5
x = 4 or x = 6

∴ x = 4, y = 15 or x = 6, y = 10
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J
J's answer
1022 answers (A Helpful Person)
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J
J
3 years ago
Alternative method :

Given : x/2 + y/5 = 5

(5x + 2y) / 10 = 5

(5x + 2y) / 60 = 5/6 ①


Given : 2/x + 5/y = 5/6

(2y + 5x)/xy = 5/6 ②


Comparing ① and ②, we can see that :

xy = 60
x = 60/y ③


Sub ③ into x/2 + y/5 = 5,

(60/y)/2 + y/5 = 5

30/y + y/5 = 5

Multiply by 5y,

150 + y² = 25y

y² - 25y + 150 = 0

(y - 10)(y - 15) = 0

y = 10 or y = 15


So,

x = 60/10 or x = 60/15
x = 6 or x = 4

∴ x = 4, y = 15 or x = 6, y = 10