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secondary 2 | Maths
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Pls help with all three! Asap tysm!
Date Posted: 9 months ago
Views: 190
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hi this is the worked out solution for your problem:
1a y=1 , x= 6
First, you substitude x into your first equation since x=5+y. This gives you 5+y=y=7. Then, you solved for y and then x.
2b. Similar to the first question, you need to make it so that x is on one side. Sicne x-2y=7, I can move -2y to the other side of the equation, giving me x=7+2y. Then i substitude x into the first equation and find y. Then x.
3a. This is a bit trickier. First you bring the x into one side of the equation which give you 2x=12-5y. Sincec 4 is a multiple of 2 (2x2 = 4) , this means I can put my third equation into the second equation , giving me 2(12-5y) +3y = -4 . Then i solve for y and then x.
If you have any other question, feel free to ask!
1a y=1 , x= 6
First, you substitude x into your first equation since x=5+y. This gives you 5+y=y=7. Then, you solved for y and then x.
2b. Similar to the first question, you need to make it so that x is on one side. Sicne x-2y=7, I can move -2y to the other side of the equation, giving me x=7+2y. Then i substitude x into the first equation and find y. Then x.
3a. This is a bit trickier. First you bring the x into one side of the equation which give you 2x=12-5y. Sincec 4 is a multiple of 2 (2x2 = 4) , this means I can put my third equation into the second equation , giving me 2(12-5y) +3y = -4 . Then i solve for y and then x.
If you have any other question, feel free to ask!
Date Posted:
9 months ago
Okkk thnaks!
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