# Simultaneous equation help!!

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Thread starter 5 years ago
#1
Can anyone tell me how to solve this using substitution method?
y-x/2=1
x/3+2y=10
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5 years ago
#2
(Original post by neza99)
Can anyone tell me how to solve this using substitution method?
y-x/2=1
x/3+2y=10
You could rearrange the equations like this..

-x/2+y=1
X/3+2y=10

And then Multiply the first equation by two so the first equation becomes -x+2y=2 and then subtract the two equations from each other,,.
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Thread starter 5 years ago
#3
(Original post by GalGirl101)
You could rearrange the equations like this..

-x/2+y=1
X/3+2y=10

And then Multiply the first equation by two so the first equation becomes -x+2y=2 and then subtract the two equations from each other,,.
i did that before but the i got x = 12... which didn't fit in the equation
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5 years ago
#4
(Original post by neza99)
Can anyone tell me how to solve this using substitution method?
y-x/2=1
x/3+2y=10
GalGirl101's method is the most efficient for solving these. However, to solve them by substitution, you need to rearrange one so that x or y is the subject and use this to solve the other equation by substituting it in so that you have an equation with only x or y. The first is easier to rearrange so should start there
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5 years ago
#5
Multiply the first equation by 2 to get 2y - x = 2. Rearrange to get x = 2y - 2.
Multiply the second equation by 3 to get x + 6y = 30. Substitute x into the second equation to get 8y - 2 = 30.
Therefore, y = 4 and further solving results in x = 6.
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Thread starter 5 years ago
#6
(Original post by DodoManiac)
GalGirl101's method is the most efficient for solving these. However, to solve them by substitution, you need to rearrange one so that x or y is the subject and use this to solve the other equation by substituting it in so that you have an equation with only x or y. The first is easier to rearrange so should start there
i think i did that, i show you
y=1+x/2 i made y the subject
x/3+2y=10
so x/3+2(1+x/2)=10
x/3+2+x=10
2x/3=8 i minused 2 from both sides.
2x=24
x=12 i did this but it didn't work :/
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5 years ago
#7
(Original post by neza99)
i think i did that, i show you
y=1+x/2 i made y the subject
x/3+2y=10
so x/3+2(1+x/2)=10
x/3+2+x=10
2x/3=8 i minused 2 from both sides.
2x=24
x=12 i did this but it didn't work :/
Right method, just an arithmetic slip. What's 1/3 + 1?
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Thread starter 5 years ago
#8
(Original post by DodoManiac)
Right method, just an arithmetic slip. What's 1/3 + 1?

1 1/3? i'm not really good in this bit
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5 years ago
#9
(Original post by neza99)
1 1/3? i'm not really good in this bit
Yeah that's right, or as a single fraction? You said x/3 + x = 2x/3

Think of it as 1/3 + 3/3 if that helps
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Thread starter 5 years ago
#10
(Original post by DodoManiac)
Yeah that's right, or as a single fraction? You said x/3 + x = 2x/3

Think of it as 1/3 + 3/3 if that helps
so it would be 4x/3?
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Thread starter 5 years ago
#11
(Original post by DodoManiac)
Yeah that's right, or as a single fraction? You said x/3 + x = 2x/3

Think of it as 1/3 + 3/3 if that helps
OMG, thank you very much it worked now! may i ask if you know what level question is this?
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5 years ago
#12
(Original post by neza99)
OMG, thank you very much it worked now! may i ask if you know what level question is this?
Great! No problem, happy to help

I'm not too sure what level question it is to be honest but would guess at higher level GCSE/starter question on a C1 paper (AS).
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Thread starter 5 years ago
#13
(Original post by DodoManiac)
Great! No problem, happy to help

I'm not too sure what level question it is to be honest but would guess at higher level GCSE/starter question on a C1 paper (AS).
alright, thanks!
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5 years ago
#14
(Original post by neza99)
Can anyone tell me how to solve this using substitution method?
y-x/2=1
x/3+2y=10
Rearrange the first equation so that x is at the front of the equation to get started.
Then, put some though into the equation and think how to get x without the divisor, if x is divided by 2, what would you multiply to it to make it just x (1x).
Remember you have to multiply the entire equation by that value and not only x.
Do this for the second equation and then subtract both of the equations to get y on it's own = to something.
Solve for y and then substitute y into the equation to solve for x.

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