y=2x
x+y=12 What is the answer in (x,y) form
y=1/3x+6
3y-x=24 What is the answer in (x,y) form
3x+2y=13
5x=2y+11 What is the answer in (x,y) form
6x-2y=6
x+2y=8 What is the answer in (x,y) form
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These are all substitution/elimination problems
y=2x
x+y=12 We know what "y" equals so put "2x" in place of the "y" in second equation
x+(2x)=12 add common terms
3x=12 divide by 3
x=4
Now we know x=4 so replace "x" with "4" in the first equation
y=2x
y=2(4)
y=8
Now we know that y=4 so to put it in (x,y) form just replace the letters with the numbers (4,8)
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y=1/3x+6
3y-x=24
Im un able to answer this question completely as I dont know if the first equation is y=(1/3)x +6 or y=1/(3x+6) there will be completely different answers so please be specific when writing the problem.
if I had to guess it would be y=(1/3)x +6.
im going to use the elimination methad as its by far easier then to substitute with fractions.
y-(1/3)x=6 id multiply this line by -3 so i can eliminate the "y"
3y-x=24
-3y+x=-24 well as we can see these 2 equasions completely cancel each other out so the only possible answers can be (8,0) or(0,-24).due to the cancelation one variable must be 0.
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3x+2y=13
5x=2y+11
Id start with the bottom equation to rearange it use the elimination method
5x=2y+11 => 5x-2y=11
3x+2y=13
5x-2y=11 add vertically
8x+0=24
x=3
Plug x=3 into one of the equations to find "y"
3(3)+2y=13
9+2y=13
2y=4
y=2
(x,y)=(3,2)
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6x-2y=6
x+2y=8
add these vertically
7x+0=14
x=2
Plug x=2 into one of the equations to find "y"
(2)+2y=8
2y=6
y=3
(x,y)=(2,3)