<span>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 ___________________________________________ 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) -------------------------------- 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. ----------------- 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) ------------------
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)</span>
