If: 3x+2y = 5x+2y = 7 Then: 3x+2y = 7 and 5x+2y = 7 Subtract the 1st equation from the 2nd equation: 2x = 0 or x = o By substitution: x = 0 and y = 3.5
If: 3x+2y = 5x+2y = 7 which is the same as 7 = 7 = 7 Then: 3x+2y = 7 and 5x+2y = 7 Subtracting equations: x = 0 Therefore by substitution: x = 0 and y = 3.5
(3, -4)
For 5x+y=1, you would subtract 5x from each side, so you would get y=1-5x For 3x+2y=2, you would subtract 3x from each side, and then divide by 2. 2y=2-3x y=1-(3/2)x
3x+2y=20 5x-2y=1 (3x+2y) + (5x-2y)= 20+1 3x+5x= 21....... 8x=21.........x=2 5/8 Subsitute x=3(2 5/8)+ 2y=20........... 7 7/8 +2y=20.........2y= 20- 7 7/8....... y=6 1/16
If: 3x+2y = 5x+2y = 7 Then: 3x+2y = 7 and 5x+2y = 7 Subtract the 1st equation from the 2nd equation: 2x = 0 or x = o By substitution: x = 0 and y = 3.5
If: 3x+2y = 5x+2y = 14 Then: 3x+2y = 14 and 5x+2y =14 Subtract the 1st equation from the 2nd equation: 2x = 0 Therefore by substitution the solutions are: x = 0 and y = 7
If: 3x+2y = 5x+2y = 7 which is the same as 7 = 7 = 7 Then: 3x+2y = 7 and 5x+2y = 7 Subtracting equations: x = 0 Therefore by substitution: x = 0 and y = 3.5
(3, -4)
For 5x+y=1, you would subtract 5x from each side, so you would get y=1-5x For 3x+2y=2, you would subtract 3x from each side, and then divide by 2. 2y=2-3x y=1-(3/2)x
3x+2y=20 5x-2y=1 (3x+2y) + (5x-2y)= 20+1 3x+5x= 21....... 8x=21.........x=2 5/8 Subsitute x=3(2 5/8)+ 2y=20........... 7 7/8 +2y=20.........2y= 20- 7 7/8....... y=6 1/16
(3x+2y)*(5x-3y) =3x*5x + 3x*(-3y) + 2y*5x + 2y*(-3y) [This is sometimes called the FOIL method - You multiply the First members of each bracket, then the Outers, Inners, Lasts] = 15x2 - 9xy + 10xy - 6y2 = 15x2 + xy - 6y2
Both of them are equations!The solution is (x, y) = (2, 1).
5x-2y+3z-2x-y-4z=3x-3y-z
If you mean: 2y+5x = 13 and 2y-3x = 5 then x = 1 and y = 4
step one: 5X+Y=1 ① 3X+2Y=2 ② step two: ① times 2 : 10X+2Y+2 ③ step three: ③-② : 7X=0 Therefore, X= 0, When X=0, Y=1
There are 2 unknowns