x + 7y = 39 ....... Eq.13x - 2y = 2 ....... Eq.2multiply Eq.1 with (-3) and so Eq.1 will be :-3x - 21y = -1173x - 2y = 2_________________ by Elimination method ( Eq.1 + Eq.2 ) the result will we :- 21y - 2y = - 115-23y = -115Y = 5to find the value of x, put the value of y in Eq.1 :x + 7y = 39x = 39 - 7yx = 39 - 7(5)X = 4___________________________________________________________you can solve it in another way by using substitution method:x + 7y = 39 ....... Eq.1 ,, you can write it in another form, x = 39 - 7y3x - 2y = 2 ....... Eq.2Put the value of Eq.1 on Eq.23(39 - 7y) - 2y = 2117 - 21y - 2y = 2117 - 23y = 223y = 115 ............ Y = 5to find the value of X , put the value of Y in Eq.1x = 39 - 7yx = 39 - 7(5)x = 39 - 35 .............. X = 4and so in both ways....you got the same answer choose the easiest way for you :)good luck my friend...
(3x-2y)2=
x+7y=16 ---------------------------1 3x- 7y=4 ---------------------------2 ------------------------------------------------------------------------------------------------------------------ 1+2 x+7y=16 3x-7y=4 4x=20 x=5 substitute x = 5 in x+7y=16 (5)+7y=16 7y=11 y=11/7
3xy-2y=0 3xy=2y y=2y (3x) y/2y=3x 1/2=3x multiply across by 2 1=6x 1/6=x therefore substituting x=1/6 into 3xy-2y; 3(1/6)y-2y=0 1/2y=2y y=2y/0.5 0.5 aka 1/2 y=1
No, it has an infinite number of solutions. The coordinates of each and every point on the line 3x + 2y + 4 = 0 is a solution.
If you mean: x+7y = 39 and 3x-2y = 2 Then by substitution: x = 4 and y = 5
x + 7y = 39 So x = 39 - 7y Substitute for x in the second equation: 3(39 - 7y) - 2y = 2 117 - 21y - 2y = 2 115 = 23y y = 5 Substitute this value for y back into the first equation: x + 7*5 = 39 x + 35 = 39 x = 4
From first equation: x = 39 - 7ySubstitute this in second equation: 3(39 - 7y) - 2y = 2ie 117 - 21y - 2y = 2ie 115 = 23yie y = 5, making x = 4.
x + 7y = 39 ....... Eq.13x - 2y = 2 ....... Eq.2multiply Eq.1 with (-3) and so Eq.1 will be :-3x - 21y = -1173x - 2y = 2_________________ by Elimination method ( Eq.1 + Eq.2 ) the result will we :- 21y - 2y = - 115-23y = -115Y = 5to find the value of x, put the value of y in Eq.1 :x + 7y = 39x = 39 - 7yx = 39 - 7(5)X = 4___________________________________________________________you can solve it in another way by using substitution method:x + 7y = 39 ....... Eq.1 ,, you can write it in another form, x = 39 - 7y3x - 2y = 2 ....... Eq.2Put the value of Eq.1 on Eq.23(39 - 7y) - 2y = 2117 - 21y - 2y = 2117 - 23y = 223y = 115 ............ Y = 5to find the value of X , put the value of Y in Eq.1x = 39 - 7yx = 39 - 7(5)x = 39 - 35 .............. X = 4and so in both ways....you got the same answer choose the easiest way for you :)good luck my friend...
x + 7y = 39 ....... Eq.13x - 2y = 2 ....... Eq.2multiply Eq.1 with (-3) and so Eq.1 will be :-3x - 21y = -1173x - 2y = 2_________________ by Elimination method ( Eq.1 + Eq.2 ) the result will we :- 21y - 2y = - 115-23y = -115Y = 5to find the value of x, put the value of y in Eq.1 :x + 7y = 39x = 39 - 7yx = 39 - 7(5)X = 4___________________________________________________________you can solve it in another way by using substitution method:x + 7y = 39 ....... Eq.1 ,, you can write it in another form, x = 39 - 7y3x - 2y = 2 ....... Eq.2Put the value of Eq.1 on Eq.23(39 - 7y) - 2y = 2117 - 21y - 2y = 2117 - 23y = 223y = 115 ............ Y = 5to find the value of X , put the value of Y in Eq.1x = 39 - 7yx = 39 - 7(5)x = 39 - 35 .............. X = 4and so in both ways....you got the same answer choose the easiest way for you :)good luck my friend...
7y + 2y = 9y 7y * 2y = 14y^2
It is: 2(3x-7y) = 6x-14y
x-2y=1 => x=2y+1 (1) 3x+y=-4 (2) sub (1) into (2) 3(2y+1)+y=-4 6y+3+y = -4 7y+3 = -4 7y = -7 y = -1 sub back into (1) x = 2(-1)+1 x = -2+1 x = -1
(3x-2y)2=
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3
x + 7y = 33 (1) --> x = 33 - 7y (1a) 3x - 2y = 7 (2) To solve the simultaneous equations, starts by substituting the value for x found in (1a) into (2): 3(33 - 7y) - 2y = 7 99 - 21y - 2y = 7 -23y = -92 y = 4 Now substitute this value for y into (1a) and solve for x: x = 33 - 7(4) x = 33 - 28 x = 5 Therefore these equations have the solution x = 5, y = 4