The elimination method only works with simultaneous equations, hence another equation is needed here for it to be solvable.
x + 4y = -14 eqn12x + 3y = 13 eqn2Using the elimination method we multiply eqn1 by 22x + 8y = -28 eqn1bSubtract eqn2 from eqn1b2x + 8y = -28 eqn1b2x + 3y = 13 eqn25y =-41y = -8 and 1/5substituting this into eqn1 we getx +4 (-41/5) = -14x = (14*5) / (41 *4)x = (35/82)
Solving these simultaneous equations by the elimination method:- x = 1/8 and y = 23/12
z=pq
Solve this simultaneous equation using the elimination method after rearraging these equations in the form of: 3x-y = 5 -x+y = 3 Add both equations together: 2x = 8 => x = 4 Substitute the value of x into the original equations to find the value of y: So: x = 4 and y = 7
y=16 x= -4
(2,-2)
By elimination: x = 3 and y = 0
16
The elimination method only works with simultaneous equations, hence another equation is needed here for it to be solvable.
Yes and it works out that x = 3 and y = 4
2x + 2y = 44x + y = 1There are many methods you can use to solve this system of equations (graphing, elimination, substitution, matrices)...but no matter what method you use, you should get x = -1/3 and y = 7/3.
Solving by the elimination method: x = 7 and y = 2
Usually elimination is used on two equations and is called linear combination. You could solve for "y." That is customary. 2x+3y=1 3y=-2y+1 y=(-2/3)x+1/3
You cannot solve one linear equation in two variables. You need two equations that are independent.
(A) x + 6y = 19 (B) -x + 6y = 17 (A)+(B): 12y = 36 so that y = 3 Substituting this value of y in (A), x + 18 = 19 so x = 1.
The answer is that it cannot be done. To solve a set of equations in k variables (in this case, 2) you need at least two independent equations.