By elimination: x = 3 and y = 0
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
x = 3 and y = 3 Form a simultaneous equation and solve by elimination.
Solving by the elimination method: x = 7 and y = 2
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.
You cannot solve one linear equation in two variables. You need two equations that are independent.
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
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.
How do you solve 4y plus x equals 8
(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.
Solving these simultaneous equations by the elimination method:- x = 1/8 and y = 23/12
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)
-2 plus 5 equals +3
No solution. The two lines are parallel and hence never intersect hence no solution.
Not enough information to solve
Solving the above simultaneous equations by means of the elimination method works out as x = 2 and y = 3
A is 14