It helps to set these up vertically.
5x +5y = -13
7x - 3y = 17
Now, I can see that it will be easiest to cancel my y's, I just need to modify the equations so that the coefficient is the same in both equations. I will multiply the top equation by 3 and the bottom equation by 5 to get,
15x + 15y = -39
35x - 15y = 85; Now, add the 2 equations and the y's will cancel.
50x = 46; x = 0.92
Plugging x back into any equation, perhaps the top one, will give
4.6 + 5y = -13
5y = -17.6
y = -3.52
ANS: (0.92,-3.52)
Solve the system by the elimination method 5x 5y-13 7x-3y17what is the solution to the system?
By elimination: x = 3 and y = 0
(2,-2)
y=16 x= -4
16
Solve the system by the elimination method 5x 5y-13 7x-3y17what is the solution to the system?
By elimination: x = 3 and y = 0
(2,-2)
y=16 x= -4
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
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.
by elimination,substitution or through the matrix method.
You can solve lineaar quadratic systems by either the elimination or the substitution methods. You can also solve them using the comparison method. Which method works best depends on which method the person solving them is comfortable with.
the answer