4 Answers | Add Yours
To do elimination, you need to make it so that one of the variable terms will cancel itself out. In other words, you need to be able to add two of one term together and get 0.
In this case, you can do this by making both one x term be 3x and the other be -3x.. The way to do that is to multiply the second equation by -1.5.
So then you would have the second equation being
3y - 3x = 12
Now you add the two equations together. That gets you
6y = -18
Which makes y = -3 when you divide both sides by 6.
Now you plug this value of y into one of the equations to solve for x. We'll do the second equation.
6 + 2x = -8
2x = -14
x = -7
First of all, we have to take a look to the both equations, to see what unknown could be canceled. In our case, it is obviously that the unknown x could be canceled easier than y, because the coefficients of the unknown x have opposite signs.
After choosing what unknown could be canceled, we have to find out which is the common denominator of the coefficients of the unknown x. In this case is 2*3=6.
The next step is to multiply the first equation with 2 and the second equation with 3, after that the both equations will be added.
2(3y+3x) + 3(-2y+2x)=-60-24
After reducing similar terms, we'll have:
After finding the unknown x, we'll substitute it into what equation we choose.
The given simultaneous linear equations are:
3y + 3x = - 30 ... (1)
- 2y + 2x = - 8 ... (2)
Multiplying equation (1) by 2 and equation (2) by 3 we get:
6y + 6x = - 60 ... (3)
- 6y + 6x = - 24 ... (4)
Adding equations (3) and (4) we get:
6y -6y + 6x + 6x = - 60 - 24
12x = - 84
Therefor x = -7
Substituting this value of x in equation (1) we get:
3y + 3*(-7) = - 30
3y - 21 = - 30
3y = - 30 + 21 = - 9
y = -9/3 = -3
x = -7, and y = -3
Both equations could be reduced. First by 3 and the second by 2.
(1) divided by 3 gives: y+x=-10..........(3)
(2)/2 gives: -y+x= -4...................(4)
(3)+(4) gives: 2x= -14. Or x=-14/2 =-4.
(3)-(4) gives: 2y = -10--4 = -6. Or y = -6/2 = -3.
We’ve answered 318,915 questions. We can answer yours, too.Ask a question