7x+3y=4
-35x-15y=-20
please explain the system of elimination for me
and if the system is inconsistent or dependent equation says so
the answer should look like { (x,y) l_____}
-35x-15y=-20
please explain the system of elimination for me
and if the system is inconsistent or dependent equation says so
the answer should look like { (x,y) l_____}
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Multiply the first equation by -5 and you will see that it becomes the second equation. So, there is really only one equation. There are infinitely many solutions, namely { (x,y) l 7x+3y=4 }.
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7x+3y:4.........1
-35x-
-35x-
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Equation A: 7x + 3y = 4
Equation B: -35x - 15y = -20
Now, we either need to eliminate "x" or "y". I will eliminate "Y" by multiplying equation A by 5 which will get me some other values and of course, 3y(5) = 15y and when I "add the equations" eliminating "Y" like so...
15(y) + (-15y) = 0 ........Now, all we have is "x's"
, I can cancel out the "Y" leaving only "x's". Also note that if you get in a jam and you have some negative value frustration you can actually "subtract" the equations [ if it is appropriate].
(Equation A)(5) ----------> (7x + 3y = 4)(5) --------> 35x + 15y = 20
Equation B [do not change] : ------------------------> -35x - 15y = -20
Yeah, the two lines are the same afterall.
Equation B: -35x - 15y = -20
Now, we either need to eliminate "x" or "y". I will eliminate "Y" by multiplying equation A by 5 which will get me some other values and of course, 3y(5) = 15y and when I "add the equations" eliminating "Y" like so...
15(y) + (-15y) = 0 ........Now, all we have is "x's"
, I can cancel out the "Y" leaving only "x's". Also note that if you get in a jam and you have some negative value frustration you can actually "subtract" the equations [ if it is appropriate].
(Equation A)(5) ----------> (7x + 3y = 4)(5) --------> 35x + 15y = 20
Equation B [do not change] : ------------------------> -35x - 15y = -20
Yeah, the two lines are the same afterall.