Algebraic System Question: Solve the system by addition?

Question

{2x-14y=8
{-x+7y=-4

The answer must be in a form similar to: (#, #) for example (–2, –5) OR (7, –3) BUT neither examples is the correct answer for this particular problem I was just giving a form example

Answer 1

There is no solution to this system of linear equations. If you try to solve using the standard method (invert the coefficient matrix) then the determinant of the matrix is 0 meaning no unique solution.

Now, if you select any value for y then calculate x as equal to 7y+4, then the resulting (x,y) pair is a solution to both equations. This of course means that there are infinitely many solutions (x,y) in the Real numbers.

Answer 2

PLUG IN!!!

Choose a problem, get the variable to one side, then plug it into the other problem.

2x - 14y = 8
becomes
y = (-4 + x) / 7

Plug it into the other equation.

-x + 7y = -4
becomes
-x + 7[(-4 + x)/7] = -4
work it out
-x - 4 + x = -4

-4 = -4

That's as far as I'm going, because on paper, both variables cancel out but the final simplification of the two are equal as shown above while looking for the y variable. x reacted the same way.

Answer 3

To solve by additions this is what you do

2x-14y=8
-x+5y=-4

We need to solve for one variable then solve for the other. To use the addition method, whichever variable you want to get rid of has to be matched negatively in the other equation. Getting rid of the x's, multiply the bottom equation by 2 then add them together.

2x-14y=8
-2x+14y=-8

as you can see here, these equations alone can not be used to solve for X and Y because they are not independent. Meaning that another equation will be necessary to solve this sytem.

Answer 4

I agree with Jay, u cannot solve this equation..
2x-14y=8 eqn 1
-x+7y=-4 eqn 2
multiply eqn 2 by 2
we get
-2x+14y=-8 eqn 3
adding eqn 1 and 3 you will get
0x+0y=0 which means they are dependent.

Answer 5

2x - 14y = 8
-x + 7y = -4

(2x - 14y) + (-x + 7y) = 8 + (-4)
2x - 14y - x + 7y = 8 - 4
x - 7y = 4
x = 7y + 4

2x + 14y = 8
2(7y + 4) + 14y = 8
14y + 8 + 14y = 8
28y + 8 = 8
28y = 0
y = 0

x = 7(0) + 4
x = 0 + 4
x = 4

x = 4
y = 0

Answer 6

First one make y = to 0 so it is 2X = 8 divide both sides by 2 so it would be X = 4 so (4,0)
Then make X = 0 and do the same.

Answer 7

Two equations, two variables (x & y).

What you do is multiply the second equation by 2 to get:
-2x + 14y = -8

then add that to the first equation getting:

0x + 28y = 0

thus y must be 0.

if you put y=0 into either original equation you get x=4.

so your answer is (4,0).

Answer 8

sure, you go with the sector of rationals in case you employ branch (or fractions). i think of you're utilising greater steps than you go with. For "removal" i might upload the two equations in the present day, acquiring 2x = 6 Then divide the two area by ability of two acquiring x =3, then plug into different eqn to get y = 2. Your "substitution" steps are actually not overdone, probable. BTW i think of the duty you have been given is remarkably stupid (and that i'm a PhD scientist), yet there is not something you're able to do approximately that, i think.

Answer 9

Multiplying both sides of the 2nd equation by -2 gives a "new" equation which is identical to the first one -- which means, really, that you don't have enough info to come up with a unique ordered-pair solution. (In other words, the system is "dependent.")

Answer 10

well if u reverse the sighns of the second equation and double it well then u get 2 of the same line ie. they are parallel in that case u cannot find the x and y values as they intersect at every point therefore my answer is
(infinate points, infinate points)