x = 4y
3x - 2y = 20
Substituting the value of x into the second equation,
3*(4y) - 2y = 20 ie 12y - 2y = 20 or 10y = 20 so that y = 2
Then x = 4y = 4*2 = 8.
So (x , y) = (8 , 2)
the system of equations 3x-6y=20 and 2x-4y =3 is?Well its inconsistent.
28+28=56
It looks like you have 2 simultaneous equations with 2 variables:4x + 8y = 20 and -4x + 2y = -30. Solution is {x=7, y = -1}.One way to solve:Add the two equations together, combining like terms: (4x - 4x) + (8y + 2y) = 20-30 --> 0 + 10y = -10 --> y = -1. Substitute this into either of the original equations and solve for x=7, then check in the other equation to make sure you calculated correctly.
The answer is 20.
Linear algebra is used to analyze systems of linear equations. Oftentimes, these systems of linear equations are very large, making up many, many equations and are many dimensions large. While students should never have to expect with anything larger than 5 dimensions (R5 space), in real life, you might be dealing with problems which have 20 dimensions to them (such as in economics, where there are many variables). Linear algebra answers many questions. Some of these questions are: How many free variables do I have in a system of equations? What are the solutions to a system of equations? If there are an infinite number of solutions, how many dimensions do the solutions span? What is the kernel space or null space of a system of equations (under what conditions can a non-trivial solution to the system be zero?) Linear algebra is also immensely valuable when continuing into more advanced math topics, as you reuse many of the basic principals, such as subspaces, basis, eigenvalues and not to mention a greatly increased ability to understand a system of equations.
The solutions are: x = 4, y = 2 and x = -4, y = -2
the system of equations 3x-6y=20 and 2x-4y =3 is?Well its inconsistent.
There are two equations in the question, both of which are wrong. There is no single fraction which will make both equations correct.
The two equations represent the same straight line.
28+28=56
Let the two numbers be x and y. Our two equations can then be written as: xy = 20 and x + y = 1 These equations have no real solutions. However, they do have complex solutions. x= 0.5 + 4.44i and y = (0.5 - 4.44i) were i is the square root of negative 1.
x = 4 or x = -5
There are infinite possible solutions: For example: 5/20 + 5/20 = 10/20 3/20 + 7/20 = 10/20
This cannot be solved for numerical answers. There must be 2 equations if there are 2 variables.
5x2 - 2y2 = -20 7x2 - y2 = 152 Eq1 - 2*Eq2: -9x2 = -324 so that x2 = 36 Substituting the value of x2 in Eq1: 2y2 = 200 so that y2 = 100 The four solutions are (-6, -10), (-6, 10), (6, -10) and (6,10)
Add the two equations you get 5y = 60 so y = 12 and x = 2
This cannot be solved. If you have 2 unknowns (a,b) then you must have 2 equations.