If the solution contains one variable which has not been fixed then there are infinitely many solution.
Such an equation has an infinite set of solutions. You can solve the equation for one variable, in terms of the other. Then, by replacing different values for one of the variables, you can get different solutions.
This equation has infinite solutions. For example x = 6, y = 5 satisfies the equation. (30 - 5 = 25)
No. There is not enough information in the equation x + 2y = 2, by itself, to solve it. There are an infinite number of solutions. A second equation, or information to allow a second equation to be derived, must be given to find a solution.
It is not appropriate to talk about "the" three solutions for this equations; the equation describes a straight line, and has an infinite number of solutions. Solve the equation for "y", substitute any value for "x", and calculate the corresponding value for "y", to get one opf these solutions.
This has an infinite number of solutions; choose any two numbers for your first two numbers, then solve for the third.This has an infinite number of solutions; choose any two numbers for your first two numbers, then solve for the third.This has an infinite number of solutions; choose any two numbers for your first two numbers, then solve for the third.This has an infinite number of solutions; choose any two numbers for your first two numbers, then solve for the third.
You cannot solve one equation in two unknowns. The given equation defines a line in the 2-dimensional plane and every point on the line is a solution. There are, therefore, an infinite number of solutions.
A linear equation in two variables will give an infinite number of solutions. These will comprise the coordinates of points along a straight line in 2-dimensional space.
Yes, that is often possible. It depends on the equation, of course - some equations have no solutions.
It very much depends on the equation. The procedure for solving an equation with just one variable is so very different from the procedure for finding solutions to non-linear equations in several variables.
Ya can't. Ya got two unknowns there. To solve for two unknowns, ya gotta have two equations. Widout anudder equation, ya got a infinite number of solutions.
The question contains an algebraic expression but, since there is no equality sign, there is no equation to solve.
-- Pick a number out of a hat or a telephone book, or ask the person standing next to you to give you a number. -- Assign that number to one of the variables. -- Solve the equation for the other variable. -- This gives you one "ordered pair" solution of the equation. -- Repeat, as many times as you want. You will never run out of solutions, and you will never find all of them, as there are an infinite number of them.
There are many kinds of differential equations and their solutions require different methods.
An "extraneous solution" is not a characteristic of an equation, but has to do with the methods used to solve it. Typically, if you square both sides of the equation, and solve the resulting equation, you might get additional solutions that are not part of the original equation. Just do this, and check each of the solutions, whether it satisfies the original equation. If one of them doesn't, it is an "extraneous" solution introduced by the squaring.
There are an infinite number of solutions to a single equation in two variables. So, in order to save myself time, the first thing I would do is go out and look for another equation to put along with the first one.
A linear equation is that of a straight line. Any one of the infinitely many points on the line will be solutions. If the equation is in terms of the variables x and y, just pick any two values of x, solve for y and the results will be the coordinates of two solutions.
it is actually an equation of a straight line. so it will give infinite values of ordered pairs, which are the solution of this equation.
It really depends on the type of equation. Sometimes you can know, from experience with similar equations. But in many cases, you have to actually do the work of trying to solve the equation.
Since there are 2 unknowns, it takes 2 independent equations to solve for unique values. So far, from the one given equation, all we know is that [ x = 5 - y ] and [ y = 5 - x ]. There are an infinite supply of number-pairs ... solutions ... that satisfy those conditions.
x + 52 + 2y + 35 = 3 x + 2y = 3 - 52 -35 x + 2y = -84 Infinite number of solutions, are you sure you posted the correct equation?
The equation has no solution. Since if you combine like terms, you get 0=-10 which isn't true, there are no solutions to the equation.
When trying to solve an equation and you end up with the exact same number on both sides , like 10=10 then the equation has infinitely many solutions. If you end up with 2 different number on both side of the equation, like 3=5 then the equation has no solution. If there is a variable on one side and a number on the other, then there is one solution, e.g. x=4. In the equation 10=10 there is no variable such as x or y that we are trying to find the solution for. The equation x=x might be said to have an infinite number of solutions, because you can choose any value you like for x. More often you would say that "x is indeterminate". So if your equation always turns out to be satisfied for any x you choose, then there is an infinity of solutions and the equation does not represent anything useful. Or, for example, it may have a result such as "true for all even numbers", and you may not be aware in advance that this might happen. Or another example might be sin(x)=0 which has solutions for all values for those x which are integer multiples of 180 degrees. The only way is to solve the equation and see what appears.
Sure. You can always 'solve for' a variable, and if it happens to be the only variable in the equation, than that's how you solve the equation.
w^2 +/- 28w - 1 is an expression, not an equation. Expressions do not have solutions.
you don't answer an equation, you solve an equation
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