Substitute the values for the two variables in the second equation. If the resulting equation is true then the point satisfies the second equation and if not, it does not.
If it has infinite number of solutions that means that any ordered pair put into the system will make it true. I believe the relationship of the graphs question your asking is that tooth equations will probably be the same line
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.
If you're talking about two linear equations, make sure they are not parallel. If you're talking about quadratics, make sure that b2-4ac is not negative.
Without any equality signs the expressions given can't be considered to be equations
The answer depends on whether they are linear, non-linear, differential or other types of equations.
Without any equality sign and not knowing whether or not 2 is + or they cannot be classed as equations
The idea is to replace one variable in the equation by the first number in the ordered pair, the other variable with the second number in the ordered pair, do the calculations, and see whether the resulting expressions are indeed equal.
Tell whether the ordered pair (5, -5) is a solution of the system
If it has infinite number of solutions that means that any ordered pair put into the system will make it true. I believe the relationship of the graphs question your asking is that tooth equations will probably be the same line
There are various methods which depend on the nature of the equation(s) and whether or not the equations can be solved analytically.
The answer will depend on the form of the equation. Whether it is an equation in one or more variables, whether it is linear or polynomial, there are different standard forms for exponential equations.
You substitute the coordinates of the point in the equation. If the result is true then the point is a solution and if it is false it is not a solution.
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.
the expression "b2-4ac" with respect to quadratic equations is called the discriminant. the discriminant of the equation tells whether or not the roots will be real numbers or not. If the discriminant is negative, then the roots are imaginary.
Checking your solution in the original equation is always a good idea,simply to determine whether or not you made a mistake.If your solution doesn't make the original equation true, then it's wrong.
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution. Furthermore, it is not clear whether there is meant to be a symbol between 6x and 7y. Well the guy failed at asking a question but im sure he means 6x+7y=96 Whoever can solve that would be nice. Answer should be in a ordered pair. * * * If s/he means 6x + 7y = 96, then this equation, of a line in two dimensional space is the set of solutions. Each and every one of the infinite number of points on that line is a solution.
If you're talking about two linear equations, make sure they are not parallel. If you're talking about quadratics, make sure that b2-4ac is not negative.