The coefficients and constant in one of the equations are a multiple of the corresponding coefficients and constant in the other equation.
They are a set of equations in two unknowns such that any term containing can contain at most one of the unknowns to the power 1. A system of linear equations can have no solutions, one solution or an infinite number of solutions.
Yes. In fact, you can have any number of solutions you want. Here is an example of an algebraic equation with three solutions:(x - 2)(x - 5)(x + 4) = 0 The solutions are 2, 5, 4, since for each of these, one of the factors will be zero, and therefore, the product will also be zero.
You need two independent equations to solve for two unknowns or variables (x and y).
There is only one type of solution if there are two linear equations. and that is the point of intersection listed in (x,y) form.
You have used an equation which is actually the same and tried to use that to solve for your two unknowns. The two equations must be different - you should not be able to transpose one to be equal to another, so therefore if you graphed the two equations for example on the number plane they would be different graphs. The point(s) of intersection would be your solutions.
By substitution or elimination in simultaneous equations.
They are a set of equations in two unknowns such that any term containing can contain at most one of the unknowns to the power 1. A system of linear equations can have no solutions, one solution or an infinite number of solutions.
You are given a system of n or more simultaneous linear equations involving n unknowns. Pick one of the unknowns, called the pivot variable. Find an equation in which it appears, called the pivot equation.
Yes. In fact, you can have any number of solutions you want. Here is an example of an algebraic equation with three solutions:(x - 2)(x - 5)(x + 4) = 0 The solutions are 2, 5, 4, since for each of these, one of the factors will be zero, and therefore, the product will also be zero.
That equation cannot be solved since there are 2 unknown in the equation (x and y) but only 1 equation. The number of unknowns must be equal to the number of equations (for simultaneous equations)
No i believe that with three unknowns you must have three equal equations. Hope this helps! -dancinggirl25
You need two independent equations to solve for two unknowns or variables (x and y).
There is only one type of solution if there are two linear equations. and that is the point of intersection listed in (x,y) form.
You have used an equation which is actually the same and tried to use that to solve for your two unknowns. The two equations must be different - you should not be able to transpose one to be equal to another, so therefore if you graphed the two equations for example on the number plane they would be different graphs. The point(s) of intersection would be your solutions.
A pair of simultaneous equations in two unknowns which are inconsistent - in the sense that there is no solution that simultaneously satisfies both equations. Graphically, the equations are those of two parallel lines (slope = 2). Since, by definition, they cannot meet there is no solution to the system.
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.
If there are less distinct equations than there are variables then there will be an infinite number of solutions.For example, you may have 3 equations with 3 unknowns, but if one of those equations is a multiple of another there there are only 2 distinct equations:2x + 3y + 5z = 1x + y - 2z = 104x + 6y + 10z = 2Equation (3) here is twice equation (2) so there are effectively only 2 distinct equations for 3 unknowns and thus there will be an infinite number of solutions. If any two equations are parallel then there is no solution; if equation (3) above was 2x + 3y + 5z = 2, then there are no solutions - subtracting equation 1 from (the new) equation 3 would result in 0 = 1 which is nonsense.