From first equation, -y = 3x + 3.
Substitute in second equation: -3x + 5(3x + 3) = -21
ie 12x = -36 so x = -3 and y = -(-9 + 3) = 6.
Easier method: subtract first equation from second giving -4y = -24 so y = 6, this in first equation gives -6 = 3x + 3, ie 3x = -9 so x = -3
There are no disadvantages. There are three main ways to solve linear equations which are: substitution, graphing, and elimination. The method that is most appropriate can be found by looking at the equation.
The answer depends on the nature of the equations. For a system of linear equations, the [generalised] inverse matrix is probably simplest. For a mix of linear and non-linear equations the options include substitution, graphic methods, iteration and numerical approximations. The latter includes trail and improvement. Then there are multi-dimensional versions of "steepest descent".
Substitution is used in Systems of Linear Equations. When you have two equations, such as y=3x+1 and 2y+5=9x+1, then substitution can be used. 2y could be substituted with 2(3x+1). You would have now the equation 6x+2=9x-1, which is more solvable.
Simultaneous equations have at least two unknown variables.
A system of linear equations is two or more simultaneous linear equations. In mathematics, a system of linear equations (or linear system) is a collection of linear equations involving the same set of variables.
There are no disadvantages. There are three main ways to solve linear equations which are: substitution, graphing, and elimination. The method that is most appropriate can be found by looking at the equation.
Elimination and substitution are two methods.
In that instance, it means that the lines never touch.
By the substitution method By the elimination method By plotting them on a graph
There is no simple answer. Sometimes, the nature of one of the equations lends itself to the substitution method but at other times, elimination is better. If they are non-linear equations, and there is an easy substitution then that is the best approach. With linear equations, using the inverse matrix is the fastest method.
If you already know that x = -3 and y = 5 what linear equations are you wanting to solve?
Elimination and substitution are two methods.
The answer depends on the nature of the equations. For a system of linear equations, the [generalised] inverse matrix is probably simplest. For a mix of linear and non-linear equations the options include substitution, graphic methods, iteration and numerical approximations. The latter includes trail and improvement. Then there are multi-dimensional versions of "steepest descent".
Yes, because non-linear equations involve x2.
You would solve them in exactly the same way as you would solve linear equations with real coefficients. Whether you use substitution or elimination for pairs of equations, or matrix algebra for systems of equations depends on your requirements. But the methods remain the same.
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Substitution is used in Systems of Linear Equations. When you have two equations, such as y=3x+1 and 2y+5=9x+1, then substitution can be used. 2y could be substituted with 2(3x+1). You would have now the equation 6x+2=9x-1, which is more solvable.