To be a linear equation, the equation must be set equal to Y. Also, it can't have any square roots, or any variables on the bottom of a fraction.
In general, the terms of a linear equation must be either first-degree polynomials with respect to the variables, constants, or products of the two. This disallows terms involving trigonometric, logarithmic, exponential, hyperbolic, and power expressions (except for the power of 1) and their inverses.
By definition, there cannot be a simultaneous equation that cannot be solved, there must be a set of simultaneous equations. It is important to realise that simultaneous equations need not be linear.It is simple to devise a pair of linear equations that are inconsistent:x + y = 1 and x + y = 2There is no solution. Graphically, the two lines are parallel.Another possibility isx + y = 1 and 2x + 2y = 2In this case there are an infinite number of solutions. Graphically, the two lines are coincidet, so that every point on the common line is a solution. There is, therefore, no unique solution.Yet another situation can arise when the domain of the equations is restricted.For example,x2 + y2 = -1 where x and y are real along with any other equation in x and y.
linear?
Well I cant graph it for you but its a linear line some points would be (0,-1) (1,-2) (2,-3)(3,-4) (-1,0) (-2,1) (-3,2) and so on
The equation you provided, 3x - y = 1, cannot be solved for any specific values of x and y without additional information. It is a linear equation in two variables, so it represents a straight line on a graph. To find a solution, you would need another equation or constraint.
For vertical lines, when you try to figure out the slope, you get zero in the denominator - in other words, a division by zero.
you cant
By definition, there cannot be a simultaneous equation that cannot be solved, there must be a set of simultaneous equations. It is important to realise that simultaneous equations need not be linear.It is simple to devise a pair of linear equations that are inconsistent:x + y = 1 and x + y = 2There is no solution. Graphically, the two lines are parallel.Another possibility isx + y = 1 and 2x + 2y = 2In this case there are an infinite number of solutions. Graphically, the two lines are coincidet, so that every point on the common line is a solution. There is, therefore, no unique solution.Yet another situation can arise when the domain of the equations is restricted.For example,x2 + y2 = -1 where x and y are real along with any other equation in x and y.
Because linear lines can't intersect in two seperate places. They either intersect at one specific coordinate, or the lines are on top of each other and they intersect at every point.
You CANT convert a linear measurement to weight !
linear?
Because there is no such thing as "aliner" equation.
Well I cant graph it for you but its a linear line some points would be (0,-1) (1,-2) (2,-3)(3,-4) (-1,0) (-2,1) (-3,2) and so on
The equation you provided, 3x - y = 1, cannot be solved for any specific values of x and y without additional information. It is a linear equation in two variables, so it represents a straight line on a graph. To find a solution, you would need another equation or constraint.
You cant square feet is area and miles are linear.
For vertical lines, when you try to figure out the slope, you get zero in the denominator - in other words, a division by zero.
Not all linear functions have defined slope. In two dimension it is definet but in three dimensions it cant be defined; For that direction ratios are defined in mathematics.
You cant solve it unless it is an equation. To be an equation it must have an equals sign.