You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.
It is simply: 7y whereas y is the unknown variable
If this involves two equations both containing the same two unknowns, then multiply (or divide) one of the equations so that the absolute value of one of the unknowns is now the same in both equations. For example, x + 2y = 11 : 3x - 57 = 13 : Multiply the first equation by three, 3x + 6y = 33 so that the 'x' terms in this and the second equations are equal. In this example they both have the same (positive) sign - see below.If this unknown has identical signs (both are positive or both are negative) then subtract one equation from the other to eliminate that unknown.If this unknown has different signs (it is positive in one equation and negative in the other equation) then add the equations together to eliminate that unknown.This will enable the value of the remaining unknown to be determined and by substitution the value of the eliminated unknown can then be found.
1 term = Monomial2 term = Binomial3 term = trinomialNo standard for 4, or any larger fixed number of terms, but "polynomial" is used when the number of terms is unknown.
The answer depends on the level of your knowledge. The High level, simple answer is first. The Low level slog follows:HIGH LEVEL, SIMPLESuppose you have n equations of the forma11x1 + a12x2 + ... + a1nxn = bn wherethe as are coefficients,x1, x2, ... xn are the unknown variablesandb1, b2, ... bn are the constants.Write the n linear equations in n unknowns in the form Ax= bwhereA is an n*n matrix of coefficientsx is the n*1 matrix of the unknown variablesandb is the n*1 matrix of the constants.Find the inverse of A.Then x = A-1b.The above method works if the system has a unique solution. If the n equations are not independent, you will need to use a generalised inverse and that starts to get rather complicated. If they are inconsistent, then neither the inverse nor generalised inverse will be found.LOW LEVEL SLOGUse the first equation to express x1 in terms of the other variables. Substitute this value for x1 in the remaining n-1 equations. You now have n-1 equations in n-1 unknown variables.Use the first of the new equations to express x2 in terms of the other variables. Substitute in remaining equations. You now have n-2 equations in n-2 unknown variables.Continue until you have 1 equation in 1 unknown.That will be of the form pxn = q so that xn = q/p.Substitute this value into one of the equations at the 2-equations-in-2-unknowns stage. That will give you xn-1.Work your way back to the top.The two methods are equivalent. There are shortcuts available for matrix inversion (eg using determinants), but these are too complicated to go into here.
You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.You find, or construct, an equation or set of equations which express the unknown variable in terms of other variables. Then you solve the equation(s), using algebra.
In simple terms all that it means that there are more solutions than you can count!If the equations are all linear, some possibilities are given below (some are equivalent statements):there are fewer equations than variablesthe matrix of coefficients is singularthe matrix of coefficients cannot be invertedone of the equations is a linear combination of the others
Equations always contain an
Letters and symbols are used in equations to represent quantities, sometimes known or unknown. For example the symbol π (pi) represents the constant 3.14… and is unchanging. X or Y in an equation represent unknown terms that may vary depending on the equation, and other variables in the equation. There are numerous examples of both symbols and letters used in mathematical equations, but almost always they are either referring to a known constant term that is more simply expressed as a symbol, or an unknown or general term yet to be concretely determined.
It is simply: 7y whereas y is the unknown variable
If this involves two equations both containing the same two unknowns, then multiply (or divide) one of the equations so that the absolute value of one of the unknowns is now the same in both equations. For example, x + 2y = 11 : 3x - 57 = 13 : Multiply the first equation by three, 3x + 6y = 33 so that the 'x' terms in this and the second equations are equal. In this example they both have the same (positive) sign - see below.If this unknown has identical signs (both are positive or both are negative) then subtract one equation from the other to eliminate that unknown.If this unknown has different signs (it is positive in one equation and negative in the other equation) then add the equations together to eliminate that unknown.This will enable the value of the remaining unknown to be determined and by substitution the value of the eliminated unknown can then be found.
1 term = Monomial2 term = Binomial3 term = trinomialNo standard for 4, or any larger fixed number of terms, but "polynomial" is used when the number of terms is unknown.
The answer depends on the level of your knowledge. The High level, simple answer is first. The Low level slog follows:HIGH LEVEL, SIMPLESuppose you have n equations of the forma11x1 + a12x2 + ... + a1nxn = bn wherethe as are coefficients,x1, x2, ... xn are the unknown variablesandb1, b2, ... bn are the constants.Write the n linear equations in n unknowns in the form Ax= bwhereA is an n*n matrix of coefficientsx is the n*1 matrix of the unknown variablesandb is the n*1 matrix of the constants.Find the inverse of A.Then x = A-1b.The above method works if the system has a unique solution. If the n equations are not independent, you will need to use a generalised inverse and that starts to get rather complicated. If they are inconsistent, then neither the inverse nor generalised inverse will be found.LOW LEVEL SLOGUse the first equation to express x1 in terms of the other variables. Substitute this value for x1 in the remaining n-1 equations. You now have n-1 equations in n-1 unknown variables.Use the first of the new equations to express x2 in terms of the other variables. Substitute in remaining equations. You now have n-2 equations in n-2 unknown variables.Continue until you have 1 equation in 1 unknown.That will be of the form pxn = q so that xn = q/p.Substitute this value into one of the equations at the 2-equations-in-2-unknowns stage. That will give you xn-1.Work your way back to the top.The two methods are equivalent. There are shortcuts available for matrix inversion (eg using determinants), but these are too complicated to go into here.
In order to find the unknown term in a number sequence, you first need to calaculate the advantage of the numbers.
You are supposed to: 1) Write two equations, one for the number of heads (express it in terms of the number of cows and horses, which of course are expressed as variables), and one for the number of legs (also expressed as an equation in terms of the number of cows and horses. 2) Solve the two equations simultaneously.
You cannot solve one linear equation in two unknown variables (x and y), although some non-linear equations will suffice. You need two independent linear equations. All you can do is express one of the variables in terms of the other, but that is not solving the equation.
Start by collecting like terms...