x + y = 1
x + y = 2
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Simultaneous equation is nothing: it cannot exist.A system of simultaneous equations is a set of 2 or more equations with a number of variables. A solution to the system is a set of values for the variables such that when the variables are replaced by these values, each one of the equations is true.The equations may be linear or of any mathematical form. There may by none, one or more - including infinitely many - solutions to a system of simultaneous equations.
Simultaneous Equations are very helpful because it can help u solve problems in real life. There are 2 ways to approach a simultaneous equation, Substitution and elimination method. As a good practice it is always good to practice your substitution method first. I wont go too advance for now but consider this question;Find two numbers whose sum is 21 and difference is 9.This question requires 2 equation to solve; thus it is call simultaneous equation.Solve: Let x be a number, and Let y be another number.x + y = 21 equation 1x - y = 9 equation 2Rearrange equation 2 to make equation 3(Equation 3 is just to sub into the other eqs)x = 9 + y equation 3Sub equation 3 into 1(9 + y) + y = 219 + 2y = 212y = 12y = 6 First solution!Sub y = 6 into equation 2x - 6 = 9x = 15 Second Solution!Therefore, the numbers are 15 and 6.In a simultaneous equation (with 2 variable) there will always be 2 answers.This is copied from my other worked examples. I do not really understand your question. If you have a simultaneous equation that you can't solve. Post it up and i will help.* * * * *Good answer, but spoiled by the last-but-one paragraph. Simultaneous linear equations with two variables can have no solutions (if the corresponding graphs are distinct parallel lines) or infinitely many solutions (if they are, in effect, the same line). And then, there are always simultaneous non-linear equations. Two quadratics, for example, can have 0, 1, 2 or infinitely many solutions.
Equations of the form z^4+az^2+a_0 are known as biquadratic equations. They are quartic equations. In general they can be solved by reducing them to a quadratic equation where x=z^2 is the variable. Then you can use the quadratic formula or factor. So plugging in x to the biquadratic giives us x^2+ax+a_0.
1 If: 2x+5y = 16 and -5x-2y = 2 2 Then: 2*(2x+5y =16) and 5*(-5x-2y = 2) is equvalent to the above equations 3 Thus: 4x+10y = 32 and -25x-10y = 10 4 Adding both equations: -21x = 42 or x = -2 5 Solutions by substitution: x = -2 and y = 4
z + 8 - 2 is an expression, not an equation. An expression cannot be solved. At best, you can simplify it to z + 6.