Assume something (e.g. equations) using k then prove k+1 using k.
There is no quadratic equation that is 'linear'. There are linear equations and quadratic equations. Linear equations are equations in which the degree of the variable is 1, and quadratic equations are those equations in which the degree of the variable is 2.
Linear Equations are equations with variable with power 1 for eg: 5x + 7 = 0 Simultaneous Equations are two equations with more than one variable so that solving them simultaneously
Say you have a pair of equations like a+b=1 and 2a+3b=5. To solve using substitution you would set one equation against one variable (say a=1-b) then plug that into the other (2(1-b)+3b=5), then solve for the one variable you have there and then plug that result into the other equation (in this case b=3 and a=-2).
To solve for 2 unknown variables you need 2 independent equations. You have only 1.
just create a mathematical equation using 1964 such as 1x1964 2x1964 3x1964 4x1964 or 1964/1 1964/2 etc
100
((9/9)-(9/9)+(9/9)) = 1
If you mean: x2+2x+1 = 0 then it is a quadratiic equations whose solutions are equal because x = -1 and x = -1
Assume something (e.g. equations) using k then prove k+1 using k.
All of them, but it does depend on what operators are allowed.
100-1 101-2 102-3
(2+3)(5)=(4)(6)+1 Is this what you mean?
3(5x-2y)=18 5/2x-y=-1
1= 3/3 2=(3+3)/3 3= 3+3-3 4=(3+3+3+3)/3 5=(3+3+3+3+3)/3 6=(3+3+3+3+3+3)/3 and so on
If you mean: 6x-3y = -33 and 2x+y = -1 Then solving the simultaneous equations by substitution: x = -3 and y = 5
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution. .