Wiki User
∙ 13y agoNO!
Wiki User
∙ 13y agoThere is no such thing as "solving integers". You can solve an equation, which means finding all the unknowns in that equation, but you can't solve an integer.
You solve the equation.
You first have to get rid of the numbers that don't have variables. then you divide by the variable and solve for it.
an equation is a number sentence with and = sign
two algebraic steps (not counting simplification steps) must be used to solve the problem. 1st step is adding or subtracting some quantity from both sides of the equation. 2nd step is multiplying or dividing by a number to solve.
You're allowed to do that, yes. Whether or not it "solves" any particular equation depends on the equation itself.
by adding, subtracting, dividing, and multiplying.
There is no such thing as "solving integers". You can solve an equation, which means finding all the unknowns in that equation, but you can't solve an integer.
In general, if you apply the same operation to both sides of an equation, you get an equivalent equation - at least if you do simple things like adding, subtracting, multiplying by a non-zero number, and dividing by some number.
You solve the equation.
You first have to get rid of the numbers that don't have variables. then you divide by the variable and solve for it.
two algebraic steps (not counting simplification steps) must be used to solve the problem. 1st step is adding or subtracting some quantity from both sides of the equation. 2nd step is multiplying or dividing by a number to solve.
an equation is a number sentence with and = sign
It depends on the problem. An integer subtraction can be one number, take away another number.
Given a number n, solve the equation x^2 + x - 2n = 0.If n is a triangular number then one solution for x will be a positive integer. Then n is the xth triangular number.
To solve this question we would need to know if you are adding or subtracting. Unless the equation is set up to equal 360 then the equation has no rationale answer.
ok this is what it looks like to me...I definitely could be wrong. one integer is three more than twice another integer if one integer is three more than another number (x+3=y) Then you just multiply "the other number" by 2. (x+3=2*y) Sum of two integers is 36....so add them (x+y=36) Solve that equation for one number so...(y=36-x) Then you would plug that equation into the first equation where there is a y: (x + 3 = 2*(36-x))