Let A and B be any number. Let's say you have (5-A)(10-B) = 0.
For this to be true, either (5-A) or (10-B) must be zero for the the answer to equal zero. Essentially what you are saying is 0(10-B)=0 or (10-A)0=0. This let's you solve for A and B.
You can just set each "factor" to zero and solve for the variable.
5-A=0
A=5
10-B=0
B=10
Another "un-factored" example:
x2 - 10x = -25
x2 - 10x + 25 = 0 <--- the factors have to be equal to zero for this to work
You need two numbers that add to negative 10 and multiply to 25.
(-5) + (-5) = -10
(-5) x (-5) = 25
Therefore:
(x-5)(x-5)=0
And x can only be equal to 5, x=5
Solve the equation - by whatever means available to you: factorising, graphical, numerical approximations, etc.
This is a quadratic equation in the form of: 52 - 25x + 30 = 0 Solve by using the quadratic equation formula: x = 2 or x = 3 Or by factorising: (5x-10)(x-3) = 0, which means that x = 10/5 or x = 3
solve it
If you solve such an equation for "y", you get an equation in the slope-intercept form.
you can only solve for one in an equation so it can equal something
Solve the equation - by whatever means available to you: factorising, graphical, numerical approximations, etc.
no bloody clue
It means you are required to "solve" a quadratic equation by factorising the quadratic equation into two binomial expressions. Solving means to find the value(s) of the variable for which the expression equals zero.
This is a quadratic equation in the form of: 52 - 25x + 30 = 0 Solve by using the quadratic equation formula: x = 2 or x = 3 Or by factorising: (5x-10)(x-3) = 0, which means that x = 10/5 or x = 3
Sure. You can always 'solve for' a variable, and if it happens to be the only variable in the equation, than that's how you solve the equation.
you don't answer an equation, you solve an equation
solve it
If you solve such an equation for "y", you get an equation in the slope-intercept form.
It is not an equation if it does not have an equals sign. You could simplify it but not solve it.
How do you use division to solve a multiplication equation?Answer this question…
you can only solve for one in an equation so it can equal something
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.