What is this crap? I don't understand. Create your own equation written in standard form. Determine any point that is a solution and will be on this line. Justify your answer algebraically.
@Luigi0210 HALP :d
Linear equation? or polynomial? etc.?
Linear equation thats what lesson thing its in
Is this for Algebra 2 ?
Alg 1
Oh okay . Do you know the formula ?
not really
@kdog771998 can u help
If it's a linear, then standard form is something like: Ax+By=C And @kdog771998 that's slope-intercept form
Lol oppsss .
i have no clue so i just want anyone to help :d
I was confused with algebra 2 standard form thats ax^2 + bx + c
For quadratic equations, yes.
Okay so then the formulas Ax + By=C
so could the equation be 10x + 14b = c?
No.
C is just a constant, any real number.
Ax+By = C A and B are coefficients, x and y are variables, C is a constant.
Dang it sham, how did you get here? ლ(■_■ლ)
In standard form you want all your values in whole numbers, (so no decimals/fractions).
For example. 4x + 5y = 10 This is an example of line in standard-form.
if you had something like 2/3 x + 2/3 y = 2 you would want to multiply the equation by 3 to get whole numbers 3(2/3x+2/3y=2) becomes 2x + 2y = 6
Ok so we have the equation right?
so what would i need to do next to get my answer?
for the second half
Let's use 2x + 2y = 6 What values make this equation true? Let's make y = 0 so 2x + 2(0) = 6 2x = 6 x = 3 so one solution is (3,0)
Another solution could be when x = 0 so 2(0) + 2y = 6 2y = 6 y = 3 so (0,3) is another solution
Ohh i get it now, so you sub both for 0 to get the points. Thanks sham
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