The system of equations is coincident. What are the missing values? http://static.k12.com/bank_packages/files/media/mathml_f6e7082b73d24e3df1288c9d50654c077c65189a_1.gif 4x + _____ y = _____
i think its 4x + 4y = 7
@AbbySmith12
@NormaValenzuela
@sleepyjess
I can't get into the link,,,it keeps saying file not found. Can you just write the problem ?
@texaschic101 here is the real link: http://static.k12.com/bank_packages/files/media/mathml_f6e7082b73d24e3df1288c9d50654c077c65189a_1.gif
yes thats the link
wait....I kinda messed up on the first part....you do not have to put it in standard form, it already is in standard form x + 3y = 5 --- multiply by 4 4x + 12y = 20 check.. 4x + 12y = 20 -- divide by 4 x + 3y = 5 (correct) same answer : 4x + 12y = 20
how'd u get 12
@texaschic101
ok...we have x + 3y = 5.....we need 4x + ?y = ?.....so we multiplied the first equation by 4. This will make the lines coincident (the same) 4(x + 3y = 5) = 4x + 12y = 20 and to see if it is the same line...we can reduce it by dividing each term by 4, and we would get : x + 3y = 5 Its kinda the same as equivalent fractions.. 2/3...multiply it by 3/3 and we get 6/9....they equal the same, therefore are equivalent. Same thing as the equations....x + 3y = 5 and 4x + 12y = 20...they are the same line
x + 3y = 5 would also be the same line as 3x + 9y = 15....I just multiplied it by 3
does this make any sense to you ?
times 4 by what
4 by 3?
and get 12
x + 3y = 5......we want the x to be the same as the 4x in the other equation. To do this, we multiply the entire 1st equation by 4. 4(x + 3y = 5) = 4x + 12y = 20
4 * x + 4 * 3y = 4 * 5 4x + 12y = 20
If your problem would have been : find a coincident line for x + 3y = 5 and you were given : 2x + ?y = ? Then we would multiply 1st equation by 2 and get : 2(x + 3y = 5) = 2x + 6y = 10 understand ?
ya
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