3x + 5y = 78 2x - y = 0 The point of intersection of the lines has an x-coordinate of _____. 78 6 -6 okay def lied , this question stumped me. so far I did the elimination method and got x+6y=78 Where do I take it from there?
well to get the answer for this... We have to multiply one equation by a factor so that when we add them we can get rid of one variable. It wants the x-coordinate. So we need to eliminate the y's. So we can multiply the whole second equation by 5. And then we add both of the equations. \(\sf 3x + 5y = 78\\ 2x - y = 0\) So that becomes \(\sf 3x + 5y = 78\\ 5\times (2x - y = 0)\) And when we multiply by 5 to the second equation we get \(\sf3x + 5y = 78\\ 10x - 5y = 5\) And now we can use elimination to solve it. To get rid of the y we need to add the equations \(~~~\sf 3x + 5y = 78\\+\\ ~~~10x - 5 = 5\\--------\)
thank you , sorry I'm cooking
but I cant add 5y to 5 they arent like terms
my bad. Typo it should be 5y.
at the end it should just be eliminated. :) :P
oh okay so , 13x = 83 ?
oh hold on lol missed a step
Nope, 13x= 83 is correct :)
oh, so I dont divide it out ? but I thought I was looking for "x"
One sec.. I think I made a mistake.
ok
my bad. *facepalm* I did \(5\times 0 =5\)
well to get the answer for this... We have to multiply one equation by a factor so that when we add them we can get rid of one variable. It wants the x-coordinate. So we need to eliminate the y's. So we can multiply the whole second equation by 5. And then we add both of the equations. \(\sf 3x + 5y = 78\\ 2x - y = 0\) So that becomes \(\sf 3x + 5y = 78\\ 5\times (2x - y = 0)\) And when we multiply by 5 to the second equation we get \(\sf3x + 5y = 78\\ 10x - 5y = 0\) And now we can use elimination to solve it. To get rid of the y we need to add the equations \(~~~\sf 3x + 5y = 78\\+\\ ~~~10x - 5 = 0\\--------\)
Just look at the ending and add them together. We should get 13x=78 and divide 13 and get the value of x :)
oh okay
6 !
6 is correct :)
Yes I'm happy I'm actually getting some help most people just throw me the answer !
That is what I'm here for-- to explain you how to do it. :) If you ever need any help, tag me and/or message me. If I can't help you, I'll find someone who can :)
Wheres my answer? D:<
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