In the solution of the system of equations, what is the value of x? 2x-4y=14 x+2y=7 a) -7 b) 7 c) 0 d) no solution
what happens when we test them out?
I don't really understand what the question is asking of me because I don't know what the system of equations is, but I put in the answers for x and for a) y=-7, 7 for b) y=0 I'm not sure how to go about this at all, I don't think I'm finding the right thing...
In the first equation, we have -4y In the second equation, we have 2y Add them to get -4y + 2y = -2y this is not 0y like we want, but we can easily fix that ------------------------------------------------------- The coefficients of the y terms must be equal but opposite in sign so when we add the y terms, they cancel out so we can double the 2y to get 4y. Notice how -4y + 4y = 0y = 0 so the y terms go away now we have to multiply EVERY term in the second equation by 2 to make sure things balance out
So multiply everything in x+2y=7 by 2 to get 2x+4y = 14
we have this new system 2x-4y=14 2x+4y=14 Now add the equations 2x-4y=14 2x+4y=14 ---------- 4x+0y = 28 If 4x = 28, then x = ??
if we get the same y value for a given x value, then the solution to the system is (x,y) notice that when x=7 2(7) -4y=14 (7) +2y=7 14 -4y=14 7 +2y=7 -4y=0 2y=0 then for each equation y=0, they have the solution point (7,0) tht satisfies both of them at the same time.
hence: "I put in the answers for x and for a) y=-7, 7 y has 2 values, so its not a solution for b) y=0 y has 1 value. it is common to both
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when x = x1, y=y1 or y2, there are 2 different points for that value since the lines are not crossing there when x=x2, y=y3 , there is only one solution that satisifies BOTH lines at the same time.
@amistre64 @jim_thompson5910 Thank you both for taking the time to explain the process in detail! Now it makes a little more sense! :) Appreciate it, you two are my favorite openstudy teachers here haha :)
jims pretty good, ill admit that :)
You're welcome. I'm glad it's making a bit more sense now.
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