Solve the given linear system 5x+10y=70 5x+25z=270 10y+25z=300
There are a few ways you can solve this, either by substituion or by straight addition/substraction. I always thought substitutionw as easier to understand, so I'll go with that. You can simplify the 2nd equation to be x= 54-5z, I assume you can do the algebra there.
If you take that, and apply the same principle to a few others, you can get values for y and z as well. for example, the first equation you can get x= 14-2y
Just do that for a couple of values, solving for x,y,z and z, and then resubstitue htose values back in to different equations when necesary.
Does that make sense?
Cause you're taking some value, finding what it is in other's terms, and just replugging it in
woah..
very confusing at all.. :(
What're you finding confusing? Go ahead and try rearranging some of the equations to get some simpler equations such as y=7-1/2x
That's the first step, should be a simple step, basic algebra getting the variable on one side
uh.. ive never try like this.. i only tried like this. y=5x-3 y=3x-1
Okay, using that example, you know that y= both of those things right? So they have to equal eachother. Cause if y=n and y=m, then n must = m, right? So if you set those equal to each other 5x-3=3x-1 Then just simplify 2x = 2 x =1 You just need to do that, but with bigger numbers and a z thrown in
Does that make sense or no?
but.. what if there are three? could you give me an example?
It's just a lot more tedious work. If you solve both y and z for x, the plug them in to the third equation, then you can get x For example, if y= 2x and z= 2x and y+z =4 Then you can plug those values in 2x +2x =4 4x=4 x= 1 And then you can use the fact that x =1 to find the original 2, so if y=2x then y=2.
You can do the same with your problem, it just looks like its going to have fractions and/or big numbers and I don't have a calculator or pencil + paper handy, but just follow the same steps
Just solve y and z for x, plug whatever those 2 equations are in to the 3rd, solve for x, and then use that value of x to find the original 2.
thanks :)
No problem. Let me know if you need any more help. You can post what you did and I can double check that you did everything right. This is just how I do those problems by the way. There are other methods that your teacher may prefer, this is just the one that I think makes the most sense.
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