math
March 24, 2011 10:27 AM Subscribe
Stupid math question:
How to solve (not trail and error) a problem set up like:
2x*3y=278
if
x+y=119
I am not sure what to call this type of problem so can not effectively google the steps. Not looking for an answer (per se) of the above problem, but the steps taken to solve it.
I feel foolish for having to ask
I feel foolish for having to ask
Check this page out: How to Solve Systems of Algebraic Equations Containing Two Variables
posted by Perplexity at 10:31 AM on March 24, 2011
posted by Perplexity at 10:31 AM on March 24, 2011
To check your work, you can plug this in to Wolfram Alpha to get the solution. There are two solutions because when you combine the equations, you get a quadratic.
posted by lukemeister at 10:35 AM on March 24, 2011
posted by lukemeister at 10:35 AM on March 24, 2011
Best answer: You just need to solve for x in terms of y , substitute, and solve.
Solve for x
x+y=119
x=119-y
Substitute
2x*3y=278
2(119-y)*3y=278
...
y=?
Substitute
x=119-y
x=...
You could solve for either x in the first step, or solve for y.
posted by K5 at 10:35 AM on March 24, 2011 [1 favorite]
Solve for x
x+y=119
x=119-y
Substitute
2x*3y=278
2(119-y)*3y=278
...
y=?
Substitute
x=119-y
x=...
You could solve for either x in the first step, or solve for y.
posted by K5 at 10:35 AM on March 24, 2011 [1 favorite]
You can also plug the equations into WolframAlpha, and it will solve it for you.
posted by grouse at 10:35 AM on March 24, 2011
posted by grouse at 10:35 AM on March 24, 2011
I haven't worked through the whole problem, but I think you'll also have to use the quadratic equation here:
After you do the substitution required for the first part of simultaneous equations, you're going to want to end up with an equation that looks like ay2+by+c = 0 (where a, b and c are constants), and then you're going to use the quadratic equation:
(-b (+ or -) sqrt(b2-4ac))/2a
because you have to both add and subtract that second part, you'll end up with 2 answers for y (and then 2 answers for x, and the answers come in pairs).
posted by brainmouse at 10:39 AM on March 24, 2011
After you do the substitution required for the first part of simultaneous equations, you're going to want to end up with an equation that looks like ay2+by+c = 0 (where a, b and c are constants), and then you're going to use the quadratic equation:
(-b (+ or -) sqrt(b2-4ac))/2a
because you have to both add and subtract that second part, you'll end up with 2 answers for y (and then 2 answers for x, and the answers come in pairs).
posted by brainmouse at 10:39 AM on March 24, 2011
You can also use matrix multiplication to solve equations like these with more variables, should you need to.
posted by empath at 10:41 AM on March 24, 2011
posted by empath at 10:41 AM on March 24, 2011
oh, wait.. the x*y multiplication makes that not work.
posted by empath at 10:43 AM on March 24, 2011
posted by empath at 10:43 AM on March 24, 2011
Response by poster: got, it. thanks.
man I need some WD40 for my high school math skills, but it all makes sense now. thanks
posted by edgeways at 10:53 AM on March 24, 2011 [1 favorite]
man I need some WD40 for my high school math skills, but it all makes sense now. thanks
posted by edgeways at 10:53 AM on March 24, 2011 [1 favorite]
Did you maybe mean 2x + 3y rather than 2x * 3y? Because the former would give you whole numbers for x and y, whereas the latter does not. Here's a link to the former on Wolfram Alpha.
posted by greenmagnet at 12:13 PM on March 24, 2011 [1 favorite]
posted by greenmagnet at 12:13 PM on March 24, 2011 [1 favorite]
Math teacher here - yes, substitution. It's called using a system of equations to find the values of two variables whose values will answer both equations. So you may end up with 2 answers for x or y, but only one of them may actually be a solution because the other one doesn't fit your equations.
Hope that helps!
posted by garnetgirl at 1:51 PM on March 24, 2011
Hope that helps!
posted by garnetgirl at 1:51 PM on March 24, 2011
Response by poster: I think if it just had been (+) and (-) involved I may have been able to dredge up the relevant process by myself, the inclusion of (*) was what gave me a little hiccup.
Once I realized how to get to 2(119-y)*3y=278 the following => (238-2y)*3y=278 => y=278-238, or y=40 (thus x=79) came pretty quick.
I wouldn't say I'm bad at math, just that I have forgotten a lot of the processes to make it work, simply because I don't have daily use for them.
posted by edgeways at 4:59 PM on March 24, 2011
Once I realized how to get to 2(119-y)*3y=278 the following => (238-2y)*3y=278 => y=278-238, or y=40 (thus x=79) came pretty quick.
I wouldn't say I'm bad at math, just that I have forgotten a lot of the processes to make it work, simply because I don't have daily use for them.
posted by edgeways at 4:59 PM on March 24, 2011
edgeways: "Once I realized how to get to 2(119-y)*3y=278 the following => (238-2y)*3y=278 => y=278-238, or y=40 (thus x=79) came pretty quick."
Actually, that's solving (238−2y)+3y=278. Since it's multiplied, it becomes:
(238−2y)*3y=278
238(3y)−2y(3y)=278
714y−6y2=278
0=6y2−714y+278
If you don't have something like Wolfram|Alpha handy, you'd have to use the quadratic equation or do some fancy mathwork to solve for x and y (and there will be two solutions).
posted by Rickalicioso at 5:46 PM on March 24, 2011
Actually, that's solving (238−2y)+3y=278. Since it's multiplied, it becomes:
(238−2y)*3y=278
238(3y)−2y(3y)=278
714y−6y2=278
0=6y2−714y+278
If you don't have something like Wolfram|Alpha handy, you'd have to use the quadratic equation or do some fancy mathwork to solve for x and y (and there will be two solutions).
posted by Rickalicioso at 5:46 PM on March 24, 2011
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posted by 0xFCAF at 10:30 AM on March 24, 2011 [2 favorites]