I NEED HELP PLEASE!!!!!!!!!!!!! Solving system of equations with fractional coefficients
@TheSmartOne
@SamsungFanBoy
what do you need help with? I might be able to give you some tips. :)
Solve the following system of equations -3/5x+10y=-1/5 1/3x-1/2y=-10
x= y= ????????????????
@Owl_Eyes03
@geerky42 @gleem @Kainui @sleepyhead314 @confluxepic
@mathmath333
the principle is the same as with integer coefficients. Let's try elimination Multiply the whole of the second equation by 20 Post the new equation here
47x+100y=49 31x-19y=-100
?????????????? @MrNood
what is that? it bears no relation to the original question as far as I can see Multiply the whole of the second equation by 20 Post the new equation here
i dont understand
same as with intergers we need to get the same coefficen in both equations . In the first you have 10y So to get 10 y in th second you need ot multiply it by 20
im not understanding how to change them from fractions
do you know the method of elimination to solve systems of equations? you do not necessarily need to get rid of the fractions - fractions are just numbers
yes, i just need to be explained how to do this with fractions
\[\frac{ -3x }{ 5 } +10y = \frac{ -1 }{ 5 }\] \[\frac{ x }{ 3} - \frac{ y }{ 2 } = -10\] is this the correct layout?
no theres 1s where the x and y RE THE X AND Y ARE ON THE RIGHT OF THE FRACTIONS
yes but that is the same : 1x is the same as x \[\frac{ 1 }{ 3 }x is the same as \frac{ x }{ 3 }\] OK?
OK
So - to eliminate a variable you need to make it so that the coefficient is th esame in 2 equations. You have 10 y in the first equation You have -y/2 in the second. So - if oyu multiply the whole second equation by 20 you will see that you get -10y in the new equation So - multiply the second equation by 20 and pot the answer here
60x+40y=-200?
no 20*1/3 x - 20 * 1/2 y = 20 * -10 write that in a simpler form
im not understanding, thats what i got when i multiplied the equation?
6x-10y=-200?
that is a bit better the first term is not correct - the other 2 are
and how did you know to multiply by 20?
8x-10y=-200?
20 * x/3 same as with integers we need to get the same coefficent in both equations . In the first you have 10y So to get 10 y in the second you need to multiply it by 20 (that's how I knew)
ok, was my new equation correct?
no - how do you get 8 (or 6) from 20 * 1x/3? leave it as a fraction
i dont understand
20/60?
what is 20 * 1/3 leave it as a fraction
20*1=20 and 20*3=60
but if oyu multiply top AND bottom by 20 then oyu have multiplied by 20/20 you only multiply th eTOP so what is 20 * 1x/3
20
you did it right for 20 * y/2 do the same for 20 *x/3
60
\[20 \times \frac{ x }{ 3 }\]
1/60?
it's 20x/3 so your new equations are -3/5x+10y=-1/5 -20x/3 -10y = -200 SO you can see that if you add them together you will eliminate y add them togheter and post eh result here
i dont know how to do it with fractions, thats why i wanna make the fractions whole numbers
OK try this Multiply the first equation by 5 multiply the second equation by 6 If you can do that then I will help Post BOTH new equations here
i do not know how to do anything with fractions. how do i make them whole numbers?
you make them whoe by doing what I said above Multiply the first equation by 5 multiply the second equation by 6
i dont know how.
Then I'm afraid you cannot answer this questin. Maybe you should go back to your text book and revise fractions and come back to this..sometimes there are certain prerequisites to have before you can progress on to a different topic
i dont have a text book. its an online class, thats why im having such a hard time
BUt oyu need some other skills before you can answer this. You cannot expect to answer a question on fractions if you cannot multiply a fraction by 5
do i multiply the top of the fraction as well
You ONLY multiply the top. But the fact that you have ot ask this implies that oyu don't have the background skills to do THIS question. You should return to your calss on fractions and multiplying them and simplifiying them until you can do that , THEN come back to this
ok thanks for your help
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