Solving Simultaneous Equations: 3x + y = 17 and 4x - 2y = 6

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To solve the simultaneous equations 4x - 2y = 6 and 3x + y = 17, isolate one variable in one equation and substitute it into the other. For example, rearranging the second equation to y = 17 - 3x allows substitution into the first equation. The elimination method can also be used, where one equation is multiplied to align coefficients for easy addition or subtraction. For the second set of equations, 7a - 3b = 17 and 2a + b = 16, multiplying the second equation helps eliminate one variable when combined with the first. Once one variable is found, the other can be determined using basic algebra.
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hi i am stuck on these simulateous equations -

4x - 2y = 6
3x + y = 17


and this one

7a - 3b = 17
2a + b = 16

please please help me
thanks
 
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What have you done so far?
 
do you know the basics of simuelteanous equations? (ie how to eliminate one of the values, or substiting one into the other?)
 
"4x - 2y = 6
3x + y = 17"

Take one of those equations, isolate one of the variables, then plug it into the other equation. That will figure out one variable, and with that one variable, you can plug into either original formulas to get the other variable.

Example, change the 2nd to y=17-3x. Plug that y into 4x-2y=6.. that should be 4x-2(17-3x)=6..and follow the rest of my instructions.
 
The other method you can use is elimination, meaning adding or subtracting the equations.


multiply one of the equations (sometimes it isn't necessary)

7a - 3b = 17
3(2a + b = 16)

if you multiply this equation by 3, you will have "3b" in both equations.



now you can subtract or add the equations together to eliminate one of the variables

7a - 3b = 17
+(6a + 3b = 48)

so, 13a = 65


you can take it from here. once you figure out one variable, you can figure out the other with simple algebra

Alex
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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